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Page No 5.10:

Question 1:

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:
(i) A=5200-1

(ii) A=-1423

(iii) A=1-324-12352

(iv) A=1abc1bca1cab

(v) A=026150371

(vi) A=ahghbfgfc

(vii) A=2-101-301-211-112-150

Answer:

(i)
        M11=-1M21= 20Cij= -1i+jMijC11= -11+1-1 = -1C21 = -11+220 = -20D = -1×5-20×0=-5

(ii)
        M11= 3M21= 4Cij = -1i+jMijC11=-11+1M11= 3C21=-12+1M21=-4 = -4D=3 × -1 - 4 × 2 = -3 - 8 = -11

(iii)
      M11= -1252 =-2 - 10=-12M21 = -3252 = -6 - 10 = -16M31 = -32-12 = -6 + 2 = -4C11= -11+1M11= -12C21= -12+1M21 = --16 = 16C31 = -13+1M31= -4D=1-12 + 38 - 6 + 220 + 3=-12 + 6 + 46 = 40

(iv)
      M11 = bcacab= ab2 - c2a = ab2 - c2M21 = abccab = a2b - c2b = ba2 - c2M31 = abcbca = a2c - b2c = ca2 - b2C11=-11+1M11 =ab2-c2C21=-12+1M21 =-ba2-c2C31=-13+1M31 = ca2-b2D=1.ab2 - c2 - aab - ca + b.cc - b=ab2 - ac2 - a2b + a2c + c2b - b2c=a2c - b  + b2a - c + c2b - a

(v)
       M11= 5071 = 5 - 0 = 5M21= 2671 = 2 - 42 = -40M31= 2650 = 0 - 30 = -30C11= -11+1M11 = 5C21=-12+1M21 =--40C31=-13+1M31 =-30D=05 - 0 -21 - 0 + 67 - 15 = -2 - 48 =-50

(vi)
       M11= bffc =bc - f2M21 = hgfc = hc - fgM31 = hgbf = hf - gbC11= -11+1M11 = bc - f2C21=-12+1M11=-hc - fg = fg - hcC31=-13+1M11= hf - gbD = abc - f2 - hhc - fg + gfh - bg    =  abc - af2 - h2c + fgh + fgh - bg2    =  abc + 2hfg - af2 - bg2 - ch2

(vii)

      M11= 00 - 5 -10 + 1 - 25 - 1 = -1 - 8= -9M21=-10 - 5 + 1(5 - 1) = 5 + 4 = 9M31=-10 + 10 + 1(0 + 1) = -10 + 1 = -9M41=-1(1 - 2) + 10 - 1 = 1 - 1 = 0C11=-11+1M11= -9C21=-12+1M21= -1 × 9C31=-13+1M31=-9C41=-14+1M41= 0D=201-21-11-150 + 1-31-21-1 1250 -1-30111-12-15=-18 - 27 + 15 = 30

Page No 5.10:

Question 2:

Evaluate the following determinants:

(i) x-7x5x+1

(ii) cos θ-sin θsin θcos θ

(iii) cos 15°sin 15°sin 75°cos 75°

(iv)  a+ibc+id-c+ida-ib

Answer:

(i)
 = x(5x + 1) + 7x = 5x2 + x + 7x=5x2 + 8x

(ii)
 = cos2θ - - sin2θ    = cos2θ + sin2θ = 1

(iii)
=cos15°cos75°- sin15°sin75°=cos15°cos75°-sin(90°- 75°)sin(90° - 15°)               sin90° - θ = cosθ=cos15°cos75° - cos75°cos15°=cos15°cos75° - cos15°cos75°= 0

(iv)
=a2 - i2b2 - i2d2 - c2=a2 - i2b2 - i2d2 + c2=a2 + c2 - i2b2 + d2                       i2 =-1=a2 + c2 + b2 + d2

Page No 5.10:

Question 3:

Evaluate 237131751520122.

Answer:

Let A=23713175152012= 2 204 - 100 -3 156 - 75 + 7 260 - 255A=2104 - 381 + 75A=208 - 243 + 35A=243 - 243 = 0 23713175152012 = 0237131751520122=02 = 0           det A2 = det A2

Page No 5.10:

Question 4:

Show that sin 10°-cos 10°sin 80°cos 80°=1

Answer:

Let =sin10°-cos10°sin80°cos80° = sin10°cos80° + cos10°sin80°=sin10°cos(90°-10°) + cos10°sin(90° - 10°)            cosθ = sin90 -  θ = sin10°sin10° + cos10°cos10°         = sin210° + cos210°           sin2θ + cos2θ = 1 = 1

Page No 5.10:

Question 5:

Evaluate 23-571-2-341 by two methods.

Answer:

Let  = 23-571-2-341

First method

=-11+1 21 + 8 + -11+2 37 - 6 + -11+3 -528 + 3= 21 + 8 - 37 - 6 - 528 + 3= 18 - 3 - 155=-140

Second method is the Sarus Method, where we adjoin the first two columns to the right to get

23-52371-271-341-34=2 × 1 × 1 + 3 × -2 × -3 - 5 × 7 × 4 - -5 × 1 × -3 + 2 × -2 × 4 + 3 × 7 × 1=2 + 18 - 140 - 15 - 16 + 21=-120 - 20=-140

Page No 5.10:

Question 6:

Evaluate =0sin α-cos α-sin α0sin βcos α-sin β0

Answer:

Let  = 0sinα-cosα-sinα0sinβcosα-sinβ0


=-11+1 0 0+sin2β + -11+2 sinα0 - sinβcosα + -11+3  -cosαsinαsinβ - 0     Expanding along R1=00 + sin2β - sinα0 - sinβcosα - cosαsinαsinβ - 0=sinαsinβcosα - sinαsinβcosα=0

Page No 5.10:

Question 7:

=cos α cos βcos α sin β-sin α-sin βcos β0sin α cos βsin α sin βcos α

Answer:

Given: =cosαcosβcosαsinβ-sinα-sinβcosβ0sinαcosβsinαsinβcosα


=-11+1 cosα cosβcosα cosβ - 0  +  -11+2 cosα sinβ-sinβ cosα  -  0  +  -11+3 -sinα-sin2β sinα - sinα cos2β         Expanding along R1=cosα cosβcosα cosβ - 0 - cosα sinβ-sinβ cosα - 0 - sinα-sin2β sinα - sinα cos2β=cos2α cos2β + cos2α sin2β + sin2α sin2β + sin2α cos2β=cos2αcos2β + sin2β + sin2αsin2β + cos2β         = cos2α + sin2α       sin2θ + cos2θ = 1 = 1                          sin2θ + cos2θ = 1

Page No 5.10:

Question 8:

If A=2521 and B=4-325, verify that |AB| = |A| |B|.

Answer:

Consider LHSAB=25214-325=8 + 10-6 + 258 + 2-6 + 5 = 181910-1AB = -18 - 190 = -208Consider RHSA =2 - 10 =-8B=20 - -6 = 26AB=-8 × 26 =-208 LHS= RHS

Page No 5.10:

Question 9:

If A 101012004, then show that |3 A| = 27 |A|.

Answer:

A=1010120043A=3030360012             Multiplying each element of A by 33A = -11+1 336 - 0 + -11+2 00 - 0 + -11+3 30 - 0 = 336 - 0 - 00 - 0 + 30 - 0    Expanding along R1=3 × 36 = 108           ...1A = -11+1 14 - 0 + -11+2 00 - 0 + -11+3 10 - 0 = 14 - 0 - 00 - 0 + 10 - 0 = 4    Expanding along  R127A = 27 × 4 = 108      ...2 3A = 27 A     From eqs. (1) and (2)

Page No 5.10:

Question 10:

Find the values of x, if

(i) 2451 = 2x46x

(ii) 2345=x32x5

(iii) 3xx1=3241

(iv) If 3x724=10, find the value of x.

(v) x+1x-1x-3x+2=4-113

(vi) 2x58x=6583

Answer:

(i)
 Given:2451 = 2x46x2 - 20 = 2x2 - 24-18 = 2x2 - 242x2 = 6x2= 3x = ±3

(ii)
Given: 2345 = x32x510 - 12 = 5x - 6x-2 = -x    x = 2

(iii)
Given: 3xx1 = 32413 - x2 = 3 - 8-x2  = -8x2 = 8x = ±22

(iv)
Given: 3x724 = 1012x - 14 = 1012x = 24x = 2

(v) 
Given: x+1x-1x-3x+2=4-113x+1x+2 - x-3x-1 = 12+1x2+3x+2-x2+4x-3 = 137x-1 = 137x = 14x = 2

(vi)
Given: 2x58x=65832x2 - 40 = 18 - 402x2 = 18x2 = 9x=±3



Page No 5.11:

Question 11:

Find the integral value of x, if x2x1021314=28.

Answer:

 Given: x2x1021314 = 28x28 - 1 - x0 - 3 + 10 - 68x2 - x2 + 3x - 6 = 287x2 + 3x - 6 = 287x2 + 3x - 34 = 07x + 17 x - 2 = 0x = 2

Integral value of x is 2. Thus, x = -177 is not an integer.

Page No 5.11:

Question 12:

For what value of x the matrix A is singular?

(i) A=1+x73-x8

 ii A=x-1111x-1111x-1 

Answer:

(i) Matrix A will be singular if

A = 0
A=1+x73-x8=08+8x-21+7x=015x-13=015x=13x=1315

(ii) Matrix A will be singular if

A = 0
x - 1x - 12 - 1 -1x - 1 - 1 + 11 -x - 1 = 0x - 1x2 - 2x -1x - 2 + 12 - x = 0x3 - 2x2 - x2 + 2x - x + 2 - x + 2 = 0x3 - 3x2 + 4 = 0x - 22 x + 1 = 0x = 2  or  x = -1



Page No 5.57:

Question 1:

Evaluate the following determinant:
(i) 1352610311138

(ii) 671921391314812426

(iii) ahghbfgfc

(iv) 1-324-12352

(v) 149491691625

(vi) 6-322-12-1052

(vii) 13927392719271327139

(viii) 

Answer:

(i) =1352610311138= 1 6101138  - 32103138 + 5263111=1228 - 110 - 376 - 310 + 522 - 186=1(118) - 3( -234) + 5(-164)=118 + 702 - 820=0(ii)=671921391314812426= 67338 - 336 -191014 - 1134  + 21936 - 1053= 67(2) - 19( -120)  +  21( -117)= 134 + 2280 - 2457=-43(iii) =ahghbfgfc=abffc -hhfgc +ghbgf=a(bc - f2) -h(hc - fg) + g(hf - gb)=abc - af2 - h2c + fgh + fgh - g2b=abc + 2fgh - af2 - ch2 - bg2(iv) =1-324-12352=1-1252  -(-3)4232  + 24-135=1-2 - 10 + 38 - 6 + 220 + 3=(-12) + 6 + 46= 40(v)= 149491691625=19161625 -4416925 + 949916=1225 - 256 -4100 - 144 + 964 - 81=1(-31) - 4(-44) + 9(-17)=-31 + 176 - 153=-8(vi) =6-322-12-10 52=6(-2 - 10) - (-3)(4 + 20) + 2(10 - 10)=-72 + 72 + 0=-72 + 72=0(vii) =13927392719271327139=192712713139 - 332719132739 + 939192732719 - 27392792712713= 19(9 - 9) - 27(243 - 3) + 1(81 - 1) -33(9 - 9) -27(81 - 81) + 1(27- 27) + 93(243 - 3) - 9(81 - 81) + 1(9 - 729) - 27(81 - 1) - 9(27 - 27) + 27(9 - 729)= 10 - 6480 + 80 - 30 - 0 + 0 + 9720 - 0 - 720 - 2780 - 0 - 19440= - 6400 + 522720= 516320

(viii)

Page No 5.57:

Question 2:

Without expanding, show that the values of each of the following determinants are zero:
(i) 82712351643

(ii) 6-322-12-1052

(iii) 23713175152012

(iv) 1/aa2bc1/bb2ac1/cc2ab

(v) a+b2a+b3a+b2a+b3a+b4a+b4a+b5a+b6a+b

(vi) 1aa2-bc1bb2-ac1cc2-ab

(vii) 491639742623

(viii) 0xy-x0z-y-z0

(ix) 143673543172

(x) 12223242223242523242526242526272

(xi) abca+2xb+2yc+2zxyz

(xii) 2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21

(xiii) sinαcosαcos(α+δ)sinβcosβcos(β+δ)sinγcosγcos(γ+δ)

(xiv) sin223°sin267°cos180°-sin267°-sin223°cos2180°cos180°sin223°sin267°

(xv) cosx+y-sinx+ycos2ysinxcosxsiny-cosxsinx-cosy

(xvi) 23+35515+465103+115155

(xvii) sin2AcotA1sin2BcotB1sin2CcotC1, where A, B, C are the angles of ABC.

Answer:

​(i) =82712351643=027035043             Applying C1C1-4C2=0(ii) =6-322-12-1052=0-320-12052         Applying C1C1+2C2=0

​(iii) =23713175152012=2371317513175     Applying R3R3-R1=0(iv) =1aa2bc1bb2ac1cc2ab=1a3abc1b3abc1c3abc      Applying R1aR1, R2bR2 and R3cR3=abc1a311b311c31=0(v) =a+b2a+b3a+b2a+b3a+b4a+b4a+b5a+b6a+b=aaa2a2a2a4a+b5a+b6a+b       Applying R1R2-R1 and R2R3-R2=2aaaaaa4a+b5a+b6a+b=0(vi) =1aa2-bc1bb2-ac1cc2-ab=0a-ba2-bc-b2+ac0b-cb2-ac-c2+ab1cc2-ab        Applying R1R1-R2, R2R2-R3=0a-ba-ba+b+ca-b0b-cb-cb+c+ab-c1cc2-ab=a-bb-c01a+b+c01a+b+c1cc2-ab=0(vii) =491639742623=116774223        Applying C1C1-8C3=0(viii) =0xy-x0z-y-z0=xyzxyz0xy-x0z-y-z0=1xyz0xzyz-xy0zy-yx-zx0=1xyz-2xy02yz-xy0zy-yx-zx0           Applying R1R1+R2+R3=1xyz000-xy0zy-yx-zx0=0        Applying R1R1-2R2(ix)=143673543172=116774332=0       Applying C2C2-7C3

x)=12223242223242523242526242526272=14916491625916253616253649=1491649162557911791113           Applying R3R3-R2 and R4R4-R3=14916491625791113791113=0          Applying R32+R3xi) =abca+2xb+2yc+2zxyz=a+2xb+2yc+2za+2xb+2yc+2zxyz         Applying R1R1+2R3=000a+2xb+2yc+2zxyz=0      Applying R1R1-R2


(xii)
 2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21=22x+2-2x+222x+2-2x-2132x+3-2x+232x+3-2x-2142x+4-2x+242x+4-2x-21=422x+2-2x-21432x+3-2x-21442x+4-2x-21       Applying C1C1-C2=4122x+2-2x-21132x+3-2x-21142x+4-2x-21=0

(xiii)
sinαcosαcos(α+δ)sinβcosβcos(β+δ)sinγcosγcos(γ+δ)=sinαsinδcosαcosδcos(α+δ)sinβsinδcosβcosδcos(β+δ)sinγsinδcosγcosδcos(γ+δ)        Applying C1sinδ C1 and C2cosδ C2=sinαsinδcos(α+δ)cos(α+δ)sinβsinδcos(β+δ)cos(β+δ)sinγsinδcos(γ+δ)cos(γ+δ)        Applying C2C2-C1=0

(xiv)
sin223°sin267°cos180°-sin267°-sin223°cos2180°cos180°sin223°sin267°=sin223°sin290-23°-1-sin290-23°-sin223°1-1sin223°sin290-23°=sin223°cos223°-1-cos223°-sin223°1-1sin223°cos223°=sin223°+cos223°cos223°-1-cos223°-sin223°-sin223°1-1+sin223°sin223°cos223°           Applying C1C1+C2=11-1-1-sin223°1-cos223°sin223°cos223°=-1-11-11-sin223°1cos223°sin223°cos223°=0

(xv)
cosx+y-sinx+ycos2ysinxcosxsiny-cosxsinx-cosy=1sinycosycosx+y-sinx+ycos2ysinxsinycosxsinysin2y-cosxcosysinxcosy-cos2y           Applying R2siny R2 and R3cosy R3=1sinycosycosx+y-sinx+ycos2ysinxsiny-cosxcosycosxsiny+sinxcosysin2y-cos2y-cosxcosysinxcosy-cos2y           Applying R2R2+R3=-1sinycosycosx+y-sinx+ycos2ycosx+y-sinx+ycos2y-cosxcosysinxcosy-cos2y=0

(xvi)
23+35515+465103+115155=355155103155+235546510115155=315555103155+2315525105155=3×51155510335+23×51512525155=0+0=0

(xvii)
sin2AcotA1sin2BcotB1sin2CcotC1=sin2A-sin2BcotA-cotB0sin2BcotB1sin2C-sin2BcotC-cotB0          Applying R1R1-R2 and R3R3-R2=sinA+BsinA-BcosAsinB-cosBsinAsinAsinB0sin2BcotB1sinC+BsinC-BcosCsinB-cosBsinCsinBsinC0=sinπ-CsinA-B-sinA-BsinAsinB0sin2BcotB1sinπ-AsinC-B-sinC-BsinBsinC0           A+B+C=π=sinCsinA-B-sinA-BsinAsinB0sin2BcosBsinB1sinAsinC-B-sinC-BsinBsinC0=sinA-BsinC-BsinBsinC-1sinA0sin2BcosB1sinA-1sinC0=sinA-BsinC-BsinBsinAsinCsinCsinA-10sin2BcosB1sinAsinC-10                  Applying R1sinA R1 and R3sinC R3=sinA-BsinC-BsinBsinAsinC000sin2BcosB1sinAsinC-10                  Applying R1R1-R3=0



Page No 5.58:

Question 3:

Evaluate :

ab+ca2bc+ab2ca+bc2

Answer:

=ab+ca2bc+ab2ca+bc2 When a=b, the first two rows become identical. Hence, a-b is a factor. Similarly, when b=c the second and third rows become identical. So, b-c is also a factor. Also, when c=a, the third and first rows become identical. Hence, c-a is also a factor. The product of diagonal elements, a(c + a) c2 is  4. So, the other factor should be a linear in a, b and c. It should also remain unaltered when any two letters are changed. Let this factor be λ(a+b+c).Here, λ is a constant.  To find this, we havea=0, b=1, c=2032121214= λ(a - b)(b - c)(c - a)(a + b + c)032121214 = λ(0 - 1)(1 - 2)(2 - 1)(0 + 1 + 2)-6 = 6λλ = -1Thus, ab + ca2bc + ab2ca + bc2 =-((a + b + c))(a - b)(b - c)(c - a)

Page No 5.58:

Question 4:

Evaluate :

1abc1bca1cab

Answer:

=1abc1bca1cab     When a=b, the first two rows become identical. Hence, a-b is a factor. Similarly, when b=c and c=a, the second and third and third and first rows become identical. Hence, b-c and c-a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors. 1abc1bca1cab=λ(a-b)(b-c)(c-a)     Where λ is a constant102110120=2λ     Putting a=0, b=1 and c=2 to find λ2=2λλ=1Hence,1abc1bca1cab =(a-b)(b-c)(c-a)

Page No 5.58:

Question 5:

Evaluate :

x+λxxxx+λxxxx+λ

Answer:

=x + λxxxx + λxxxx + λ=  λ   0x-λ   λx0-λ x + λ          ApplyingC1C1 -C2,  C2C2-C3=    λ   0x -λ   0   2x +λ 0-λ  x + λ          Applying R1R2+R3= λ02x +λ -λx + λ  + x-λ   0  0-λ= λ[λ(2x + λ)] + xλ2= λ2 (2x + λ + λ2x)= 3λ2x + λ3= λ2 (3x + λ )

Page No 5.58:

Question 6:

Evaluate :

abccabbca

Answer:

=abccabbca=a(a2 - bc) - b(ca - b2) + c(c2 - ba)=a3 - abc - bca + b3 + c3 - abc=a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

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Question 7:

Evaluate the following:

x111x111x
 

Answer:

Let =x111x111x.
=x111x111x   =x-11-x01x101-xx-1      Applying R1R1-R2 and R3R3-R2   =x-121-101x10-11   =x-121-101x+11001      Applying C2C2+C3   =x-12(x+1+1)        Expanding along last row   =x-12(x+2) =x-12(x+2)

Page No 5.58:

Question 8:

Evaluate the following:

0xy2xz2x2y0yz2x2zzy20

Answer:

Let =0xy2xz2x2y0yz2x2zzy20.
=0xy2xz2x2y0yz2x2zzy20   =x2y2z20xxy0yzz0      Taking x2 common from C1, y2 common from C2 and z2 common from C3   =x3y3z3011101110      Taking x common from R1, y common from R2 and z common from R3   =x3y3z30011-11110      Applying C2C2-C3   =x3y3z31+1        Expanding along first row   =2x3y3z3 =2x3y3z3

Page No 5.58:

Question 9:

Evaluate the following:

a+xyzxa+yzxya+z

Answer:

Let =a+xyzxa+yzxya+z.
=a+xyzxa+yzxya+z   =a+x+y+zyza+x+y+za+yza+x+y+zya+z      Applying C1C1+C2+C3   =a+x+y+z1yz1a+yz1ya+z      Taking a+x+y+z common from C1   =a+x+y+z1yz0a000a      Applying R2R2-R1 and R3R3-R1   =a+x+y+za2        Expanding along first column =a+x+y+za2

Page No 5.58:

Question 10:

If =1xx21yy21zz2, 1=111yzzxxyxyz, then prove that +1=0.

Answer:

+1=1xx21yy21zz2+111yzzxxyxyz           =1xx21yy21zz2+1yzx1zxy1xyz      Interchanging rows and coloumns in 1           =1xx21yy21zz2-1xyz1yzx1zxy      Applying C2C3 in 1           =1xx20y-xy2-x20z-xz2-x2-1xyz0y-xzx-yz0z-xxy-yz      Applying R2R2-R1 and R3R3-R1           =y-xz-x1xx201y+x01z+x-y-xz-x1xyz01-z01-y      Taking y-x common from R2 and z-x common from R3           =y-xz-xz+x-y-x-y-xz-x-y+z      Expanding along first column           =y-xz-xz-y1-1           =0 +1=0.

Page No 5.58:

Question 11:

Prove that :

abca-bb-cc-ab+cc+aa+b=a3+b3+c3-3abc

Answer:

=abca-bb-cc-ab+cc+aa+b=abca-bb-cc-aa+b+cc+a+ba+b+c          Applying R3R3+R2=(a+b+c)abca-bb-cc-a111          Taking (a+b+c) common=(a+b+c)abcbca111                           Applying R2R1-R2=(a+b+c)a-bb-ccb-cc-aa001              C1C1-C2 and C2C2-C3=(a+b+c)-1a-bc-a-b-c2=(a+b+c)-ac-bc-a2+ab-b2-c2+2bc=(a+b+c)a2+b2+c2-ab-bc-ca=a3+b3+c3-3abc

Page No 5.58:

Question 12:

Prove that :

b+ca-bac+ab-cba+bc-ac=3abc-a3-b-c3

Answer:


 Let LHS =Δ = b + c    a - b     ac + a    b - c     b a + b    c - a     c Δ=b + c   b - c     b   c - a     c - a - b   c + a    b  a + b   c + a  c + a    b - c  a + b    c - a     Expanding =b + c bc - c2 - bc + ab  - a - bc2 + ac - ab - b2 + ac2 - a2 - ab + ac - b2 + bc=bc2 - c3 + abc - ac2 - a2c + a2b + ab2+ bc2+ abc - ab2 - b3+ ac2 - a3 - a2c - ab2 + abcΔ=3abc-a3-b3-c3      Simplyfying =RHS

Page No 5.58:

Question 13:

Prove that :

a+bb+cc+ab+cc+aa+bc+aa+bb+c=2abcbcacab

Answer:

 Let Δ=a+b   b+c   c+a b+c   c+a    a+bc+a    a+b   b+c Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we getΔ =a   b   c b   c    ac    a   b  + b   c   a c   a    ba    b   c     =a   b   c b   c    ac    a   b + -1a   c   b b   a   cc   b   a          Applying C1 C3 in second determinant to get negative value of the deteminant=a   b   c b   c   ac    a  b + -1-1 a   b   c b   c    ac   a    b                Applying C2C3=  2 a   b   c b   c    ac    a   b = RHS

Page No 5.58:

Question 14:

Prove that :

a+b+2cabcb+c+2abcac+a+2b=2 a+b+c3

Answer:

Let LHS= Δ=a+b+2c        a                 b     c          b+c+2a           b     c               a              c+a +2b   =  2a+2b+2c               a                              b        2a+2b+2c          b+c+2a                       b 2a+2b+2c               a                         c+a +2b    Applying C1 C1+C2+C 3  =  2 a+b+c  1              a                              b       1         b+c+2a                       b1              a                         c+a +2b        Taking out 2(a+b+c) common from C1 =2 a+b+c  1                a                           b       0           b+c+a                     00        -b-c-a                c+a +b       Applying R2 R2-R1 and R2R2 -R3=2a+b+ca+b+ca+b+c1          a          b  0          1          00        -1         1             Taking out (a+b+c) common from R2 and R3=2a+b+c3 11-0         Expanding along C1=2a+b+c3 =RHS



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Question 15:

Prove that :

a-b-c2a2a2bb-c-a2b2c2cc-a-b= a+b+c3

Answer:

 Let LHS==a-b-c        2a              2a    2b         b-c-a          2b    2c             2c            c-a-b=  a+b+c       a+b+c             a+b+c     2b           b-c-a                   2b    2c               2c                   c-a-b         Applying R1R1+R2+R3=a+b+c     1                 1                       1           2b           b-c-a                 2b    2c               2c                  c-a-b=a+b+c          0                      1                      1        b +c+a           b-c-a                 2b         0                     2c                  c-a-b          Applying C1 C1-C2= a+b+c a+b+c×  1                1      2c           c-a-b          Expanding along C1=a+b+c3=RHS

Page No 5.59:

Question 16:

Prove that :

1b+cb2+c21c+ac2+a21a+ba2+b2=a-b b-c c-a

Answer:

Let LHS == 1    b+c     b2+c21    c+a     c2+a21    a+b     a2+b2=0      b+c-c+a           b2+c2-c2+a20      c+a - a+b        c2+a2-a2+b21              a+b                              a2+b2   Applying R1 R1-R2 and R2 R2-R3 = 0    b-a     b2-a20   c-b     c2-b21    a+b     a2+b2=   -12  0      a-b     a2-b20       b-c     b2-c21       a+b     a2+b2          Taking out -1 common from R1 and R2=a-bb-c  0        1              a+b0       1                b+c1      a+b          a2+b2 =  a-bb-c1×1         a+b1         b+c         Expanding along C1=a-bb-cc-a=RHS

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Question 17:

Prove that :

aa+ba+2ba+2baa+ba+ba+2ba=9 a+b b2

Answer:

 Let LHS== a           a+b      a+2ba+2b     a          a+ba+b      a+2b     a  Δ =3a+3b     3a+3b    3a+3b a+2b       a              a+ba+b         a+2b            a               Applying R1R1+R2+R3=3 a+b1                1                  1 a+2b       a              a+ba+b         a+2b            a         Taking out 3 a+b common from R1= 3 a+b 0            0             12b       -b              a+b-b       2b            a            Applying C1 C1-C2and C2C2-C3 = 3 a+b b2  0           0             1 2       -1           a+b-1       2              a            Taking out b common from C1 and C2= 3 a+b b2×3=9a+b b2=RHS

Page No 5.59:

Question 18:

Prove that :

1abc1bca1cab=1aa21bb21cc2

Answer:

Let  LHS ==1   a   bc1   b   ca 1   c   ab=1abca      a2       abcb      b2       bca c       c2       abc     Applying R1a R1, R2b R2 and R3c R3 and then dividing it by abc =abcabca        a2      1b       b2          1c        c2       1       Taking out abc common from C3=  -1 1       a2     a1      b2         b1      c2      c         Interchanging C3 and C1 to get-ve value of original determinant=-1-11     a     a21    b         b21    c      c2         Applying C2C3=  1     a     a21    b         b21    c      c=RHS

Page No 5.59:

Question 19:

Prove that :

zxyz2x2y2z4x4y4=xyzx2y2z2x4y4z4=x2y2z2x4y4z4xyz=xyz x-y y-z z-x x+y+z.

Answer:

 Let Δ1 =z      x     yz2     x2    y2 z4    x4    y4, Δ2=x     y      zx2     y2    z2 x4    y4     z4, Δ3=x2     y2      z2x4      y4     z4 x      y       z and Δ4=xyzx-y y-z z-x x+y+zNow,Δ1 =z      x     yz2     x2    y2 z4    x4    y4 Using the property that if two rows ( or columns ) of a determinant are interchanged, the value of the determinant  becomes negetive, we get Δ1 = -1   x     z     yx2     z2    y2x4    z4    y4        C1 C2   =-1-1x        y       zx2      y2      z2x4      y4       z4       C2 C3=   x        y       zx2      y2      z2x4      y4       z4 = Δ2              ...(1)=  -1  x2        y2       z2x         y         zx4        y4        z4                 Applying R1 R2=  -1 -1  x2         y2         z2x4         y4          z4x          y             z                  Applying R2  R3      =  x2         y2         z2x4         y4          z4x          y             z =Δ3               ...(2)Thus,Δ1 =  Δ2 =  Δ3                      From eqs. (1) and (2) 


2 = x     y      zx2     y2    z2 x4    y4     z4= xyz 1      1     1x      y       z x3    y3     z3            Taking out common factor x from C1 , y from C2 and z from C3=xyz  0                  0             1  x-y        y-z            z x3-y3      y3  -z3         z3           Applying C C1-C2 and C2 C2 -C3=xyz x-y  y-z         0                           0                            1        1                          1                            z x2+2xy+y2        y2  +2yz+z2                      z3              a3-b3=a-ba2+ab+b2   Taking out common factor x-y from C1 and y-z from C2=xyz x-y  y-z1×    1                          1                x2+xy+y2        y2  +yz+z2             Expanding along R1  =xyz x-y  y-zy2  +yz+z2-x2-xy-y2=xyz x-y  y-zyz-xy +z2-x2=xyz x-y  y-zyz-x+z-xz+x=xyz x-y  y-zz-xy+x+z=xyz x-y  y-zz-xx+y+z=4 Thus, 1 =2 =3=4 

Page No 5.59:

Question 20:

Prove that :

b+c2a2bcc+a2b2caa+b2c2ab=a-b b-c c-a a+b+c a2+b2+c2

Answer:

Let LHS =Δ=b+c2   a2     bcc+a2   b2     caa+b2   c2     ab= b+c2-c+a2      a2- b2       bc-cac+a2-a+b2       b2  -c2     ca-aba+b2                         c2           ab        Applying R1 R1-R2 and R2R2-R3=  b-ab+2c+a      a+b a-b              cb-a c-b b+2a+c     b-c b+c              ac-b a+b2                         c2                           ab=a-bb-c-b+2c+a      a+b          -c -b+2a+c      b+c         -a      a+b2         c2             ab               Applying x2-y2=x+y x-y and taking out a-b common from R1 and b-c from R2= a-bb-c   -2b+c+a        a+b          -c -2b+a+c        b+c         -a  a+b2 -c2         c2             ab        Applying C1 C1-C2= a-bb-c -2b+c+a                 a+b              -c -2b+a+c                 b+c             -a  a+b+c a+b-c           c2             ab        Applying x2-y2=x+y x-y in C1= a-bb-ca+b+c -2                    a+b              -c -2                   b+c               -a  a+b-c           c2                 ab              Taking out a+b+c common from C1= a-bb-ca+b+c-2                     a+b                -c  0                      c-a                 c-aa+b-c           c2                     ab           Applying  R2R2-R1= a-bb-ca+b+cc-a-2                     a+b                -c  0                     1                         1a+b-c           c2                     ab       Taking out c-a common from R2= a-bb-ca+b+cc-a-2                     a+b +c               -c  0                          0                        1a+b-c           c2-ab                     ab     Applying C2 C2-C3= a-bb-ca+b+cc-a  -1-2                     a+b +c  a+b-c           c2-ab          Expanding along R2=  - a-bb-ca+b+cc-a-2c2+2ab -a2-b2-2ab+c2= - a-bb-ca+b+cc-a-a2-b2-c2= a-bb-ca+b+cc-aa2+b2+c2=RHS

Page No 5.59:

Question 21:

Prove that :

a+1 a+2a+21a+2 a+3a+31a+3 a+4a+41=-2

Answer:

Let  LHS= Δ = a+1a+2   a+2      1a+2a+3   a+3      1 a+3a+4   a+4      1= a+1a+2-a+2a+3      a+2 -a+3      0a+2a+3-a+3a+4      a+3-a+4       0 a+3a+4                                a+4              1        Applying C1C1-C2 and C2C2-C3=-2a+2        -1          0-2a+3        -1          0a+3a+4    a+4      1=1×-2a+2          -1  -2a+3         -1       Expanding along C3 =4+2a-2a-6=-2=RHS 

Hence proved.

Page No 5.59:

Question 22:

Prove that :

a2a2-b-c2bcb2b2-c-a2cac2c2-a-b2ab=a-b b-c c-a a+b+c a2+b2+c2

Answer:

Let LHS =Δ=a2    a2-b-c2     bcb2      b2-c-a2     cac2     c2- a-b2    ab 

=a2    -b-c2      bcb2    -c-a2      cac2    -a-b2      ab       Applying C2C2-C1=-1a2    b-c2      bcb2    c-a2      cac2    a-b2      ab=-a2    b2 +c2         bcb2    c2 + a2       cac2    a2 +b2        ab           Applying C2C2-2C1


=-a2 +b2 +c2       b2 +c2      bcb2+ c2 + a2     c2 + a2     cac2 +a2 +b2       a2 +b2      ab   Applying C1C1+C2=-a2 +b2 +c21    b2 +c2      bc1    c2 + a2     ca1    a2 +b2      ab


=-a2+b2 +c2 1              b2 +c2                              bc0            c2 + a2 -b2 +c2        ca-bc0             a2 +b2  -b2 +c2       ab-bc   Applying R2R2-R1 and R3R3-R1=a2+b2 +c2 1              b2 +c2             bc0              a2- b2             ca-b0              a2-c2              b a-c=-a2+b2 +c2 a-b   a-c1              b2 +c2             bc0              a+b                  c0              a+c              b       Taking a-b common from R2  and a-c common from R3=a2+b2 +c2 a-b  c-a×1×    a+b                  ca+c              b      c-a=-a-c      Expanding along C1=a2+b2 +c2 a-b   c-a ab+b2-ac-c2   


=a2+b2 +c2 a-bc-aab-c+b+cb-c=a-bc-ab-ca+b+ca2+b2 +c2 

 = RHS

Hence proved.
 

Page No 5.59:

Question 23:

Prove that :

1a2+bca31b2+cab31c2+abc3=-a-b b-c c-a a2+b2+c2

Answer:

Let LHS =Δ=1    a2+bc     a31    b2+ca     b31    c2+ab     c3Δ =0     a2+bc-b2+ca        a3-b30      b2+ca-c2+ab        b3-c31       c2+ab                         c      Applying R1R1-R2    and R2R2- R3

=0        a2-b2-ca+ bc       a3-b30        b2-c2-ab+ ca       b3-c31        c2+ab                         c3=0       a-b a+b-c              a-ba2+ab+b20        b-cb+ c-a              b-cb2+bc+a21        c2+ab                               c3

=a-bb-c0           a+b-c             a2+ab+b20          b+c-a             b2 +bc+c21           c2+ab                    c3         Taking out a-b common from R1  and b-c from R2= a-bb-c0           a+b-c                        a2+ab+b20       b+c-a- a+b-c       b2 +bc+c2-a2+ab+b21           c2+ab                           c3   Applying R2R2-R1

=a-bb-c0           a+b-c                        a2+ab+b20       2  c-a                   bc-a+c2-a21           c2+ab                           c3=a-bb-cc-a 0           a+b-c         a2+ab+b20               2                  a+b+c1           c2+ab                    c3=a-bb-cc-a×1×  a+b-c         a2+ab+b2     2                  a+b+c               Expanding along C1


=a-bb-cc-a× a+b2-c2 -   2a2+2ab+2b2 =a-bb-cc-aa+b2-c2 -a+b2-a2+b2=-a-bb-cc-aa2+b2+c2=RHS

Hence proved.

Page No 5.59:

Question 24:

Prove that :

a2bcac+c2a2+abb2acabb2+bcc2=4a2b2c2

Answer:

Let LHS =Δ=a2             bc                ac+c2a2+ab      b2                  acab             b2+bc             c2Δ=abc a            c           a+ca+b         b            ab             b+c         c             Taking out a, b and c common from C1, C2 and C3 

 =abc a          c              0a+b     b         -2bb         b+c     -2b        Applying C3C3-C2-C1 =abc-2b a          c              0a+b     b             1b         b+c         1      Taking (-2b)  common from C3=abc-2b a          c                0a       -c               0b         b+c            1      Applying R2R2-R3  =abc-2b×1 a          c   a       -c              Expanding along C3=abc-2b-2ac=4a2b2c2=RHS

Page No 5.59:

Question 25:

Prove that :

x+4xxxx+4xxxx+4=16 3x+4

Answer:

Let LHS =Δ=x+4      x           xx         x+4        xx           x          x+4=3x+4      3x+4           3x+4x             x+4                   xx                x               x+4      Applying R1 R1+R2+R3 =3x+4 1               1                     1x             x+4                   x x                x                x+4        Taking out 3x+4  common from R1=3x+4 1           0            0x           4             0x           0            4      Applying C2C2-C1 and C3C3-C1=3x+4 42    Expanding along R1=163x+4 =RHS

Page No 5.59:

Question 26:

Prove that :

11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q=1

Answer:

 Let LHS= Δ=1    1+p       1+p+q2    3+2p     4+3p+2q3    6+3p      10+6p+3q =1    1   1+p2    3    4+3p3    6   10+6p+1    p       q2    2p     2q3    3p      3q =1    1   12    3    43    6   10+1    1   p2    3   3p3    6   6p+pq 1   1       12    2      23    3      3            Taking out pq common from last determinant=1    1   12    3    43    6   10+p1    1   12    3   33    6   6+0         Taking out p common from second determinant =1    1   12    3    43    6   10+0       Value of determinant with two identical columns is zero=1    0   02    1   23   3   7              Applying C2C2-C1 and C3C3-C1=1×1237         Expanding along R1=7-6=1  =RHS

Page No 5.59:

Question 27:

Prove that :

ab-cc-ba-cbc-aa-bb-ac= a+b-c b+c-a c+a-b

Answer:

 Let  LHS =Δ=a         b-c     c-ba-c     b         c-aa-b     b-a      cΔ=a         0                       c-b+aa-c     b+c-a                 0a-b     b+c-a             c+a-b         Applying C2C2+C3 and C3C1+C3 = b+c-ac+a-ba           0        1a-c      1       0a-b      1       1       Taking out common factor from C2 and C3= b+c-ac+a-ba  ×1       01       1+1×  a-c      1a-b      1      Expanding along R1=a+b-c  b+c-ac+a-b=RHS



Page No 5.60:

Question 28:

Prove that a22abb2b2a22ab2abb2a2= a3+b32

Answer:

Let LHS==a2   2ab    b2  b2     a2     2ab 2ab    b2      a2= a2a2     2ab  b2      a2  -2ab  b2       2ab 2ab      a2  +b2  b2     a2  2ab    b2    Expanding=a2a4-2ab3 -2ab b2a2-4a2b2+b2b4-2a3b=a6-2a3b3-2a3b3+8a3b3 +b6-2a3b3=a6+2a3b3 +b6=a32+2a3b3 +b32=a3+b32=RHS

Hence proved.

Page No 5.60:

Question 29:

Prove that a2+1abacabb2+1bccacbc2+1=1+a2+b2+c2

Answer:

Let LHS = Δ=a2+1       ab      acab          b2+1     bcca             cb      c2+1=abc a+1a          b                 ca              b+1b           ca                   b            c+1c              Taking out a, b and c common from R1 , R2 and R3= abc  a+1a             b              c-1 a           1b              0 -1 a            0              1c             Applying R2R2-R1 and R3R3-R1 = abc  1abca2+1        b2         c2-1           1          0 -1           0           1         Applying C1 aC1, C2bC2 and C3cC3 =a2+1        b2         c2-1           1          0 -1           0           1=  -1 b2         c2 1          0  +  1 a2+1        b2 -1           1          Expanding along R3=-1 -c2  +a2+1+ b2=a2+1+ b2+c2=a2+ b2+c2+1=RHS

Page No 5.60:

Question 30:

1aa2a21aaa21= a3-12

Answer:

Let LHS=Δ= 1     a    a2a2    1    aa      a2  1Δ= 1 +a2+a    1 +a2+a    1 +a2+aa2                       1                   aa                        a2                   1       Applyng R1R1+R2+R2=  1 +a2+a 1     1        1 a2    1        aa     a2        1           Applying C2C2-C1 and C3C3-C1= 1 +a2+a  1         0             0 a2       1-a2      a-a2a         a2-a        1-a= 1 +a2+a 1         0                        0 a2        1-a1+a        a1-aa        aa-1                 1-a= 1 +a2+aa-1a-1 1           0               0a2      -1+a      -aa           a             -1     Taking out (a-1) common from C2 and C3= a3-1a-1 1           0                  0a       -1+a        -aa           a              -1                    1 +a2+aa-1=a3-1= a3-1a-11+a +a2=a3-1a3-1=a3-12= RHS 

Hence proved.

Page No 5.60:

Question 31:

a+b+c-c-b-ca+b+c-a-b-aa+b+c=2a+b b+c c+a

Answer:

Let LHS ==a+b+c         -c            -b-c              a+b+c           -a-b              -a                   a+b+c= a                -c                  -bb             a+b+c               -ac              -a                   a+b+c                  Applying C1C1+C2+C3= a+b              a+b                 -a+b  b+c             b+c                   b+c c                 -a                   a+b+c         Applying R1R1+R2 and R2R2+R3=a+bb+c    1            1                  -1  1            1                     1 c          -a           a+b+c      Taking out common factor from R 1 and R2=  a+bb+c  0        0              -2 1       1                 1     c       -a         a+b+c              Applying R1 R1- R2=a+bb+c -2-a-c     Expanding along R1=2 a+bb+c  c+a  =RHS

Hence proved.

Page No 5.60:

Question 32:

b+caabc+abcca+b=4abc

Answer:

=b+caabc+abcca+b=0-2c-2bbc+abcca+b                   Applying R1R1-(R2+R3)=0-2c-2bbc+a-b0c0a+b-c      Applying C2C2-C1 and C3C3-C1 =0c+a-b00a+b-c-(-2c)b0c  a+b-c-2bbc+a-bc0  Expanding along R1 =2c[b(a+b-c)-0]-2b[0-c(c+a-b)]=2bc[a+b-c]-2bc[b-c-a]=2bc[(a+b-c)-(b-c-a)]=4abc

Hence proved.

Page No 5.60:

Question 33:

b2+c2abacbac2+a2bccacba2+b2=4a2b2c2

Answer:


=b2+c2abacbac2+a2bccacba2+b2

a(b2+c2)a2ba2cb2ab(c2+a2)b2cc2ac2bc(a2+b2)   Multiplying the three rows by a, b and c 


=abcabcb2+c2a2a2b2c2+a2b2c2c2a2+b2        Taking out a, b and c common from the three columns=2(b2+c2)2(a2+c2)2(a2+b2)b2c2+a2b2c2c2a2+b2     Applying R1R1+R2+R3=2b2+c2a2+c2a2+b2-c20-a2-b2-a20   Taking out 2 common from the three columns and then applying R2R2-R1 and R3R3-R1 =20c2b2-c20-a2-b2-a20                            Applying R1R1+R2+R3=2{[-c2(-a2b2)]+[b2(c2a2)]}         Expanding along R1=4a2b2c2

Page No 5.60:

Question 34:

0b2ac2aa2b0c2ba2cb2c0=2a3b3c3

Answer:

=0  b2a  c2aa2b0c2ba2c  b2c0=1abc0b3ac3aa3b0c3ba3cb3c0     Multiplying the three columns by a, b and c =abcabc0b3c3a30c3a3b30             Taking out a, b and c common from the three rows =b3a3c3a30+c3a30a3b3=2a3b3c3        Expanding along R1

Page No 5.60:

Question 35:

Prove that a2+b2cccab2+c2aabbc2+a2b=4abc

Answer:

=a2+b2cccab2+c2aabbc2+a2b=1abca2+b2c2c2a2b2+c2a2b2b2c2+a2                           Multiplying R1, R2 and R3 by c,a and b and then dividing by abc=1abca2+b2 c2-a2-b2  c2-a2-b2a2b2+c2-a20b20  c2+a2-b2       Applying C2C2-C1 and C3C3-C1=1abc0 -2b2  -2a2a2b2+c2-a20b20  c2+a2-b2                Applying R1R1-R2-R3=1abc[-a2-2b2-2a20c2+a2-b2+b2-2b2-2a2b2+c2-a20    Expanding along C1=1abc-a2-2b2(c2+a2-b2)+b20+2a2b2+c2-a2=1abc-a2-2b2c2-2b2a2+2b4+b22a2b2+2a2c2-2a4=1abc2a2b2c2+2a4b2-2a2b4+2a2b4+2a2b2c2-2a4b2=1abc4a2b2c2=4abc

Hence proved.

Page No 5.60:

Question 36:

Prove that -bcb2+bcc2+bca2+ac-acc2+aca2+abb2+ab-ab=ab+bc+ca3

Answer:

=-bcb2+bcc2+bca2+ac-acc2+aca2+abb2+ab-ab=1abc-abcab2+abcac2+abca2b+abc-abcc2b+abca2c+abcb2c+abc-abc   Applying R1aR1, R2bR2 and R3cR3 and then dividing by abc=abcabc-bcab+acac+abab+bc-accb+abac+bcbc+ac-ab      Taking out a, b and c common from the three columnsab+bc+caab+bc+caab+bc+caab+bc-accb+abac+bcbc+ac-ab           Applying R1R1+R2+R3=(ab+bc+ca)111ab+bc-accb+abac+bcbc+ac-ab=(ab+bc+ca)0010-(ab+bc+ac)cb+abac+bc+abbc+ac+ab-ab    Applying C1C1-C3 and C2C2-C3=(ab+bc+ca)0-(ab+bc+ac)ac+bc+abbc+ac+ab=(ab+bc+ca)(ab+bc+ac)2=(ab+bc+ca)3

Hence proved.

Page No 5.60:

Question 37:

Prove the following identities:
x+λ2x2x2xx+λ2x2x2xx+λ=5x+λλ-x2

Answer:

LHS:x+λ2x2x2xx+λ2x2x2xx+λ=x+λ2x2x2x-x-λx+λ-2x02x-x-λ0x+λ-2x      Applying R2R2-R1 and R3R3-R1=x+λ2x2x-(λ-x)λ-x0-(λ-x)0λ-x=λ-x2x+λ2x2x-110-101      Taking λ-x common from R2 and λ-x common from R3=λ-x2-1-2x+1x+λ+2x      Expanding along last row=λ-x2λ+5x=RHS x+λ2x2x2xx+λ2x2x2xx+λ=λ-x2λ+5x

Page No 5.60:

Question 38:

Using properties of determinants prove that

x+42x2x2xx+42x2x2xx+4=5x+4 4-x2

Answer:

=x+42x2x2xx+42x2x2xx+4=5x+45x+45x+42xx+42x2x2xx+4              Applying R1R1+R2+R3=5x+41112xx+42x2x2xx+4              Take out 5x+4 common from R1=5x+41002x4-x02x04-x              Applying C2C2-C1 and C3C3-C1=5x+4(4-x)2                              Expanding along R1

Hence proved.

Page No 5.60:

Question 39:

Prove the following identities:

y+zzyzz+xxyxx+y=4xyz

Answer:

LHS:y+zzyzz+xxyxx+y=y+z-z-yz-z-x-xy-x-x-yzz+xxyxx+y      Applying R1R1-R2-R3=0-2x-2xzz+xxyxx+y=-2x011zz+xxyxx+y      Taking -2x common from R1=-2x001zzxy-yx+y      Applying C2C2-C3=-2x-zy-zy      Expanding along first row=4xyz=RHS y+zzyzz+xxyxx+y=4xyz



Page No 5.61:

Question 40:

-a b2+c2-a22b32c32a3-b c2+a2-b22c32a32b3-c a2+b2-c2=abc a2+b2+c23

Answer:

=-a(b2+c2-a2)2b32c32a3-b(c2+a2-b2)2c32a32b3-c(a2+b2-c2)=abc-b2-c2+a22b22c22a2-c2-a2+b22c22a22b2-a2-b2+c2            Taking out a, b and c common from C1, C2 and C3=abca2+b2+c22b22c2a2+b2+c2-c2-a2+b22c2a2+b2+c22b2-a2-b2+c2               Applying C1C1+C2+C3=abc(a2+b2+c2)12b22c21-c2-a2+b22c212b2-a2-b2+c2          Taking out a2+b2+c common from C1=abc(a2+b2+c2)12b22c20-c2-a2-b2000-a2-b2-c2         Applying R2R2-R1 and R3R3-R1=abc(a2+b2+c2)3     Expanding

Hence proved.

Page No 5.61:

Question 41:

Evaluate the following determinant:

(i) 1+a1111+a1111+a=a3+3a2

(ii) a2+2a2a+112a+1a+21331=a-13

Answer:

(ii) To Prove: a2+2a2a+112a+1a+21331=a-13


LHS=a2+2a2a+112a+1a+21331Applying R1R1-R2      =a2+2a-2a-12a+1-a-21-12a+1a+21331      =a2-1a-102a+1a+21331Taking a-1 common from R1      =a-1a+1102a+1a+21331Applying R2R2-R3      =a-1a+1102a+1-3a+2-31-1331      =a-1a+1102a-2a-10331Taking a-1 common from R2      =a-12a+110210331Expanding through C3      =a-1211a+1-2      =a-121a+1-2      =a-12a-1      =a-13=RHS


Hence, a2+2a2a+112a+1a+21331=a-13.

Page No 5.61:

Question 42:

Prove the following identities:

2yy-z-x2y2z2zz-x-yx-y-z2x2x=x+y+z3

Answer:

LHS=2yy-z-x2y2z2zz-x-yx-y-z2x2x=2y+2z+x-y-zy-z-x+2z+2x2y+z-x-y+2x2z2zz-x-yx-y-z2x2x      Applying R1R1+R2+R3=x+y+zx+y+zx+y+z2z2zz-x-yx-y-z2x2x=x+y+z1112z2zz-x-yx-y-z2x2x      Taking x+y+z common from R1=x+y+z01102zz-x-y-x-y-z2x2x      Applying C1C1-C2=x+y+z201102zz-x-y-12x2x      Taking x+y+z common from C1=x+y+z2-1z-x-y-2z      Expanding along first column=x+y+z3=RHS 2yy-z-x2y2z2zz-x-yx-y-z2x2x=x+y+z3

Page No 5.61:

Question 43:

Show that y+zxyz+xzxx+yyz=x+y+z x-z2

Answer:

 Let =y+z     x    yz+x     z    x x+y     y    z =2x+y+z     x+y+z    x+y+z      z+x             z             x       x+y            y              z      Applying R1 R1 +R2+ R3=x+y+z   2          1       1   z+x       z       x  x+y      y        z= x+y+z 0         1    10         z    x x-z   y    z                Applying C1C1-C2-C3= x+y+z x-z×11zx         Expanding along C1= x+y+z x-z2

Page No 5.61:

Question 44:

Prove the following identities:

a+xyzxa+yzxya+z=a2a+x+y+z

Answer:

LHS=a+xyzxa+yzxya+z=a+x+y+zyza+x+y+za+yza+x+y+zya+z      Applying C1C1+C2+C3=a+x+y+z1yz1a+yz1ya+z      Taking a+x+y+z common from C1=a+x+y+z1yz0a000a      Applying R2R2-R1 and R3R3-R1=a+x+y+za2      Expanding along first column=a2a+x+y+z=RHS a+xyzxa+yzxya+z=a2a+x+y+z

Page No 5.61:

Question 45:

Prove the following identities:

a32ab32bc32c=2a-bb-cc-aa+b+c

Answer:

LHS=a32ab32bc32c=a32ab3-a30b-ac3-a30c-a      Applying R2R2-R1 and R3R3-R1=-a-bc-aa32ab2+a2+ab01c2+a2+ac01      Taking b-a common from R2 and c-a common from R3=-a-bc-aa32ab2-c2+ab-ac00c2+a2+ac01      Applying R2R2-R3=-a-bc-aa32ab-ca+b+c00c2+a2+ac01=-a-bc-ab-ca+b+ca32a100c2+a2+ac01      Taking b-ca+b+c common from R2=-a-bc-ab-ca+b+c-2      Expanding along second column=2a-bc-ab-ca+b+c=RHS a32ab32bc32c=2a-bb-cc-aa+b+c

Page No 5.61:

Question 46:

Without expanding, prove that

abcxyzpqr = xyzpqrabc = ybqxapzcr

Answer:

xyzpqrabc R2R3=-xyzabcpqr R1R2=abcxyzpqrybqxapzcr =yxzbacqpr C1C2=-xyzabcpqr R1R2=abcxyzpqr

Hence proved.

Page No 5.61:

Question 47:

Show that x+1x+2x+ax+2x+3x+bx+3x+4x+c=0 where a, b, c are in A.P.

Answer:

Given: a, b, c are in A.P.

2b=a+c

=x+1x+2x+ax+2x+3x+bx+3x+4x+c              Applying R2=2R2=12x+1x+2x+a2x+42x+62x+2bx+3x+4x+c =12x+1x+2x+a000x+3x+4x+c          2b=a+c     Applying R2R2-R1+R3=0

Page No 5.61:

Question 48:

Show that x-3x-4x-αx-2x-3x-βx-1x-2x-γ=0, where α, β, γ are in AP.

Answer:

Given:α, β, γ areinA.P.

Now,
2β=α+γ

=x-3x-4x-αx-2x-3x-βx-1x-2x-γ =12x-3x-4x-α2x-42x-62x-2βx-1x-2x-γ           Applying R22R2=12x-3 x-4 x-α00 -2β+α+γx-1 x-2x-γ          2β=α+γ       Applying R2R2-R1+R3=12x-3 x-4   x-α000x-1x-2x-γ=0

Page No 5.61:

Question 49:

If a, b, c are real numbers such that b+cc+aa+bc+aa+bb+ca+bb+cc+a=0, then show that either a+b+c=0 or, a=b=c.

Answer:

  Let  Δ =b+c  c+a  a+bc+a  a+b b+ca+b b+c  c+a =2a+b+c  2a+b+c  2a+b+c   c+a            a+b          b+c   a+b            b+c          c+a              Applying R1R1+R2+R3=2a+b+c   1      1       1c+a  a+b  b+ca+b  b+c  c+a =2a+b+c   1      0       0c+a  b-c  b-aa+b  c-a  c-b              Applying C2C2-C1 and C3C3-C1=2a+b+c1b-cb-ac-ac-b=2a+b+cb-cc-b-b-ac-a=-2a+b+ca2+b2+c2-ab-bc-ca=-a+b+c2a2+2b2+2c2-2ab-2bc-2ca=-a+b+ca-b2+b-c2+c-a2But Δ=0         Given-a+b+ca-b2+b-c2+c-a2=0Either  a+b+c=0 or a-b2+b-c2+c-a2=0a+b+c=0 or a=b=cHence proved.

Page No 5.61:

Question 50:

If pbcaqcabr=0, find the value of pp-a+qq-b+rr-c, pa, qb, rc.

Answer:

Let =pbcaqcabr.
Now,
=pbcaqcabr   =pbc0q-bc-rabr        Applying R2R2-R3   =prq-b-bc-r+abc-r-cq-b        Expanding along first column   =prq-b+pbr-c-abr-c-acq-b   =pr-acq-b+bp-ar-cSince, =0. pr-acq-b+bp-ar-c=0pr-acp-ar-c+bq-b=0pr-ar+ar-acp-ar-c+bq-b=0rp-a+ar-cp-ar-c+bq-b=0rr-c+ap-a+bq-b=0pp-a+qq-b+rr-c=pp-a+qq-b-ap-a-bq-bpp-a+qq-b+rr-c=p-ap-a+q-bq-bpp-a+qq-b+rr-c=2Hence, the value of pp-a+qq-b+rr-c is 2.

Page No 5.61:

Question 51:

Show that x = 2 is a root of the equation

x-6-12-3xx-3-32xx+2=0 and solve it completely.

Answer:

Let =x-6-12-3xx-3-32xx+2=x-6-12-3xx-3-3-x2x+6x+3         Applying R3R3-R1=x+3x-6-12-3xx-3-121 =x+3x-23x-6-x+22-3xx-3-121              Applying R1R1-R2=x+3x-213-12-3xx-3-121 =x+3x-21302-3xx-1-120           Applying C3C3+C1=x+3x-2x-11302-3x1-120              =x+3x-2x-1-113-12       Expanding along C3=-5x+3x-2x-1x=2,-3, 1

Page No 5.61:

Question 52:

​Solve the following determinant equations:

(i) x+abcax+bcabx+c=0

(ii) x+axxxx+axxxx+a=0, a0

(iii) 3x-83333x-83333x-8=0

(iv) 1xx21aa21bb2=0, ab

(v) x+1352x+2523x+4=0

(vi) 1xx31bb31cc3=0, bc

(vii) 15-2x11-3x7-x111714101613=0

(viii) 11xp+1p+1p+x3x+1x+2=0

(ix) 3-2sin3θ-78cos2θ-11142=0

(x) 4-x4+x4+x4+x4-x4+x4+x4+x4+x=0

Answer:

(x) Given: 4-x4+x4+x4+x4-x4+x4+x4+x4+x=0



LHS=4-x4+x4+x4+x4-x4+x4+x4+x4+xApplying R2R2-R1      =4-x4+x4+x4+x-4+x4-x-4-x4+x-4-x4+x4+x4+x      =4-x4+x4+x2x-2x04+x4+x4+xTaking 2x common from R2      =2x4-x4+x4+x1-104+x4+x4+xApplying R3R3-R1      =2x4-x4+x4+x1-104+x-4+x4+x-4-x4+x-4-x      =2x4-x4+x4+x1-102x00Expanding through C3      =2x4+x1×0+1×2x      =2x4+x2x      =2x24+xThus, 4-x4+x4+x4+x4-x4+x4+x4+x4+x=2x24+xBut it is given that, 4-x4+x4+x4+x4-x4+x4+x4+x4+x=02x24+x=02x2=0 or 4+x=0x=0 or x=-4

Hence, x = 0, −4.

 



Page No 5.62:

Question 53:

If a,b and c are all non-zero and 1+a1111+b1111+c=0, then prove that 1a+1b+1c+1=0.

Answer:

We have,

1+a1111+b1111+c=0

C1 C1 -C2a11-b1+b1011+c =0C2 C2 -C3a01-bb10-c1+c =0Expanding along R1, we geta(b+bc+c)+1(bc)=0ab+abc+ac+bc=0Dividing by abc, we get1c+1+1b+1a=0 1a+1b+1c+1=0

Page No 5.62:

Question 54:

If ab-yc-za-xbc-za-xb-yc=0, then using properties of determinants, find the value of ax+by+cz, where x,y,z0.

Answer:

 ab-yc-za-xbc-za-xb-yc=0

R1 R1 - R2x-y0a-xbc-za-xb-yc =0R2 R2 - R3x-y00y-za-xb-yc =0Expanding along first row, we getx(yc+zb-zy)+y(0-za+zx) = 0xyc+xzb-xyz+zya-xyz =0 xyc+xzb-2xyz+zya=0Dividing by xyz, we getcz+by-2+ax=0 ax+by+cz=2
 

Page No 5.62:

Question 55:

Using properties of determinants, prove that 111+3x1+3y1111+3z1=93xyz+xy+yz+zx.

Answer:



Let =111+3x1+3y1111+3z1Applying R2R2 - R1,  R3R3 - R1 =111+3x3y0-3x03z-3x
Expanding along R1 ,we get
=1(0+9xz) - 1(-9xy-0)+(1+3x)(9yz-0)= 9xz+9xy+9yz+27xyz= 93xyz +xy +yz+zx



Page No 5.71:

Question 1:

Find the area of the triangle with vertices at the points:
(i) (3, 8), (−4, 2) and (5, −1)
(ii) (2, 7), (1, 1) and (10, 8)
(iii) (−1, −8), (−2, −3) and (3, 2)
(iv) (0, 0), (6, 0) and (4, 3).

Answer:

(i)
=12381-4215-11 =12381-7-605-11       Applying R2R2-R1=123 81-7-602-90      Applying R3R3-R1=12-7-62-9=1263 + 12= 1275 = 752square units

(ii)
=122711111081 =12271-1-601081      Applying R2R2-R1=12 2  71-1-60  8  10       Applying R3R3-R1=12-1-681=12-1 + 48=1247 = 472square units

(iii)
=12-1-81-2-31321 =12-1-81-150321             Applying R2R2-R1=12-1-81-1504100             Applying R3R3-R1=12-15410=12-10-20=1230=15 square units

(iv)
=12001601431 =12001600431         Applying R2R2-R1=12001600430         Applying R3R3-R1=126043=1218-0=1218=9 square units

Page No 5.71:

Question 2:

Using determinants show that the following points are collinear:
(i) (5, 5), (−5, 1) and (10, 7)
(ii) (1, −1), (2, 1) and (4, 5)
(iii) (3, −2), (8, 8) and (5, 2)
(iv) (2, 3), (−1, −2) and (5, 8)

Answer:

(i) If the points  (5, 5), (−5, 1) and (10, 7) are collinear, then

=551-5111071=0=551-10-401071      Applying R2R2-R1= 5 51-10-40520      Applying R3R3 - R1=-10-452=-20 + 20 = 0


Thus, these points are colinear.

(ii) If the points (1, −1), (2, 1) and (4, 5) are collinear, then

=1-112  114 51 = 0=1-111  204 51         Applying R2R2-R1=1-11120360         Applying R3R3-R1=1236 = 6 - 6 = 0

Thus, these points are collinear.

(iii) If the points (3, −2), (8, 8) and (5, 2)  are collinear, then

=3-21881521 = 0=3-215100521       Applying R2R2-R1=3-215100240       Applying R3R3-R1=51024=20 - 20 = 0

Thus the points are colinear.

(iv) If the points (2, 3), (−1, −2) and (5, 8) are collinear, then

=231-1-21581=0=231-3-50581                Applying R2R2-R1=231-3-50350                Applying R3R3-R1=-3-535=-15+15=0

Thus the points are colinear.

Page No 5.71:

Question 3:

If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.

Answer:

If the points (a, 0), (0, b) and (1, 1) are collinear, then

a010b1111=0a01-ab0111=0        Applying R2R2-R1a01-ab01-a10=0      Applying R3R3-R1=-ab1-a1=0-a-b1-a=0a+b=ab

Page No 5.71:

Question 4:

Using determinants prove that the points (a, b), (a', b') and (aa', bb') are collinear if ab' = a'b.

Answer:

ab1a'b'1a-a'b-b'1=ab1a'-ab'-b0a-a'b-b'1              Applying R2R2-R1=ab1a'-ab'-b0-a'-b'0              Applying R3R3-R1=a'-ab'-b-a'-b'=-b'a'-a + a'b'-b=-b'a' + b'a + a'b' - a'b=b'a - a'b

If the points are collinear, then ∆ = 0. So,
ab' − a'b = 0

Thus, ab' = a'b

Page No 5.71:

Question 5:

Find the value of λ so that the points (1, −5), (−4, 5) and λ, 7 are collinear.

Answer:

If the points (1, −5), (−4, 5) and λ, 7 are collinear, then

1-51-451λ71=0  1-51-5100  λ71 = 0           Applying R2R2-R11-51-5100λ-1120 = 0          Applying R3R3-R1=-510λ-112 = 0-60-10λ-1 = 0-60-10λ+10=0-10λ=50λ=-5

Page No 5.71:

Question 6:

Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).

Answer:

=12x412-61541=±35=12x412-x-100541=±35          Applying R2R2-R1=12x412-x-1005-x00=±35           Applying R3R3-R1=122-x-105-x0=±35=0 + 105 - x = ±7050 - 10x = 70 or 50 - 10x = -70-10x = 20 or -10x=-120x=-2 or x =12

Page No 5.71:

Question 7:

Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?

Answer:



=12141231-5-31=121  4 11-10-5-31               Applying R2R2-R1=121411-10-6-70                Applying R3R3-R1=121-1-6-7=12-7 - 6=132square units        Area cannot be negative

Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because, 141231-5-31 is not equal to 0.

Page No 5.71:

Question 8:

Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).

Answer:

Given:
Vertices of triangle: (− 3, 5), (3, − 6) and (7, 2)

Area of the triangle==12-3  513-617 21=12-3516-110721        Applying R2R2-R1=12-351 6-11010-30         Applying R3R3-R1=126-1110-3=12-18 + 110=46 square units

Page No 5.71:

Question 9:

Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.

Answer:

If the points (k, 2 − 2 k), (− k + 1, 2k) and (− 4 − k, 6 − 2k) are collinear, then

=k2-2k1-k+12k1-4-k6-2k1=0k2-2k1-2k+14k-20-4-k6-2k1=0          Applying R2R2-R1k2-2k1-2k+14k-20-4-2k40=0            Applying R3R3-R1-2k+14k-2-4-2k4=0-8k + 4 + 16k - 8 + 8k2 - 4k = 0 8k2 + 4k - 4 = 08k - 4k + 1 = 0k = -1 or k = 12

Page No 5.71:

Question 10:

If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.

Answer:

If the points (x, −2), (5, 2), (8, 8) are collinear, then

x-21521881=0=x-21521881=x-215-x40881         Applying R2R2-R1=x-215-x 408-x100            Applying R3R3-R1=5-x48-x10=50 - 10x - 32 + 4x=18 - 6x=18 - 6x= 0          Given18 - 6x = 0x = 3



Page No 5.72:

Question 11:

If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.

Answer:

If the points (3, −2), (x, 2) and (8, 8) are collinear, then

3-21x21881 = 0=3-21x21881=3-21x-340881        Applying R2R2-R1=3-21x-3405100           Applying R3R3-R1=x-34510=10x - 30 - 20=10x - 50=0            Given10x - 50 = 010x = 50x = 5

Page No 5.72:

Question 12:

Using determinants, find the equation of the line joining the points
(i) (1, 2) and (3, 6)
(ii) (3, 1) and (9, 3)

Answer:

(i)
Given: A  =  (1, 2) and B  =  (3, 6)

Let the point P be (x, y).  So,
Area of triangle ABP = 0

=12121361xy1 = 016 - y - 23 - x + 13y - 6x = 06 - y - 6 + 2x + 3y - 6x = 02y - 4x = 0y = 2x

(ii)
Given: A = (3, 1) and B = (9, 3)

Let the point P be (x, y). So,

Area of triangle ABP = 0

=12311931xy1=033-y-19-x+19y-3x=09-3y-9+x+9y-3x=0-2x+6y=0x=3y

Page No 5.72:

Question 13:

Find values of k, if area of triangle is 4 square units whose vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)

Answer:

(i)  If the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then Δ=12k   0   14   0   10   2   1 =12 2  × k   14   1             Expanding along C2=k-4Since area is always+ve, we take its absolute value, which is given as 4 square units.( k-4 )=±4(k-4)=4 or (k-4 )=-4k-4=4 or k-4 =-4k=8 or k=0k=8, 0(ii)If the area of a triangle with vertices (-2, 0) (0, 4) and (0, k) is 4 square units, then 1 =12-2        0      1   0       4       1   0        k      1=12  -2×4 1k 1          Expanding along C1=-4-kSince area is always+ve, we  take its absolute value, which is given as 4 square units.-4-k=±4-4-k=±4-4-k=4 or -4-k=-4k=4+4 or k =-4+4 k=8 or k=0



Page No 5.84:

Question 1:

x − 2y = 4
−3x + 5y = −7

Answer:

Given:     x-2y = 4           -3x + 5y = -7Using the properties of determinants, we getD=   1   -2-3      5   = 5 - 6 = -1  0D1=  4      -2 -7       5 = 20 - 14 =  6D2 =  1         4-3    -7 = -7 + 12 = 5Using Cramer's Rule, we getx=D1D =   6-1 = -6y= D2D = 5-1  = -5 x=-6 and y =-5

Page No 5.84:

Question 2:

2xy = 1
7x − 2y = −7

Answer:

Given:2x - y = 1           7x - 2y = -7Using Crammer's Rule, we getD =2      -17     -2 =-4 + 7 =  3D1=  1      -1-7    -2=-2 - 7 = -9D2=2        1 7    -7 = -14  - 7 = -21Now,x=D1D = -93= -3y=D2D= -213 = -7 x=-3 and y=-7

Page No 5.84:

Question 3:

2xy = 17
3x + 5y = 6

Answer:

Given: 2x - y = 17             3x + 5y = 6Using Cramers Rule, we get D= 2   -1 3      5 = 10 +  3 = 13D1=17   -1  6       5 = 85 + 6 = 91D2= 2    17  3     6 = 12 - 51 = -39Now,x= D1D = 9113 = 7y=D2D = -39  13=-3 x = 7 and y = -3

Page No 5.84:

Question 4:

3x + y = 19
3xy = 23

Answer:

Given:3x + y = 19             3x - y = 23Using Cramer's Rule, we getD=3      1 3  -1= -3 - 3 = -6D1=19      123   -1  =-19 - 23 = -42D2=3    19 3    23= 3 × 23 - 3 × 19 = 3 × 4 = 12Now,x=D1D=-42-6 = 7y=D2D= 12-6 = -2 x=7 and y=-2

Page No 5.84:

Question 5:

2xy = − 2
3x + 4y = 3

Answer:

Given: 2x - y = -2              3x + 4y = 3Using Cramer's Rule, we get D=2  -1 3     4= 8 + 3 = 11D1 =-2   -1  3       4 =-8 + 3 =-5D2=2   -2 3      3 = 6 + 6 = 12Now,x=D1 D = -511y=D2D = 1211 x =-511 and y = 1211

Page No 5.84:

Question 6:

3x + ay = 4
2x + ay = 2, a ≠ 0

Answer:

Given: 3x+ay = 4             2x+ay=  2  Using Cramer's rule, we get D =3   a2   a=3a-2a=aD1 =4   a2   a  =4a-2a =2aD2=3   4 2   2 =6-8=-2Now,x=D1 D=2aa=2y=D2D=-2a=-2a x=2 and y=-2a

Page No 5.84:

Question 7:

2x + 3y = 10
x + 6y = 4

Answer:

Given: 2x + 3y =10             x + 6y = 4Using Cramer's Rule, we get D =2    31    6 =12 - 3 = 9D1 =10   34    6=60 - 12 = 48D2= 2   101     4 = 8 - 10 =-2Now,x= D1 D = 489 = 163y=D2D = -29 x = 163 and y = -29

Page No 5.84:

Question 8:

5x + 7y = − 2
4x + 6y = − 3

Answer:

Given:5x + 7y = -2            4x + 6y = -3Using Cramer's Rule, we get D= 5    7  4    6= 30 - 28 = 2D1 =-2    7-3   6=-12 + 21 = 9D2=5  -2  4  -3= -15 + 8 = -7Now,x= D1 D =  92y = D2D = -72 x = 92 and y = -72

Page No 5.84:

Question 9:

9x + 5y = 10
3y − 2x = 8

Answer:

Given: 9x + 5y = 10               3y - 2x = 8   Rearranging the second equation, the two equations can be written as   9x + 5y = 10-2x + 3y = 8Now,D =   9     5-2    3 = 27 + 10 = 37D1 = 10  5 8   3 = 30 - 40 = -10D2 =   9      10 -2    8  = 72 + 20 = 92Using Cramer's rule, we getx= D1D=-1037 y= D2D=9237 x=-1037  and y=9237

Page No 5.84:

Question 10:

x + 2y = 1
3x + y = 4

Answer:

Given: x + 2y = 1
            3x + y = 4

D=1231 = -5D1=1241 = -7D2=1134 = 1Now,x = D1D = 75y = D2D = -15 x = 75 and y = -15

Page No 5.84:

Question 11:

3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11

Answer:

Given: 3x + y + z = 2
            2x − 4y + 3z = − 1
            4x + y − 3z = − 11


D=3112-4341-3= 312 - 3 - 2-3 - 1 + 43 + 4= 27 + 8 + 28= 63D1=211-1-43-111-3=212 - 3 + 1-3 - 1 - 113 + 4=18 - 4 - 77=-63D2=3212-134-11-3=33 + 33 - 2-6 + 11 + 46 + 1=108 - 10 + 28=126D3=3122-4-141-11=344 + 1 - 2-11 - 2 + 4-1 + 8=135 + 26 + 28=189Now,x=D1D=-63  63=-1y=D2D=12663=2z=D3D=18963=3 x=-1, y=2 and z=3

Page No 5.84:

Question 12:

x − 4yz = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

Answer:

Given: x − 4yz = 11
           2x − 5y + 2z = 39
           − 3x + 2y + z = 1


D=1-4-12-52-321=1-5-4-(-4)(2+6)+(-1)(4-15)=1(-9)-(-4)(8)+(-1)(-11)=34D1=11-4-139-52121=11(-5-4)-(-4)(39-2)+(-1)(78+5)=11(-9)-(-4)(37)+(-1)(83)=-34D2=111-12392-311=1(39-2)-11(2+6)+(-1)(2+117)=1(37)-11(8)+(-1)(119)=-170D3=1-4112-539-321=1(-5-78)-(-4)(2+117)+11(4-15)=1(-83)-(-4)(119)+11(-11)=272Now,x=D1D=-3434=-1y=D2D=-17034=-5z=D3D=27234=8 x = −1, y= −5and z=8

Page No 5.84:

Question 13:

6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

Answer:

Given: 6x + y − 3z = 5
            x + 3y − 2z = 5
           2x + y + 4z = 8


D=61-313-2214=6(12+2)-1(4+4)-3(1-6)=6(14)-1(8)-3(-5)=91D1=51-353-2814=5(12+2)-1(20+16)-3(5-24)=5(14)-1(36)-3(-19)=91D2=65-315-2284=6(20+16)-5(4+4)-3(8-10)=6(36)-5(8)-3(-2)=182D3=615135218=6(24-5)-1(8-10)+5(1-6)=6(19)-1(-2)+5(-5)=91Now,x=D1D=9191=1y=D2D=18291=2z=D3D=9191=1 x=1, y=2 and z=1

Page No 5.84:

Question 14:

x+ y = 5
y + z = 3
x + z = 4

Answer:

These equations can be written as
x + y + 0z = 5
0x + y + z = 3
x + 0y + z = 4

D=110011101=1(1-0)-1(0-1)+0(0-1)=1(1)-1(-1)+0=2D1=510311401=5(1-0)-1(3-4)+0(0-4)=5(1)-1(-1)=6D2=150031141=1(3-4)-5(0-1)+0(0-4)=1(-1)-5(-1)=4D3=115013104=1(4-0)-1(0-3)+5(0-1)=1(4)-1(-3)+5(-1)=2Now,x=D1D=62=3y=D2D=42=2z=D3D=22=1 x=3, y=2 and z=1

Page No 5.84:

Question 15:

2y − 3z = 0
x + 3y = − 4
3x + 4y = 3

Answer:

These equations can be written as
0x + 2y − 3z = 0
x + 3y + 0z = − 4
 3x + 4y + 0z = 3

D=02-3130340=0(0-0)-2(0-0)-3(4-9)=15D1=02-3-430340=0(0-0)-2(0-0)-3(-16-9)=75D2=00-31-40330=0(0-0)-0(0-0)-3(3+12)=-45D3=02013-4343=0(9+16)-2(3+12)-0(4-9)=-30Now,x=D1D=7515=5y=D2D=-4515=-3z=D3D=-3015=-2 x=5, y=-3 and z=-2

Page No 5.84:

Question 16:

5x − 7y + z = 11
6x − 8yz = 15
3x + 2y − 6z = 7

Answer:

Given: 5x − 7y + z = 11
           6x − 8yz = 15
           3x + 2y − 6z = 7

D=5-716-8-132-6=5(48+2)+7(-36+3)+1(12+24)=5(50)+7(-33)+1(36)=55D1=11-7115-8-172-6=11(48+2)+7(-90+7)+1(30+56)=11(50)+7(-83)+1(86)=55D2=5111615-137-6=5(-90+7)-11(-36+3)+1(42-45)=5(-83)-11(-33)+1(-3)=-55D3=5-7116-815327=5(-56-30)+7(42-45)+11(12+24)=5(-86)+7(-3)+11(36)=-55Now,x=D1D=5555=1y=D2D=-5555=-1z=D3D=-5555=-1 x=1, y=-1 and z=-1

Page No 5.84:

Question 17:

2x − 3y − 4z = 29
− 2x + 5yz = − 15
3xy + 5z = − 11

Answer:

Given: 2x − 3y − 4z = 29
          − 2x + 5yz = − 15
           3xy + 5z = − 11

D=2-3-4-25-13-15=2(25-1)+3(-10+3)-4(2-15)=2(24)+3(-7)-4(-13)=79D1=29-3-4-155-1-11-15=29(25-1)+3(-75-11)-4(15+55)=29(24)+3(-86)-4(70)=158D2=229-4-2-15-13-115=2(-75-11)-29(-10+3)-4(22+45)=2(-86)-29(-7)-4(67)=-237D3=2-329-25-153-1-11=2(-55-15)+3(22+45)+29(2-15)=2(-70)+3(67)+29(-13)=-316Now,x=D1D=15879=2y=D2D=-23779=-3z=D3D=-31679=-4 x=2, y=-3 and z=-4

Page No 5.84:

Question 18:

x + y = 1
x + z = − 6
xy − 2z = 3

Answer:

These equations can be written as
x+ y + 0z = 1
x + 0y + z = − 6
xy − 2z = 3


D=1101011-1-2=1(0+1)-1(-2-1)+0(-1-0)=4D1=110-6013-1-2=1(0+1)-1(12-3)+0(6-0)=-8D2=1101-6113-2=1(12-3)-1(-2-1)+0(3+6)=12D3=11110-61-13=1(0-6)-1(3+6)+1(-1-0)=-16Now,x=D1D=-84=-2y=D2D=124=3z=D3D=-164=-4 x=-2, y=3 and z=-4

Page No 5.84:

Question 19:

x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0

Answer:

These equations can be written asx + y + z = -1ax + by + cz = -da2x + b2y + x2z = -d2D=111abca2b2c2           =100aa-bb-ca2a2-b2b2-c2            Applying C2C1-C2 , C3C2-C3Taking (b-a) and (c-a) common from C1 and C2, respectively, we get =(a-b)(b-c)100a11a2a+bb+c=(a-b)(b-c)(c-a)        (1)D1=-111-dbc-d2b2c2 = -111dbcd2b2c2D1=-(d-b) (b-c) (c-d)             Replacing a by d in eq. (1)D2=1-11a-dca2-d2c2 = -111adca2d2c2D2=-(a-d)(d-c)(c-a)          Replacing b by d in eq. (1)D3=11-1ab-da2b2-d2 = -111abda2b2d2D3=-(a-b)(b-d)(d-a)           Replacing c by d in eq. (1)Thus,x = D1D = -(d-b)(b-c)(c-d)(a-b)(b-c)(c-a)y = D2D = -(a-d)(d-c)(c-a)(a-b)(b-c)(c-a)z = D3D = -(a-b)(b-d)(d-a)(a-b)(b-c)(c-a)

Page No 5.84:

Question 20:

x + y + z + w = 2
x − 2y + 2z + 2w = − 6
2x + y − 2z + 2w = − 5
3xy + 3z − 3w = − 3

Answer:

D=11111-22221-223-13-31-2221-22-13-3 - 11222-2233-3 + 11-222123-1-3 - 11-2221-23-13=1-26 - 6-2-3 + 2 + 23 - 2 -116 - 6 -2-6 - 6 + 26 + 6 + 11-3 + 2 + 2-6 - 6 + 2-2 - 3 - 113 - 2 + 26 + 6 +2-2 - 3=4 - 48 - 35 - 15=-94D1=2111-6-222-51-22-3-13-32-2221-22-13-3 -1-622-5-22-33-3 +1-6-22-512-3-1-3 -1-6-22-51-2-3-13=2-26 - 6 -2-3 + 2 + 23 - 2 -1-66 - 6 -215 + 6 +2-15 - 6 +1 -6 -3 + 2 + 215 + 6 +25 + 3 -1 -63 - 2 +2-15 - 6 +25 + 3=188D2=12111-6222-5-223-33-31-622-5-22-33-3-21222-2233-3+11-622-523-3-3-11-622-5-23-331-66 - 6 -215 + 6 +2-15 - 6  -216 - 6 -2-6 - 6 +26 + 6 + 1115 + 6 + 6-6 - 6 +2-6 + 15 -11-15 - 6 -66 + 6 +2 -6 + 15=1D3=11211-2-6221-523-1-3-31-2-621-52-1-3-3-11-622-523-3-3+21-222123-1-3-11-2-621-53-1-3=1 -215 + 6 +6-3 + 2 +2-3 - 5 -1115 + 6 + 6-6 - 6 +2-6 + 15 +21 -3 + 2 + 2-6 - 6 + 2-2 - 3 -11-3 - 5 +2-6 + 15 -6-2 - 3=-141D4=11121-22-621-2-53-13-31-22-61-2-5-13-3-112-62-2-533-3+11-2-621-53-1-3-21-2221-23-1-31-26 + 15 -2-3 -  5 -63 - 2 -116 + 15 -2-6 + 15 -66 + 6 +11- 3 - 5 +2-6 + 15 -6-2 - 3  -21 -3 - 2 +2-6 + 6 +2-2 - 3=47Thus,x = D1D = 188-94 = -2y = D2D = -282-94 = 3z= D3D = -141-94 = 1.5w = D4D = 47-94 = -0.5

Page No 5.84:

Question 21:

2x − 3z + w = 1
xy + 2w = 1
− 3y + z + w = 1
x + y + z = 1

Answer:

D=20-311-1020-31111102-102-311110-0-31-120-31110-11-100-31111= 2-10-1-00-1+2-3-1-310-1+10-1+20+3-11-3-1+10-1+00+3= -21D1= 10-311-1021-31111101-102-311110-0-31-121-31110-11-101-31111=1-10-1-00-1+2-3-1-310-1+10-1+21+3-11-3-1+10-1+21+3=-21D2=21-31110201111110=2102111110-1102011110+(-3)112011110-1110011111210-1+21-1-110-1+20-1-310-1-10-1+20-1-111-1-10-1=6D3=20111-1120-3111110=2-112-311110-0+11-120-31110-11-110-31111=2-10-1-10-1+2-3-1+110-1 +10-1+20+3-11-3-1+10-1+10+3=-6D4=20-311-1010-3111111= 2-101-311111-0-31-110-31111-11-100-31111= 2-11-1+1-3-1-31-3-1+10-1+10+3-11-3-1+10-1= 3So, by Cramer's rule , we obtain x = D1D = 2121= 1y= D2D = 6-21 = -27z = D3D = -6-21 = 27w=D4D=3-21=-17Hence, x = 1, y =  -27, z = 27,  w= -17

Page No 5.84:

Question 22:

2xy = 5
4x − 2y = 7

Answer:

Given: 2xy = 5
            4x − 2y = 7

D=2-14-2 = -4 + 4 = 0D1=5-17-2=-10 + 7 = -3D2=2547 = 14 - 20 = -6

Here, D1 and D2 are non-zero, but D is zero. Thus, the given system of linear equations is inconsistent.

Page No 5.84:

Question 23:

3x + y = 5
− 6x − 2y = 9

Answer:

Given: 3x + y = 5
          − 6x − 2y = 9

D=3   1-6-2 = -6 + 6 = 0D1=5    19-2 = -10 - 9 = -19D2=35-69 = 27 + 30 = 57

Here, D1 and D2 are non-zero, but D is zero. Thus, the system of linear equations is inconsistent.

Page No 5.84:

Question 24:

3xy + 2z = 3
2x + y + 3z = 5
x − 2yz = 1

Answer:

Given: 3xy + 2z = 3
            2x + y + 3z = 5    
            x − 2yz = 1

D=3-122131-2-1=3-1+6+1-2-3+2-4-1=0D1=3-125131-2-1=3-1+6+1-5-3+2-10-1=-15D2=33225311-1=3-5-3-3-2-3+22-5=-15D3=3-132151-21=31+10+12-5+3-4-1=-15

Here, D is zero, but D1, D2 and D3​ are non-zero. Thus, the system of linear equations is inconsistent.

Page No 5.84:

Question 25:

3xy + 2z = 6
2xy + z = 2
3x + 6y + 5z = 20.

Answer:

Given: 3xy + 2z = 6
            2xy + z = 2
            3x + 6y + 5z = 20

D=3-122-113653-5 - 6 + 110 - 3 + 212 + 3 = 4 

Since D is non-zero, the system of linear equations is consistent and has a unique solution.

D1=6-122-112065=6-5 - 6 + 110 - 20 + 212 + 20=-66-10+64=-12D2=3622213205=310-20-610-3+240-6=-30-42+68=-4D3=3-162-123620=3-20-12+140-6+612+3=-96+34+90=28Now,x=D1D=-12  4=-3y=D2D=-44=-1z=D3D=284=7 x=-3, y=-1 and z=7



Page No 5.85:

Question 26:

xy + z = 3
2x + yz = 2
x − 2y + 2z = 1

Answer:

Using the equations we get D=1-1  12  1-1-1-2  212 - 2 + 14 - 1 + 1-4 + 1 = 0D1=3-1   12   1-11-2  232 - 2 + 14 + 1 + 1-4 - 1 = 0D2=   13   1  22-1-11   214 + 1 - 34 - 1 + 12 + 2 = 0D3= 1-13  2   12-1-2111 + 4 + 12 + 2 + 3-4 + 1 = 0

Here,
D=D1=D2=D3=0
Thus, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 27:

x + 2y = 5
3x + 6y = 15

Answer:

Using the equations, we get D = 1236 = 6 - 6 = 0D1 = 52156 = 30 - 30 = 0D2= 15315 = 15 - 15 = 0

  D=D1=D2

Hence, the system of linear equation has infinitely many solutions.

Page No 5.85:

Question 28:

x + yz = 0
x − 2y + z = 0
3x + 6y − 5z = 0

Answer:

Using the equations we get D=11-11-2   136-5110 - 6 -1-5 - 3 -16 + 6 = 0D1=01-10-2   106-5010 - 6 -10 - 0 -10 + 0 = 0D2=10-110   130-510 - 0 -0-5 - 3 -10 - 0 = 0D3=1101-2036010 - 0 -10 - 0 + 06 + 6 = 0

 D=D1=D2

Hence, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 29:

2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2

Answer:

Using the equations we getD=2   1-21-2  15-5 12-2 + 5 - 11 - 5 -2-5 + 10 = 0D1=41-2-2-2 1-2-5 14-2 + 5 -1-2 + 2 -210 - 4 = 0D2=24-21-2 15-2 12-2 + 2 -41- 5 -2-2 + 10 = 0D3=2141-2-25-5-224 - 10 -1-2 + 10 + 4-5 + 10 = 0

 D=D1=D2=0

Hence, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 30:

xy + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10

Answer:

Using the equations, we get  D=1-1  313-3533=1(9 + 9) + 1(3 + 15) + 3(3 - 15)=18 + 18 - 36 = 0D1=6-13-43-31033=6(9 + 9) + 1(-12 + 30) + 3(-12 - 30)=108 + 18 - 126 = 0D2=1631-4-35103 =1(-12 + 30) -6(3 + 15) + 3(10 + 20)=18 - 108 + 90 = 0D3=1-1613-45310=1(30 + 12) + 1(10 + 20) + 6(3 - 15)= 42 + 30 - 72 = 0 D = D1= D2= D3= 0

Hence, the system of equations has infinitely many solutions.

Page No 5.85:

Question 31:

A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission
 

Month Sale of units Total commission
drawn (in Rs)
  A B C  
Jan 90 100 20 800
Feb 130 50 40 900
March 60 100 30 850

Find out the rates of commission on items A, B and C by using determinant method.

Answer:

Let x, y and z be the rates of commission on items A, B and C respectively. Based on the given data, we get

90x + 100y + 20z = 800130x + 50y + 40z = 90060x + 100y + 30z = 850

Dividing all the equations by 10 on both sides, we get

9x + 10y + 2z = 8013x + 5y + 4z = 906x + 10y + 3z = 85


D=910213546103        Expressing the equation as a determinant=9(15-40)-10(39-24)+2(130-30)=9(-25)-10(15)+2(100)=-175D1=80102905485103=80(15-40)-10(270-340)+2(900-425)=80(-25)-10(-70)+2(475)=-350D2=9802139046853=9(270-340)-80(39-24)+2(1105-540)=9(-70)-80(15)+2(565)=-700D3=910801359061085=9(425-900)-10(1105-540)+80(130-30)=9(-475)-10(565)+80(100)=-1925Thus,x=D1D=-350-175=2y=D2D=-700-175=4z=D3D=-1925-175=11

Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively.

Page No 5.85:

Question 32:

An automobile company uses three types of steel S1, S2 and S3 for producing three types of cars C1, C2 and C3. Steel requirements (in tons) for each type of cars are given below :
 

  Cars
C1
C2 C3
Steel S1 2 3 4
S2 1 1 2
S3 3 2 1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

Answer:

 Let x, y and z denote the number of cars that can be produced of each type. Then,2x + 3y + 4z = 29x + y + 2z = 133x + 2y + z = 16Using Cramer's rule, we getD=234112321=2(1 - 4) - 3(1 - 6) + 4(2 - 3)=-6 + 15 - 4= 5D1=293413121621=29(1 - 4) - 3(13 - 32) + 4(26 - 16)=-87 + 57 + 40= 10D2=229411323161=2(13 - 32) - 29(1 - 6) + 4(16 - 39)=-38 + 145 - 92= 15D3=232911133216=2(16 - 26) - 3(16 - 39) + 29(2 - 3)=-20 + 69 - 29= 20Thus,x = D1D = 105 = 2y = D2D = 155 = 3z = D3D = 205 = 4

Therefore, 2 C1 cars, 3 C2 cars and 4 C​3 cars can be produced using the three types of steel.



Page No 5.89:

Question 1:

Solve each of the following system of homogeneous linear equations.
x
+ y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

Answer:

Given: x + y − 2z = 0
            2x + y − 3z = 0              
            5x + 4y − 9z = 0

D=11-221-354-9=1(-9 + 12) - 1(-18 + 15) - 2(8 - 5)=0So, the system has infinitely many solutions. Putting z=k in the first two equations, we getx + y = 2k2x + y = 3kUsing Cramer's rule, we getx = D1D = 2k13k11121 = -k-1 = ky = D2D = 12k23k1121 = -k-1 = k z=kClearly, these values satisfy the third equation.Thus, x = y = z = k         kR

Page No 5.89:

Question 2:

Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2xy + 3z = 0

Answer:

D = 2  341  112-13         = 2 (3 + 1) -3 (3 - 2) + 4(-1 -2)     = 8 - 3 - 12     = -7So, the given system of equations has only the trivial solution i.e. x = 0, y = 0,  z= 0
 

Page No 5.89:

Question 3:

Solve each of the following system of homogeneous linear equations.
3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0

Answer:

Given: 3x + y + z = 0
            x − 4y + 3z = 0
            2x + 5y − 2z = 0

D=3   1   11-4   32  5-2=0The system has infinitely many solutions. Putting z=k in the first two equations, we get3x + y = -kx - 4y = -3kSolving these equations by Cramer's rule, we getx = D1D = -k   1-3k-43    11-4 = -7k13y=D2D=3-k1-3k3    11-4=8k13 z = k x = -7k13, y = 8k13 and z = kor  x = -7k, y = 8k and z = 13kClearly, these values satisfy the third equation.Thus, x=-7ky = 8k            kRz = 13k        

Page No 5.89:

Question 4:

Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions

2 λ x-2y+3z=0       x+λy+2z=0            2x+λ z=0

Answer:

The given system of equations can be written as2λx - 2y + 3z = 0x + λy + 2z = 02x + 0y + λz = 0The given system of equations will have non-trivial solutions if D=0.2λ-231λ220λ = 02λ(λ2) + 2(λ - 4) + 3(-2λ) =02λ3 - 4λ - 8 =0λ = 2So, the given system of equations will have non-trivial solutions if λ=2.Now, we shall find solutions for λ = 2.Replacing z by k in the first two equations, we get2λx - 2y = -3kx + λy = -2kSolving these by Cramer's rule, we getx = -3k-2-2kλ2λ-21λ = -3kλ - 4k2λ2 + 2 = -3k(2) - 4k2(2)2 + 2 = -6k - 4k10 = -ky=2λ-3k1-2k2λ-21λ = -4kλ + 3k2λ2 + 2 = -4k(2) + 3k2(2)2 + 2 = -5k10 = -k2Substituting these values of x and y in the third equation, we getLHS = 2(-k) + 0(-k2) + 2(k) = 0 = RHSThus,λ = 2, x = -k, y = -k2 and z = k        kR

Page No 5.89:

Question 5:

If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.

Answer:

The three equations can be expressed as

a - 1x - y - z = 0-x + b - 1y - z = 0-x - y + c - 1z = 0

Expressing this as a determinant, we get

=a-1-1-1-1b-1-1-1-1c-1

If the matrix has a non-trivial solution, then

a-1-1-1-1b-1-1-1-1c-1=0

a - 1b - 1c - 1 - 1 + 1-c - 1 - 1 - 11 + b - 1 = 0a - 1bc - c - b + 1 - 1 + 1-c + 1 - 1 - 1b = 0a - 1bc - b - c - c - b = 0abc - ab - ac - bc + b + c - b - c = 0ab + ac + bc = abc

Hence proved.



Page No 5.90:

Question 1:

If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when
(a) det (A) = 0 or det (B) = 0
(b) det (A) + det (B) = 0
(c) det (A) = 0 and det (B) = 0
(d) A + B = O

Answer:

(d) A + B = O

Let A= ai j and B=bi j be a square matrix of order 2.As their orders are same, A+B is defined asA + B =  ai j + bi j  A + B = ai j + bi j Now, A + B = 0ai j + bi j  = 0ai j + bi j  = 0           each corrsponding term is 0A + B =0

Page No 5.90:

Question 2:

Which of the following is not correct?
(a) |A|=|AT|, where A=aij3×3

(b) |kA|=|k3|, where A=aij3×3

(c) If A is a skew-symmetric matrix of odd order, then |A| = 0

(d) a + bc + de + fg + h = a  ce g + b dfh

Answer:

(d) a + bc + de + fg + h = a ceg + b dfh

a + b   c + de + f   g + h = a + b   ce + f   g  + a + b   de + f   h=a  ce  g  + b   cf   g  + a   d e   h + b   df   h

Page No 5.90:

Question 3:

If A=a11a12a13a21a22a23a31a32a33 and Cij is cofactor of aij in A, then value of |A| is given by

(a) a11C31 + a12C32 + a13C33
(b) a11C11 + a12C21 + a13C31
(c) a21C11 + a22C12 + a23C13
(d) a11C11 + a21C21 + a13C31

Answer:

(d) a11C11 + a21C21 + a13C31

Properties of determinants state that if A is a square matrix of the order n, then Det (A) is the sum of products of elements of a row (or a column) with the corresponding cofactor of that element.

A = a11C11 + a21C21 + a31'C31    Calculating along C1

Page No 5.90:

Question 4:

Which of the following is not correct in a given determinant of A, where A = [aij]3×3.
(a) Order of minor is less than order of the det (A)
(b) Minor of an element can never be equal to cofactor of the same element
(c) Value of determinant is obtained by multiplying elements of a row or column by       corresponding cofactors
(d) Order of minors and cofactors of elements of A is same

Answer:

(b) Minor of an element can never be equal to the cofactor of the same element.

      Ci j = -1i+jMi jSo, for even values of i+j, Ci j = Mi j.

Page No 5.90:

Question 5:

Let x2xx2x6xx6=ax4+bx3+cx2+dx+e

Then, the value of 5a+4b+3c+2d+e is equal to
(a) 0
(b) − 16
(c) 16
(d) none of these

Answer:

(d) none of these

=x    2   xx2  x   6x   x   6=x x    6 x   6 - x2 2   x x   6 + x 2   x x   6                    Expanding along C1=0 - x212 - x2  + x12 - x2=x4 - 12x2 + 12x - x3  =ax4 + bx3 + cx2 + dx + e                   Givenx4 - 12x2 + 12x - x3 = ax4 + bx3 + cx2 + dx + e        a=1, b=-1, c=-12, d = 12, e = 0Thus, 5a + 4b + 3c + 2d + e = 5 - 4 - 36 + 24 + 0 = -11

Page No 5.90:

Question 6:

The value of the determinant

a2a1cos nxcos n+1 xcos n+2 xsin nxsin n+1 xsin n+2 x is independent of

(a) n
(b) a
(c) x
(d) none of these

Answer:

(a) n

Let  A = nx, B = n + 1 x, C=  n + 2 xC - B = x, B - A = x, C - A = 2xThus, the given determinant is   a2            a              1cos A   cos B     cos Csin A     sin B       sin C=a2 cos B sin C - cos C sin B - a × cos A sin C - cos C sin A + 1 × cos A sin B - sin A cos B=a2 sin C - B - a sin C - A + sin B - A=a2 sin x - a sin 2x +  sin x     Independent of n

Page No 5.90:

Question 7:

If 1=111abca2b2c2, 2=1bca1cab1abc, then

(a) 1+2=0
(b) 1+2 2=0
(c) 1=2
(d) none of these

Answer:

(a) 1+2=0

Δ2 =1   bc   a1   ca    b1   ab   c=1abca    abc   a2b    bca    b2c    cab   c2    [R1, R2, R3 are multiplied by a, b and c respectively, therefore we divide by abc]=abcabc a    1   a2b    1   b2c    1   c2       Taking abc common from C2=-1    a    a21    b   b21    c   c2              C1C2We know that the value of a determinant remains unchanged if its rows and columns are interchanged. So,2=-1     1        1a     b       ca2   b2    c2               =-Δ1Δ1+ Δ2 = 0

Page No 5.90:

Question 8:

If Dk=1nn2kn2+n+2n2+n2k-1n2n2+n+2 and k=1n Dk=48, then n equals

(a) 4
(b) 6
(c) 8
(d) none of these

Answer:

(a) 4

Dk=   1              n                   n  2k      n2+n+2         n2  +n2k-1         n2           n2+n+2=  1              n                 n  1            n+2             - 22k-1         n2           n2+n+2              Applying R2R2-R3=1              n                          n 0              2                  - 2-n2k-1         n2           n2+n+2              Applying R2R2-R1k=1nDk=  1                    n                        n  0                    2                   - 2-n  1                   n2                  n2+n+2+  1                    n                        n  0                    2                   - 2-n  3                   n2                  n2+n+2+...+  1                    n                        n  0                    2                   - 2-n  n                   n2                  n2+n+2k=1nDk=12n2+n+2+2+nn2+1n-2-n-2n+12n2+n+2+2+nn2+2n-2-n-2n+...+12n2+n+2+2+nn2+nn-2-n-2nk=1nDk=n2n2+n+2+2+nn2+n-2-n-2n1+3+5+7+...+nk=1nDk=n2n2+n+2+2+nn2+n-2-n-2nn2k=1nDk=2n2+4n2n2+4n=48n-6n-4=0n=4

Page No 5.90:

Question 9:

Let x2+3xx-1x+3x+1-2xx-4x-3x+43x=ax4+bx3+cx2+dx+e

be an identity in x, where a, b, c, d, e are independent of x. Then the value of e is
(a) 4
(b) 0
(c) 1
(d) none of these

Answer:

(b) 0

Let Δ=x2+ 3x   x - 1     x + 3x + 1     -2x        x - 4x - 3      x + 4     3x=x2 + 3x-2x          x - 4  x + 4       3x -  x - 1 x + 1      x - 4x - 3         3x + x + 3x + 1        -2x x - 3      x + 4=x2 + 3x-6x - x2 + 16 -  x - 1 3x2 + 3x - x2 + 7x - 12 + x + 3x2 + 5x + 4 + 2x2 - 6x=-7x4 + 16x2 + 48x + 21x3 + 8x2 - 22x - 2x3 - 12 + 8x2 + x + 3x3 + 12=-7x4 + 22x3 + 32x2 + 27x + 0But x is a root of ax4 + bx3 + cx2 + dx + e.e = 0



Page No 5.91:

Question 10:

Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant

-2aa+ba+cb+a-2bb+cc+ac+b-2c

The other factor in the value of the determinant is
(a) 4
(b) 2
(c) a + b + c
(d) none of these

Answer:

(a) 4

Δ=-2a       a + b      a + cb + a   -2b         b + cc + a     c + b    -2c Let a + b = 2C, b + c = 2A and c + a = 2B.a + b + b + c + c + a = 2A + 2B + 2C2a + b + c  = 2A + B + Ca + b + c = A + B + CAlso,a = a + b + c  - b + c = A + B + C - 2A = B + C - ASimilarly,b = C + A - B,  c = A + B - CΔ = 2A - 2B - 2C             2C                        2B      2C                  2B-2C-2A                 2A      2B                       2A                 2C-2A-2B = 8 × A - B - C           C                      B      C             B - C - A              A     B                    A                C - A - B    taking out 2 common from R1 R2  R3=8 ×  A - B            C + B               B B - A           B - C               A B + A          C - B        C-A-B      Applying C1C1+C2 , C2C2 + C3=8 ×  A - B    C + B            B  0              2B           A + B 2B             0            C - B                    Applying R2 R1+ R2, R3 R2 +R3=8 × A - B    2B           A + B  0             C - B +   2B ×   C + B       B    2B       A + B          Expanding along C1=16 BA - BC - B + C + BA + B - 2B2= 32 ABC=32b + c2c + a2a + b2=4a + bb + cc + aHence, 4 is the other factor of the determinant. 

Page No 5.91:

Question 11:

If a, b, c are distinct, then the value of x satisfying 0x2-ax3-bx2+a0x2+cx4+bx-c0=0 is
(a) c
(b) a
(c) b
(d) 0

Answer:

(d) 0

When we put x = 0 in the given matrix, then it turns out to be the skew symmetric matrix of order 3 and the determinant of the skew symmetric matrix of odd order is always 0.

Page No 5.91:

Question 12:

If the determinant

ab   2a α + 3bbc 2b α + 3c2a α + 3b  2b α + 3c0 = 0, then

(a) a, b, c are in H.P.
(b) α is a root of 4ax2+12bx+9c=0 or a, b, c are in G.P.
(c) a, b, c are in G.P. only
(d) a, b, c are in A.P.

Answer:

(b) α is a root of  4ax2 + 12bx + 9c = 0 or a, b, c are in G.P.


  Let Δ=   a                   b                    2aα + 3b   b                   c                    2bα + 3c2aα + 3b       2bα + 3c          0=  a-b                               b                2aα+3b  b-c                               c                2bα+3c2aα+3b-2bα-3c       2bα+3c          0         Applying C1C1-C2=           a - b                         b                2aα + 3b           b - c                         c                2bα + 3c2a - bα + 3b - c       2bα+3c          0= a-b       b                      2aα+3b b-c       c                      2bα+3c   0          0     - 2α 2aα+3b -3 2bα+3c                    Applying R3R3-2α, R1-3R2=- 2α 2aα+3b -3 2bα+3c  a-b       b   b-c       c          Expanding along R3=-4aα2+12bα+9cac-b2 But Δ=0        Given-4aα2+12bα+9cac-b2=04aα2+12bα+9c=0 or ac-b2=0α is a root of 4ax2+12bx+9c=0or ac=b2, i.e. a, b, c are in G.P.

Page No 5.91:

Question 13:

If ω is a non-real cube root of unity and n is not a multiple of 3, then

=1ωnω2nω2n1ωnωnω2n1 is equal to

(a) 0
(b) ω
(c) ω2
(d) 1

Answer:

(a) 0
  Δ= 1        wn      w2nw2n     1        wn wn     w2n      1=1+ wn+ w2n    wn     w2nw2n+ 1+ wn    1        wn wn+ w2n+1     w2n   1          Appplying C1C1+ C2+C3Now,1+w+w2=0                                   w is a complex cube root of unity1+ wn+ w2n=0                         n is not a multiple of 3Δ=0   wn   w2n0   1     wn 0     w2n  1 = 0      

Page No 5.91:

Question 14:

If Ar=1r2r2nn2nn n+122n+1, then the value of r=1n Ar is
(a) n
(b) 2n
(c) − 2n
(d) n2

Answer:

Ar = 1               r                  2r2              n                  n2n       nn+12       2n+1r=1nAr  = r=1n1              r=1n r                r=1n  2rr=1n2              n                  n2n           nn+12             2n+1As r=1n1 = 1 + 1 + 1 ... + 1 (n times) = n  r=1n r = 1 + 2 + 3 + ...+ n= nn + 12 Let S=r=1n  2r=2 + 22 + 23= ... + 2n      2S = 22 + 23=...+ 2n + 2n+12S - S = S =r=1n  2r= 2n+1 - 2r=1nAr= n              nn+12                2n+1-22n              n                              n2 n       nn+12                        2n+1 Applying R1R1-R2r=1nAr= n -n             nn+12 - nn+12              2n+1-2- 2n+12n                                n                                         n2 n                       nn+12                                      2n+1=0                  0                             -22n              n                              n2 n       nn+12                        2n+1=-2 ×  2n                       n                           n                  nn+12         =-2n3 + n2 - n2=-2n3

Page No 5.91:

Question 15:

If a > 0 and discriminant of ax2 + 2bx + c is negative, then

=abax+bbcbx+cax+bbx+c0 is

(a) positive
(b) ac-b2 ax2+2bx+c
(c) negative
(d) 0

Answer:

(c) negativeDiscriminant D of ax2 + 2bx + c = 2b2 - 4ac < 0        Given4b2 - 4ac < 0 b2 - ac < 0, where a>0                        (1)Δ=   a          b             ax + b    b         c             bx + cax + b   bx + c         0=   ax         bx        ax2 + bx    b         c           bx + cax + b    bx + c         0                                         Applying  R1x R1=1x ax + b         bx + c        ax2 + bx + bx + c    b                c                  bx + cax + b          bx + c                0             Applying R1R1+R2=1x    0                0          ax2 + 2bx + c    b                c               bx + cax + b          bx + c                0                     Applying R1R1-R3=1xax2 + 2bx + c bcax + b   bx + c                        Expanding along  R1=1xax2 + 2bx + cb2x + bc - acx - bc=1xax2 + 2bx + c x b2 - ac = ax2 + 2bx + cb2 - ac < 0                   From eq. (1)Δ < 0  

Page No 5.91:

Question 16:

The value of 525354535455545556 is
(a) 52
(b) 0
(c) 513
(d) 59

Answer:

(b) 0

52    53   5453   54   55 54  55   56  =52 × 53 × 54 1  5   521  5   521  5   52         Taking out common factors from R1, R2, R3=52 × 53 × 54 × 5 1  1   521  1   521  1   52 =52 × 53× 54× 0=0

Page No 5.91:

Question 17:

log3 512log4 3log3 8log4 9 × log2 3log8 3log3 4log3 4

(a) 7
(b) 10
(c) 1
(d) 17

Answer:

(b) 10

log3  512   log4 3      log3 8     log4 9  ×  log2 3   log8 3  log34    log3 4=log3 29   log22 3      log3 23     log22 33  × log2 3   log23 3  log322   log3 22=9 log3 2      12log23     3 log3 2     12×2 log2 3  × log2 3          13 log2 3 2 log32         2 log3 2                                    log bm an = nm logb a=9 log3 2 × log2 3 - 3 log3 2 × 12log23 × log2 3 ×  2 log3 2  - 13 log2 3 × 2 log32                          logmn × lognm =1=9 - 32 × 2 - 23=152 × 43  = 10

Page No 5.91:

Question 18:

If a, b, c are in A.P., then the determinant

x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c

(a) 0
(b) 1
(c) x
(d) 2x

Answer:

(a) 0


x + 2  x + 3    x + 2ax + 3   x + 4   x + 2bx + 4   x + 5   x + 2c=   0           0            2a + c - 2bx + 3     x + 4      x + 2bx + 4     x + 5       x + 2c   Applying R1R1+R3-R2, R1R1 - R2  =   0        0           0x+3   x+4    x+2bx+4  x+5     x+2c             a, b, c are in A.P.= 0    



Page No 5.92:

Question 19:

If A+B+C=π, then the value of sin A+B+Csin A+Ccos C-sin B0tan Acos A+Btan B+C0 is equal to

(a) 0
(b) 1
(c) 2 sin B tan A cos C
(d) none of these

Answer:

(a) 0

A + B + C = π     A + C = π - B,  A + B = π - C and B + C = π - AThus the determinant becomes   sin π         sin π-B        cos C-sin B                 0              tan Acos π-C    tan π-A          0=     0           sin B        cos C-sin B          0           tan A-cos C   -tan A            0                    sin π=0, sin π-B=B, cos π-C=-cos C, tan π-A=-tan AIt is a skew  symmetric matrix of the odd order 3. Thus, by property of determinants, we get = 0     0               sin B        cos C-sin B          0              tan A-cos C     -tan A           0 = 0

Page No 5.92:

Question 20:

The number of distinct real roots of cosec xsec xsec xsec xcosec xsec xsec xsec xcosec x=0 lies in the interval -π4xπ4 is
(a) 1
(b) 2
(c) 3
(d) 0

Answer:

(b) 2
Let =cosec x     sec x           sec xsec x      cosec x        sec xsec x       sec x       cosec x=cosec x3         1                     sec x cosec x           sec x cosec x sec x cosec x                 1                         sec x cosec x sec x cosec x           sec x cosec x                          1=cosec x3     1               tan x              tan x      tan x               1                  tan xtan x            tan x                     1=cosec x3 1 - tan x             tan x - 1                   0         0                       1 - tanx              tan x - 1tan x                     tan x                      1           Applying R1 R1-R2, R2 R2-R3=cosec x3 1 - tan x2   1                   -1                  0         0                     1                -1tan x                 tan x             1                  Taking out 1 - tan x  common from R1 and R2=cosec x31 - tan x211-1tan x   1 + tan x-1 01-1                  Expanding along C1=cosec x3 1 - tan x2 1 + tan x + tan x=cosec x3 1 - tan x2 1 + 2 tan x = 0cosec x3 1 - tan x2 1 + 2 tan x = 01 - tan x = 0, cosec x3= 0 and 1 + 2 tan x = 0ortan x = 1, cosec x = 0 and tan x = -1  2- π4   x  π4           tan x = 1, tan x = -12 are 2 real roots as cosec x = 0 has no solutionThus, there are 2 solutions.  

Page No 5.92:

Question 21:

Let A=1sin θ1-sin θ1sin θ-1- sin θ1, where 0 θ 2π. Then,

(a) Det A=0
(b) Det A 2, 
(c) Det A 2, 4
(d) Det A 2, 4

Answer:

(d) Det A 2, 4

     1            sin θ       1- sin θ          1     sin θ   -1      -  sin θ     1=   1                sin θ      2- sin θ          1         0    -1      -  sin θ     0       Applying C3C3 + C1= 2 × - sin θ          1        -1      -  sin θ        Expanding along C3=2 sin2 θ + 1Given: 0θ2π   -1sin θ10sin2 θ1A=2sin2 θ + 1A =2 × 1 = 2            θ = 0     =2 × 2 = 4           θ = 2π Det A2, 4      

Page No 5.92:

Question 22:

If 2x58x=6-273, then x

(a) 3
(b) ± 3
(c) ± 6
(d) 6

Answer:

2x58x=6-2732x2-40=18+142x2-40=322x2=72x2=36x=±6

Hence, the correct option is (c).

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Question 23:

If fx=0x-ax-bx+a0x-cx+bx+c0, then

(a) f(a) = 0
(b) f(b) = 0
(c) f(0) = 0
(d) f(1) = 0

Answer:

Let fx=0x-ax-bx+a0x-cx+bx+c0.

Now,
fa=0a-aa-ba+a0a-ca+ba+c0      =00a-b2a0a-ca+ba+c0      =a-b2a2+2ac0fb=0b-ab-bb+a0b-cb+bb+c0      =0b-a0b+a0b-c2ab+c0      =b-a2ab-2ac0f0=00-a0-b0+a00-c0+b0+c0      =0-a-ba0-cbc0      =a(bc)-b(ac)=0

Hence, the correct option is (c).

Page No 5.92:

Question 24:

The value of the determinant a-bb+cab-cc+abc-aa+bc is

(a) a3+b3+c3
(b) 3bc
(c) a3+b3+c3-3abc
(d) none of these

Answer:

a-bb+cab-cc+abc-aa+bc=-bb+c+aa-cc+a+bb-aa+b+cc       Applying C1C1-C3 and C2C2+C3=-1a+b+cb1ac1ba1c       Taking -1 common from C1 and a+b+c common from C2=-1a+b+cb1ac-b0b-aa-b0c-a       Applying R2R2-R1 and R3R3-R1=-1a+b+c-c-bc-a+b-aa-b=-1a+b+c-c2+ac+bc-ab+ba-b2-a2+ab=-1a+b+c-a2-b2-c2+ab+bc+ac=a+b+ca2+b2+c2-ab-bc-ac=a3+ab2+ac2-a2b-abc-a2c+ba2+b3+bc2-ab2-b2c-abc+ca2+cb2+c3-acb-bc2-ac2=a3+b3+c3-3abc

Hence, the correct option is (c).

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Question 25:

If x, y, z are different from zero and 1+x1111+y1111+z=0, then the value of x−1 + y1 + z1 is

(a) xyz
(b) x−1 y1 z1
(c) − x − y − z
(d) − 1
 

Answer:

1+x1111+y1111+z=0x0-z0y-z111+z=0      Applying R2R2-R3 and R1R1-R3xy1+z+z+1yz=0      Expanding along first columnxy+yz+z+yz=0xy+xyz+xz+yz=0xy+yz+zx=-xyzxyxyz+yzxyz+zxxyz=-xyzxyz1z+1x+1y=-1x-1+y-1+z-1=-1

Hence, the correct option is (d).

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Question 26:

The determinant b2-abb-cbc-acab-a2a-bb2-abbc-cac-aab-a2 equals

(a) abcb-cc-aa-b
(b) b-cc-aa-b
(c) a+b+cb-cc-aa-b
(d) none of these

Answer:

b2-abb-cbc-acab-a2a-bb2-abbc-cac-aab-a2=bb-ab-ccb-aab-aa-bbb-acb-ac-aab-a=b-a2bb-ccaa-bbcc-aa        Taking b-a common from C1 and C3=b-a20b-cc0a-bb0c-aa        Applying C1C1-C2-C3=0

Hence, the correct option is (d).

Page No 5.92:

Question 27:

If x, y, then the determinant =cosx-sinx1sinxcosx1cosx+y-sinx+y0 lies in the interval

(a) -2, 2
(b) -1, 1
(c) -2, 1
(d) -1, -2

Answer:

=cosx-sinx1sinxcosx1cosx+y-sinx+y0   =cosx-sinx1sinxcosx100siny-cosy        Applying R3R3-cosy R1+siny R2   =siny-cosycos2x+sin2x   =siny-cosy   =212siny-12cosy   =2cosπ4siny-sinπ4cosy   =2siny-π4Therefore, -22.

Hence, the correct option is (a).



Page No 5.93:

Question 28:

The maximum value of =11111+sinθ11+cosθ11 is (θ is real)

(a) 12
(b) 32
(c) 2
(d) -32

Answer:

=11111+sinθ11+cosθ11   =1110sinθ0cosθ00       Applying R2R2-R1 and R3R3-R1   =-sinθcosθ   =-sin2θ2Now, Maximum and minimum value of sinθ is 1 and -1.So, the maximum value of -sinθ is 1.So, the maximum value of -sin2θ is 1.Therefore, the maximum value of -sin2θ2 is 12.

Hence, the correct option is (a).

Page No 5.93:

Question 29:

The value of the determinant xx+yx+2yx+2yxx+yx+yx+2yx is

(a) 9x2(x + y)
(b) 9y2(x + y)
(c) 3y2(x + y)
(d) 7x2(x + y)

Answer:

xx+yx+2yx+2yxx+yx+yx+2yx=-2yyyx+2yxx+y-y2y-y      Applying R1R1-R2 and R3R3-R2=y2-211x+2yxx+y-12-1      Taking y common from R1 and from R3=y2-2-33x+2y3x+4y-y-100      Applying C2C2+2C1 and C3C3-C1=y2-13y-9x-12y=y29y+9x=9y2y+x

Hence, the correct option is (b).

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Question 30:

Let fx=cosxx12sinxx2xsinxxx, then limx0fxx2 is equal to

(a) 0
(b) −1
(c) 2
(d) 3

Answer:

fx=cosxx12sinxx2xsinxxx      =cosxx1sinx0xsinxxx      Applying R2R2-R3      =cosxx1sinx0xsinx-cosx0x-1      Applying R3R3-R1      =-xx sinx-sinx-x sinx+x cosx      =-xx cosx-sinx limx0fxx2=limx0xsinx-x cosxx2                  =limx0sinxx-limx0cosx                  =1-1=0

Hence, the correct option is (a).

Page No 5.93:

Question 31:

There are two values of a which makes the determinant =1-252a-1042a equal to 86. The sum of these two values is

(a) 4
(b) 5
(c) −4
(d) 9

Answer:

=1-252a-1042a=8612a2+4-2-4a-20=862a2+4+8a+40=862a2+8a-42=0a2+4a-21=0a2+7a-3a-21=0aa+7-3a+7=0a+7a-3=0a=-7,3Sum of the two values of a =-7+3=-4.

Hence, the correct option is (c).

Page No 5.93:

Question 32:

If apxbqycrz=16, then the value of p+xa+xa+pq+yb+yb+qr+zc+zc+r is

(a) 4
(b) 8
(c) 16
(d) 32

Answer:

p+xa+xa+pq+yb+yb+qr+zc+zc+r=paaqbbrcc+papqbqrcr+pxaqybrzc+pxpqyqrzr+xaaybbzcc+xapybqzcr+xxayybzzc+xxpyyqzzr                            =0+0+pxaqybrzc+0+0+xapybqzcr+0+0                            =pxaqybrzc+xapybqzcr                            =2apxbqycrz                            =2×16=32

Hence, the correct option is (d).

Page No 5.93:

Question 33:

The value of 111nC1n+2C1n+4C1nC2n+2C2n+4C2 is

(a) 2
(b) 4
(c) 8
(d) n2

Answer:

111nC1n+2C1n+4C1nC2n+2C2n+4C2=111nn+2n+4nn-12n+2n+12n+4n+32=100n24nn-124n+228n+122        Applying C2C2-C1 and C3C3-C1=100n24nn-122n+14n+6=8n+12-8n-4=8

Hence, the correct option is (c).

Page No 5.93:

Question 34:

The number of distinct real root of sin xcos xcos xcos xsin xcos xcos xcos xsin x=0 in the interval -π4, π4, is
(a) 0
(b) 2
(c) 1
(d) 3

Answer:

Given: sinxcosxcosxcosxsinxcosxcosxcosxsinx=0



sinxcosxcosxcosxsinxcosxcosxcosxsinxApplying R2R2-R1=sinxcosxcosxcosx-sinxsinx-cosxcosx-cosxcosxcosxsinx=sinxcosxcosx-sinx-cosxsinx-cosx0cosxcosxsinxTaking sinx-cosx common from R2=sinx-cosxsinxcosxcosx-110cosxcosxsinxApplying C2C2+C1=sinx-cosxsinxcosx+sinxcosx-11-10cosxcosx+cosxsinx=sinx-cosxsinxcosx+sinxcosx-100cosx2cosxsinxExpanding through R2=sinx-cosx--1cosx+sinxsinx-2cosxcosx=sinx-cosx1cosxsinx+sin2x-2cos2x=sinx-cosxcosxsinx+sin2x-2cos2x=sinx-cosxsin2x+2cosxsinx-cosxsinx-2cos2x=sinx-cosxsinxsinx+2cosx-cosxsinx+2cosx=sinx-cosxsinx-cosxsinx+2cosx=sinx-cosx2sinx+2cosxThus, sinxcosxcosxcosxsinxcosxcosxcosxsinx=sinx-cosx2sinx+2cosxBut it is given that, sinxcosxcosxcosxsinxcosxcosxcosxsinx=0sinx-cosx2sinx+2cosx=0sinx-cosx2=0 or sinx+2cosx=0sinx-cosx=0 or sinx=-2cosxsinx=cosx or tanx=-2Since x-π4, π4, Thus tanx-2.sinx=cosx at x=π4 Therefore, we have 1 distinct real root.

Hence, the correct option is (c).
 

Page No 5.93:

Question 35:

If A, B and C are angles of a triangle, then the determinant -1cos Ccos Bcos C-1cos Acos Bcos A-1 is equal to
(a) 0
(b) –1
(c) 1
(d) none of these

Answer:

Given: AB and C are angles of a triangle
Therefore, A+B+C=π  ...1


-1cosCcosBcosC-1cosAcosBcosA-1Expanding through R1=-11-cos2A-cosC-cosC-cosAcosB+cosBcosAcosC+cosB=-1+cos2A+cos2C+cosAcosBcosC+cosAcosBcosC+cos2B=-1+2cosAcosBcosC+cos2A+cos2B+cos2C=-1+2cosAcosBcosC+1+cos2A2+1+cos2B2+cos2C                       2cos2θ=1+cos2θ=-1+2cosAcosBcosC+1+cos2A+1+cos2B2+cos2C=-1+2cosAcosBcosC+2+2cos2A+2B2cos2A-2B22+cos2C        cosA+cosB=2cosA+B2cosA-B2=-1+2cosAcosBcosC+2+2cosA+BcosA-B2+cos2C=-1+2cosAcosBcosC+2+2cosπ-CcosA-B2+cos2C                   A+B+C=π=-1+2cosAcosBcosC+2-2cosCcosA-B+2cos2C2 =-1+2cosAcosBcosC+1-cosCcosA-B+cos2C=-1+2cosAcosBcosC+1-cosCcosA-B-cosC=-1+2cosAcosBcosC+1-cosC-2sinA-B+C2sinA-B-C2      cosA-cosB=-2sinA+B2sinA-B2=-1+2cosAcosBcosC+1-cosC-2sinA+C-B2sinA-B+C2=-1+2cosAcosBcosC+1-cosC-2sinπ-B-B2sinA-π-A2    A+B+C=π=-1+2cosAcosBcosC+1+2cosCsinπ2-BsinA-π2=-1+2cosAcosBcosC+1+2cosCcosBsinA-π2=-1+2cosAcosBcosC+1-2cosCcosBsinπ2-A=-1+2cosAcosBcosC+1-2cosCcosBcosA=0


Hence, the correct option is (a).

Page No 5.93:

Question 36:

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(a) 9
(b) 3
(c) –9
(d) 6

Answer:

Given: Area of a triangle with vertices (–3, 0), (3, 0) and (0, k) = 9 sq. units

Area of the triangle with vertices (x1, y1), (x2y2) and (x3y3) is 12x1y11x2y21x3y31.

According to the question,
12-3013010k1=912-3013010k1=±9-3013010k1=±18-30-k-03+13k=±18-3-k+13k=±183k+3k=±186k=±18k=±3

Hence, the correct option is (b).

 



Page No 5.94:

Question 1:

If A = diag (1, 2, 3), then |A| = ____________.

Answer:

Given: A = diag (1, 2, 3)

If A = diag (a, b, c), then |A| = a × b × c.

Thus, if A = diag (1, 2, 3), then |A| = 1 × 2 × 3 = 6.

Hence, |A| = 6.

Page No 5.94:

Question 2:

If A is a skew-symmetric matrix of order 3 × 3, then |A| = ______________.

Answer:

Given: A is a skew-symmetric matrix of order 3 × 3

A=-ATTaking determinant on both sides, we getA=-ATA=-13AT          Order of A is 3×3A=-13A           AT=AA=-AA+A=02A=0A=0

Hence, |A| = 0.

Page No 5.94:

Question 3:

If the matrix A=13x+22483510 is singular, then x = _______________.

Answer:

Given: The matrix A=13x+22483510 is singular

A is singular  A=0Thus,13x+22483510=0140-40-320-24+x+210-12=010-3-4+x+2-2=012-2x-4=08-2x=02x=8x=4

Hence, x = 4.

Page No 5.94:

Question 4:

If A and B are non-singular square matrices of order n such that A = kB, then AB= ____________.

Answer:

Given:
A and B are non-singular square matrices of order n.
A
 = kB


A=kBTaking determinant on both sides, we getA=kBA=knB           Order of B is n×nAB=kn

Hence, AB=kn.

Page No 5.94:

Question 5:

The set of real values of a for which the matrix A=a224 is non-singular is ______________.

Answer:

Given: The matrix A=a224 is non-singular

A is non-singular  A0Thus,a22404a-404a4a1aR-1

Hence, the set of real values of a for which the matrix A=a224 is non-singular is R-1.

Page No 5.94:

Question 6:

If A is a 2 × 2 matrix such that |A| =5, then |5A| = ___________.

Answer:

Given: 
A is a 2 × 2 matrix
|A| = 5


Now,5A=52A                  Order of A is 2×2       =25A       =25×5                A=5       =125

Hence, |5A| = 125.

Page No 5.94:

Question 7:

If A and B are square matrices of order 3 such that |A| = –1, |B| = 3, then  |3AB| = _____________.

Answer:

Given: 
A and B are square matrices of order 3
|A| = –1
|B| = 3


Now,3AB=3AB                 AB=AB, if they are square matrices of same order         =3A×3                 B=3         =33A×3                Order of A is 3×3         =81A         =81×-1               A=-1         =-81

Hence, |3AB| = –81.

Page No 5.94:

Question 8:

If A=α22α and |A3| = 125, then α = ___________.

Answer:

Given: 
A=α22α 
|A3| = 125



Now,A3=125A3=53                 An=AnA=5α22α=5α2-4=5α2=5+4α2=9α=±3

Hence, α = ±3.

Page No 5.94:

Question 9:

If A=ln x-1-ln x2 and if det (A) = 2, then x = ___________.

Answer:

Given: 
A=lnx-1-lnx2 
det (A) = 2



Now,A=2lnx-1-lnx2=22lnx-lnx=2lnx=2x=e2

Hence, x = e2.

Page No 5.94:

Question 10:

If I is the identity matrix of order 10, then determinant of I is ___________.

Answer:

Given: 
Order of I is 10

Det(I) = 1, where I is the identity matrix of order n.

Hence, determinant of I is 1.

Page No 5.94:

Question 11:

If |A| denotes the value of the determinant of a square matrix of order 3, then |–2A| = ___________.

Answer:

Given: 
A is a 3 × 3 matrix


Now,-2A=-23A                  Order of A is 3×3         =-8A

Hence, |–2A| = –8|A|.

Page No 5.94:

Question 12:

Let A = [aij] be a 3 × 3 matrix such that |A| = 5. If Cij = Cofactor of aij in A. Then a11C11 + a12C12 + a13C13 = ________.

Answer:

Given: 
|A| = 5

As we know,
Sum of products of elements of row (or column) with their corresponding cofactors = Value of the determinant
and 
Sum of products of elements of row (or column) with the cofactors of any other row (or column) = 0


Thus, a11 C11 + a12 C12 + a13 C13 = |A| = 5

Hence, a11 C11 + a12 C12 + a13 C13 = 5.

Page No 5.94:

Question 13:

In the above question, a11C21 + a12C22 + a13C23 = __________.

Answer:

Given: 
|A| = 5

As we know,
Sum of products of elements of row (or column) with their corresponding cofactors = Value of the determinant
and 
Sum of products of elements of row (or column) with the cofactors of any other row (or column) = 0


Thus, a11 C21 + a12 C22 + a13 C23 = 0

Hence, a11 C21 + a12 C22 + a13 C23 = 0.

Page No 5.94:

Question 14:

If A = diag (2, 3, 4), then |A2| = _____________.

Answer:

Given: A = diag (2, 3, 4)

A2=A2          An=An      =2×3×42      =242      =576

Hence, |A2| = 576.

Page No 5.94:

Question 15:

Let A = [aij] be a square matrix of order 3 with |A| = 2 and let C = [cij], where cij = cofactor of aij in A. Then, |C| = ______________.

Answer:

Given: 
|A| = 2
Order of A = 3


As we know,adjA=An-1       where n is the order of AAlso, adjA=CTadjA=CTadjA=C         CT=CAn-1=CC=An-1C=A3-1           order of A=3C=A2C=22                 A=2C=4

Hence, |C| = 4.

Page No 5.94:

Question 16:

The value of the determinant =1234563x6x9x is ____________.

Answer:

Given: =1234563x6x9x

=1234563x6x9xTaking 3x common from R3=3x123456123=3x0             The value of determinant with two identicals rows is equal to zero=0

Hence, the value of the determinant =1234563x6x9x is 0.

Page No 5.94:

Question 17:

The value of the determinant =sec2θtan2θ1tan2θsec2θ-122202 is _____________.

Answer:

Given: =sec2θtan2θ1tan2θsec2θ-122202

=sec2θtan2θ1tan2θsec2θ-122202Applying C2C2+C3=sec2θtan2θ+11tan2θsec2θ-1-12220+22=sec2θsec2θ1tan2θtan2θ-122222=0             The value of determinant with two identicals columns is equal to zero

Hence, the value of the determinant =sec2θtan2θ1tan2θsec2θ-122202 is 0.

Page No 5.94:

Question 18:

The value of the determinant =0x-yy-zy-x0z-xz-yx-z0 is _____________.

Answer:

Given: =0x-yy-zy-x0z-xz-yx-z0

Let A=0x-yy-zy-x0z-xz-yx-z0AT=0x-yy-zy-x0z-xz-yx-z0T    =0y-xz-yx-y0x-zy-zz-x0    =-10x-yy-zy-x0z-xz-yx-z0    =-ASince, AT=-ATherefore, A is a skew symmetric matrix.Thus, A=00x-yy-zy-x0z-xz-yx-z0=0=0

Hence, the value of the determinant =0x-yy-zy-x0z-xz-yx-z0 is 0.



Page No 5.95:

Question 19:

Let A = [aij] and B = [bij] be a square matrices of order 3 such that bi1 = 2 ai1, bi2 = 3 ai2 and bi3 = 4 ai3, i = 1, 2, 3
If |A| = 5, then |B| = _____________.
 

Answer:

Given: 
A and B are square matrices of order 3
bi1 = 2 ai1bi2 = 3 ai2 and bi3 = 4 ai3i = 1, 2, 3
|A| = 5


Let A=a11a12a13a21a22a23a31a32a33Since, bi1=2ai1, bi2=3ai2 and bi3=4ai3Therefore,b11=2a11, b21=2a21, b31=2a31,b12=3a12, b22=3a22, b32=3a32,b13=4a13, b23=4a23, b33=4a33B=b11b12b13b21b22b23b31b32b33B=2a113a124a132a213a224a232a313a324a33Taking out 2, 3 and 4 common from C1, C2 and C3, respectivelyB=2×3×4a11a12a13a21a22a23a31a32a33B=24AB=24×5      A=5B=120

Hence,  |B| = 120.

Page No 5.95:

Question 20:

If A and B are square matrices of order 3 and that |A| = –2, |B| = 4, then |2AB| = __________.

Answer:

Given: 
A and B are square matrices of order 3
|A| = –2
|B| = 4


Now,2AB=2AB                 AB=AB, if they are square matrices of same order         =2A×4                 B=4         =23A×4                Order of A is 3×3         =32A         =32×-2               A=-2         =-64

Hence, |2AB| = –64.

Page No 5.95:

Question 21:

The value of the determinant cos x+y-sin x+ycos 2ysin xcos xsin y-cos xsin xsin y depends on __________ only.

Answer:

Let âˆ† = cosx+y-sinx+ycos2ysinxcosxsiny-cosxsinxsiny 



=cosx+y-sinx+ycos2ysinxcosxsiny-cosxsinxsinyApplying R2sinyR2 and R3cosyR3   =1sinycosycosx+y-sinx+ycos2ysinxsinycosxsinysin2y-cosxcosysinxcosysinycosyApplying R1R1+R2+R3   =1sinycosycosx+y+sinxsiny-cosxcosy-sinx+y+cosxsiny+sinxcosycos2y+sin2y+sinycosysinxsinycosxsinysin2y-cosxcosysinxcosysinycosy   =1sinycosycosx+y-cosx+y-sinx+y+sinx+ycos2y+sin2y+sinycosysinxsinycosxsinysin2y-cosxcosysinxcosysinycosy   =1sinycosy00cos2y+sin2y+sinycosysinxsinycosxsinysin2y-cosxcosysinxcosysinycosyExpanding through R1   =1sinycosycos2y+sin2y+sinycosysin2xsinycosy+cos2xsinycosy   =1sinycosycos2y+sin2y+sinycosysinycosysin2x+cos2x   =cos2y+sin2y+sinycosysin2x+cos2x   =cos2y+sin2y+sinycosy1   =cos2y+sin2y+sinycosy   =1-2sin2y+sin2y+sinycosy   =1-sin2y+sinycosy

Hence, the value of the determinant cosx+y-sinx+ycos2ysinxcosxsiny-cosxsinxsiny depends on y only.
 

Page No 5.95:

Question 22:

If x, y, zR, the value of the determinant 2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21 is equal to ________________.

Answer:

Let âˆ† = 2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21



=2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21Applying C1C1-C2    =2x+2-x2-2x-2-x22x-2-x213x+3-x2-3x-3-x23x-3-x214x+4-x2-4x-4-x24x-4-x21   =2x+2-x+2x-2-x2x+2-x-2x+2-x2x-2-x213x+3-x+3x-3-x3x+3-x-3x+3-x3x-3-x214x+4-x+4x-4-x4x+4-x-4x+4-x4x-4-x21   =2x+2x2-x+2-x2x-2-x213x+3x3-x+3-x3x-3-x214x+4x4-x+4-x4x-4-x21   =2.2x2.2-x2x-2-x212.3x2.3-x3x-3-x212.4x2.4-x4x-4-x21Taking out (4) common from C1   =42x2-x2x-2-x213x3-x3x-3-x214x4-x4x-4-x21   =412x-2-x2113x-3-x2114x-4-x21   =40                                   if two columns are identical then the value of determinant is zero   =0

Hence, the value of the determinant 2x+2-x22x-2-x213x+3-x23x-3-x214x+4-x24x-4-x21 is equal to 0.

Page No 5.95:

Question 23:

If fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47=A+Bx+Cx2+ ______________, then A = _________.

Answer:

Let fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47



fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47Taking out 1+x17, 1+x23 and 1+x41 common from R1, R2 and R3, respectively      =1+x171+x231+x4111+x21+x611+x61+x1111+x21+x6      =1+x8111+x21+x611+x61+x1111+x21+x6Taking out 1+x2 and 1+x6 common from C2 and C3, respectively      =1+x811+x21+x611111+x41+x5111      =1+x8911111+x41+x5111      =1+x890                                   if two rows are identical then the value of determinant is zero      =0

Hence, A = 0.

Page No 5.95:

Question 24:

If cos 2θ = 0, then 0cos θsin θcos θsin θ0sin θ0cos θ2=_________________.

Answer:

Given: cos2θ = 0

cos2θ=0cos2θ-sin2θ=0cos2θ=sin2θθ=±π4          ...1


Now,

0cosθsinθcosθsinθ0sinθ0cosθ2Expanding along R1 =-cosθcos2θ-0+sinθ0-sin2θ2=-cosθcos2θ+sinθ-sin2θ2=-cos3θ-sin3θ2=-cos3θ+sin3θ2=cos3θ+sin3θ2=cosθ+sinθcos2θ+sin2θ-cosθsinθ2=cosθ+sinθ1-cosθsinθ2For θ=π40cosθsinθcosθsinθ0sinθ0cosθ2=cosπ4+sinπ41-cosπ4sinπ42                            =12+121-12×122                            =221-122                            =22122                            =122                            =12For θ=-π40cosθsinθcosθsinθ0sinθ0cosθ2=cos-π4+sin-π41-cos-π4sin-π42                            =12-121-12×-122                            =01+122                            =0

Hence, if cos2θ = 0, then 0cosθsinθcosθsinθ0sinθ0cosθ2=0 or 12.

Page No 5.95:

Question 25:

If A is a matrix of order 3 × 3, then the number of minors in A is ____________.

Answer:

A is a matrix of order 3 × 3
A has 9 elements
A has 9 minors

Hence, the number of minors in A is 9.
 

Page No 5.95:

Question 26:

If x = –9 is a root of x372x276x = 0, then other two roots are ___________.

Answer:

Given: x = –9 is a root of x372x276x = 0

Let âˆ† = x372x276x  



=x372x276x Applying R1R1+R2+R3   =x+2+73+x+67+2+x2x276x    =x+9x+9x+92x276x Taking out x+9 common from R1   =x+91112x276x Applying C2C2-C1 and C3C3-C1   =x+911-11-12x-22-276-7x-7    =x+91002x-207-1x-7 Expanding through R1   =x+91x-2x-7   =x+9x-2x-7Since, =0x+9x-2x-7=0x=-9, 2, 7

Hence, other two roots are 2 and 7.

Page No 5.95:

Question 27:

0xyzx-zy-x0y-zz-xz-y0=_____________.

Answer:

=0xyzx-zy-x0y-zz-xz-y0Expanding along R1, we get   =00-y-xz-y-y-z0-x-zz-y+z-xxyzy-z-0   =0-y-x-x-zz-y+z-xxyzy-z   =y-xx-zz-y+z-xxyzy-z   =y-zz-xy-x+xyzHence, 0xyzx-zy-x0y-zz-xz-y0=y-zz-xy-x+xyz

Page No 5.95:

Question 28:

If A and B are square matrices of order 3 and |A| = 5, |B| = 5, then |3AB| = ____________.

Answer:

Given: 
A and B are square matrices of order 3
|A| = 5
|B| = 5


Now,3AB=3AB                 AB=AB, if they are square matrices of same order         =3A×5                 B=5         =33A×5                Order of A is 3×3         =135A         =135×5               A=5         =675

Hence, |3AB| = 675.

Page No 5.95:

Question 29:

If x+1x+2x+ax+2x+3x+bx+3x+4x+c=0, then a, b, c are in ____________.

Answer:

Given: x+1x+2x+ax+2x+3x+bx+3x+4x+c=0 


x+1x+2x+ax+2x+3x+bx+3x+4x+c=0Applying C1C1-C2x+1-x-2x+2x+ax+2-x-3x+3x+bx+3-x-4x+4x+c=0-1x+2x+a-1x+3x+b-1x+4x+c=0Taking out -1 common from C1-11x+2x+a1x+3x+b1x+4x+c=0Applying R2R2-R1 and R3R3-R1-11x+2x+a1-1x+3-x-2x+b-x-a1-1x+4-x-2x+c-x-a=0-11x+2x+a01b-a02c-a=0Expanding through C1-11c-a-2b-a=0c-a-2b+2a=0c+a-2b=0c+a=2ba, b, c are in A.P.

Hence, if x+1x+2x+ax+2x+3x+bx+3x+4x+c=0, then abc are in A.P..

Page No 5.95:

Question 30:

The value of the determinant sin Acos Asin A+cos Bsin Bcos Asin B+cos Bsin Ccos Asin C+cos B is ________________.

Answer:

Let âˆ† = sinAcosAsinA+cosBsinBcosAsinB+cosBsinCcosAsinC+cosB


=sinAcosAsinA+cosBsinBcosAsinB+cosBsinCcosAsinC+cosB   =sinAcosAsinAsinBcosAsinBsinCcosAsinC+sinAcosAcosBsinBcosAcosBsinCcosAcosB   =0+sinAcosAcosBsinBcosAcosBsinCcosAcosB      if any two columns are identical then the value of determinant is zero   =sinAcosAcosBsinBcosAcosBsinCcosAcosBTaking out cosA and cosB common from C2 and C3, respectively   =cosAcosBsinA11sinB11sinC11   =cosAcosB0                   if any two columns are identical then the value of determinant is zero   =0

Hence, the value of the determinant sinAcosAsinA+cosBsinBcosAsinB+cosBsinCcosAsinC+cosB is 0.

Page No 5.95:

Question 31:

If the determinant x+ap+ul+fy+bq+vm+gz+cr+wn+h splits into exactly k determinants of order 3, each element of which contains only one term, then k = __________.

Answer:

Let âˆ† = x+ap+ul+fy+bq+vm+gz+cr+wn+h


=x+ap+ul+fy+bq+vm+gz+cr+wn+h   =x+ap+uly+bq+vmz+cr+wn+x+ap+ufy+bq+vgz+cr+wh   =x+aply+bqmz+crn+x+auly+bvmz+cwn+x+ap+ufy+bq+vgz+cr+wh   =xplyqmzrn+aplbqmcrn+x+auly+bvmz+cwn+x+ap+ufy+bq+vgz+cr+wh   =xplyqmzrn+aplbqmcrn+xulyvmzwn+aulbvmcwn+x+ap+ufy+bq+vgz+cr+wh   =xplyqmzrn+aplbqmcrn+xulyvmzwn+aulbvmcwn+x+apfy+bqgz+crh+x+aufy+bvgz+cwh   =xplyqmzrn+aplbqmcrn+xulyvmzwn+aulbvmcwn+xpfyqgzrh+apfbqgcrh+x+aufy+bvgz+cwh   =xplyqmzrn+aplbqmcrn+xulyvmzwn+aulbvmcwn+xpfyqgzrh+apfbqgcrh+xufyvgzwh+aufbvgcwh

Hence, if the determinant x+ap+ul+fy+bq+vm+gz+cr+wn+h splits into exactly k determinants of order 3, each element of which contains only one term, then k = 8.

Page No 5.95:

Question 32:

The maximum value of 11111+sinθ1111+cosθ is ___________.

Answer:

Let âˆ† = 11111+sinθ1111+cosθ


=11111+sinθ1111+cosθApplying R2R2-R1 and R3R3-R1   =1111-11+sinθ-11-11-11-11+cosθ-1   =1110sinθ000cosθExpanding along C1   =1sinθcosθ   =2sinθcosθ2   =sin2θ2   But, sin2θ1sin2θ212

Hence, the maximum value of 11111+sinθ1111+cosθ is 12.



Page No 5.96:

Question 33:

If A, B, C are the angles of a triangle, then sin2Acot A1sin2Bcot B1sin2Ccot C1= _____________.

Answer:

Given: ABC are the angles of a triangle
Then, A + B + C π

Let âˆ† = sin2AcotA1sin2BcotB1sin2CcotC1


=sin2AcotA1sin2BcotB1sin2CcotC1Applying R2R2-R1 and R3R3-R1   =sin2AcotA1sin2B-sin2AcotB-cotA1-1sin2C-sin2AcotC-cotA1-1   =sin2AcotA1sinB-sinAsinB+sinAcosBsinB-cosAsinA0sinC-sinAsinC+sinAcosCsinC-cosAsinA0   =sin2AcotA12cosA+B2sinB-A22sinA+B2cosB-A2cosBsinA-cosAsinBsinAsinB02cosA+C2sinC-A22sinA+C2cosC-A2cosCsinA-cosAsinCsinAsinC0   =sin2AcotA12cosA+B2sinA+B22sinB-A2cosB-A2sinA-BsinAsinB02cosA+C2sinA+C22sinC-A2cosC-A2sinA-CsinAsinC0   =sin2AcotA1sin2A+B2sin2B-A2sinA-BsinAsinB0sin2A+C2sin2C-A2sinA-CsinAsinC0   =sin2AcotA1sinA+BsinB-AsinA-BsinAsinB0sinA+CsinC-AsinA-CsinAsinC0   =sin2AcotA1sinπ-CsinB-AsinA-BsinAsinB0sinπ-BsinC-AsinA-CsinAsinC0                    A+B+C=π   =sin2AcotA1sinCsinB-AsinA-BsinAsinB0sinBsinC-AsinA-CsinAsinC0Expanding along C3   =1sinCsinB-A×sinA-CsinAsinC-sinBsinC-A×sinA-BsinAsinB   =sinB-A×sinA-CsinA-sinC-A×sinA-BsinA   =-sinA-B×-sinC-AsinA-sinC-A×sinA-BsinA   =sinA-B×sinC-AsinA-sinC-A×sinA-BsinA   =0

Hence, if ABC are the angles of a triangle, then sin2AcotA1sin2BcotB1sin2CcotC1= 0.

Page No 5.96:

Question 34:

The determinant A=23+35515+465103+115155 is equal to is equal to ____________.

Answer:

Let A=23+35515+465103+115155


A=23+35515+465103+115155   =3+235553+22351033+523155   =3555351033155+2355223510523155   =P+Q                ...1whereP=3555351033155 and Q=2355223510523155Now,P=3555351033155Taking out 3, 5 and 5 common from C1, C2 and C3, respectively   =3×5×5111552335   =530                            C1 and C2 are identical,  determinant is zero   =0             ...2Q=2355223510523155Taking out 23, 5 and 5 common from C1, C2 and C3, respectively   =23×5×5111252535   =5230                            C1 and C3 are identical,  determinant is zero   =0               ...3From 1, 2 and 3, we getA=0+0  =0

Hence, the determinant A=23+35515+465103+115155 is equal to 0.

Page No 5.96:

Question 35:

The value of the determinant =sin233°sin257°cos180°-sin257°-sin233°cos2180°cos180°sin233°sin257° is equal to _______________

Answer:

Let =sin233°sin257°cos180°-sin257°-sin233°cos2180°cos180°sin233°sin257°


=sin233°sin257°cos180°-sin257°-sin233°cos2180°cos180°sin233°sin257°   =sin233°sin257°-1-sin257°-sin233°-12-1sin233°sin257°   =sin233°sin257°-1-sin257°-sin233°1-1sin233°sin257°Applying R1R1+R2   =sin233°-sin257°sin257°-sin233°-1+1-sin257°-sin233°1-1sin233°sin257°   =sin233°-sin257°sin257°-sin233°0-sin257°-sin233°1-1sin233°sin257°Taking out sin233°-sin257° common from R1   =sin233°-sin257° 1-10-sin257°-sin233°1-1sin233°sin257°Applying C2C2+C1   =sin233°-sin257° 1-1+10-sin257°-sin233°-sin257°1-1sin233°-1sin257°   =sin233°-sin257° 100-sin257°cos233°-1-sin257°1-1-1-sin233°sin257°   =sin233°-sin257° 100-sin257°cos233°-1-sin257°1-1-cos233°sin257°   =sin233°-sin257° 100-sin257°cos290°-57°-1-sin257°1-1-cos290°-57°sin257°   =sin233°-sin257° 100-sin257°sin257°-1-sin257°1-1-sin257°sin257°   =sin233°-sin257° 100-sin257°-11-1-sin257°sin257°Taking out -1 common from C2   =-1sin233°-sin257° 100-sin257°11-1sin257°sin257°   =-1sin233°-sin257°0                    C2 and C3 are identical,  the value of determinant is zero   =0

Hence, the value of the determinant =sin233°sin257°cos180°-sin257°-sin233°cos2180°cos180°sin233°sin257° is equal to 0.

Page No 5.96:

Question 36:

If 2-x2+x2+x2+x2-x2+x2+x2+x2-x=0, then x = ________________.

Answer:

Given: 2-x2+x2+x2+x2-x2+x2+x2+x2-x=0



2-x2+x2+x2+x2-x2+x2+x2+x2-x=0Applying R2R2-R12-x2+x2+x2+x-2+x2-x-2-x2+x-2-x2+x2+x2-x=02-x2+x2+x2x-2x02+x2+x2-x=0Taking 2x common from R22x2-x2+x2+x1-102+x2+x2-x=0Applying R3R3-R12x2-x2+x2+x1-102+x-2+x2+x-2-x2-x-2-x=02x2-x2+x2+x1-102x0-2x=0Taking 2x common from R32x22-x2+x2+x1-1010-1=0Applying C3C3+C12x22-x2+x2+x+2-x1-10+110-1+1=02x22-x2+x41-11100=0Expanding through R32x212+x1+4=02x22+x+4=02x2x+6=02x2=0 or 6+x=0x=0 or x=-6

Hence, x = 0, −6.
 

Page No 5.96:

Question 1:

If A is a singular matrix, then write the value of |A|.

Answer:

Given: A is a singular matrix.

Thus, A=0

Page No 5.96:

Question 2:

For what value of x, the following matrix is singular?

5-xx+124

Answer:

If a matrix A is singular, then A=0

 5 - xx + 124 = 0

4(5 - x) - 2(x + 1) 020 - 4x - 2x - 218 - 6x = 018 = 6xx = 3

Page No 5.96:

Question 3:

Write the value of the determinant 2342x3x4x568.

Answer:

Let Δ=  2     3      4 2x   3x    4x   5     6      8=x  2     3      4  2     3      4   5     6      8      Taking out x common from R2=0

Page No 5.96:

Question 4:

State whether the matrix 2364 is singular or non-singular.

Answer:

Let Δ = 2   3 6   4= 2 × 4 - 6 × 3 = 8 - 18 = -10A matrix is said to be singular if its determinant is equal to zero.Since Δ=-100, the given matrix is non-singular.

Page No 5.96:

Question 5:

Find the value of the determinant 4200420142054203.

Answer:

Let Δ=4200   42014202   4203 Δ=4200  14202  1         Applying C2C2 -  C1= 4200  - 4202 =-2

Page No 5.96:

Question 6:

Find the value of the determinant 101102103104105106107108109

Answer:

Let Δ = 101  102  103104  105  106 107  108  109  Δ = 101   1   2104   1   2107   1   2              Applying C2C2 - C1 and C3 C3 - C1 = 2101   1   1104   1   1107   1   1= 0             Since two columns are identitical, the value of the determinant is zero. Δ = 101  102  103104  105  106 107  108  109 = 0



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Question 7:

Write the value of the determinant a1b+cb1c+ac1a+b.

Answer:

Let  = a   1   b + cb   1   c + ac   1   a + b=a + b + c   1   b + ca + b + c   1   c + aa + b + c   1   a + b               Applying C1C1 + C3=a + b + c 1   1   b + c1   1   c + a1   1   a + b=a + b + c × 0=0

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Question 8:

If A=0ii1 and B=0110, find the value of |A| + |B|.

Answer:

A = 0   ii   1         A = 0-i2            =--1 = 1Also,B = 0   11   0       B = 0 - 1 = -1 So,A + B = 1 - 1 = 0

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Question 9:

If A=123-1 and B=10-10, find |AB|.

Answer:

A=1    23 -1             B=   1      0 -1    0         AB=1    23 -1   1      0-1     0 = 1 - 20 + 03 + 10 + 0 = -1 0 4 0AB = 0 - 0 = 0

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Question 10:

Evaluate 4785478747894791

Answer:

 Let Δ=4785   47874789   4791Δ=4785    24789    2          Applying C2C2-C1=2 × 4785    14789    1=2 × 4785  - 4789  = 2 × -4 = -84785   47874789   4791 = -8

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Question 11:

If w is an imaginary cube root of unity, find the value of 1ww2ww21w21w.

Answer:

  1     w    w2 w     w2   1 w2     1    w=1+w+w2        w    w2 w+w2+1     w2    1 w2 +1+w     1    w              Applying C1C1+ C2+C3= 0     w    w2 0     w2    1 0     1    w           1+w+w2=0, w is the imaginary cube root of unity =0

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Question 12:

If A=123-1 and B=1-43-2, find |AB|.

Answer:

A=1  23-1A = -1 - 6 = -7B=1 -43-2B=-2 + 12 = 10If A and B are square matrices of the same order, then AB=AB.AB = A B = -7 × 10=-70

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Question 13:

If A=aij is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.

Answer:

  If A=aij is a diagonal matrix of order n, then A=a11 × a22  × a33 × ... × ann.Given: a11= 1, a22 = 2 and a33= 3A = 1 × 2 × 3 = 6               Applying the above property

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Question 14:

If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.

Answer:

A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number.

Given: A = ai j  is 3 × 3 matrix, where a11=2 A = 2  0  00  2  00  0  2   A  =  2  0  0 0  2  0 0  0  2 =2 ×  2  0 0  2                  Expanding along C1=2 × 2 × 2 = 8

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Question 15:

If I3 denotes identity matrix of order 3 × 3, write the value of its determinant.

Answer:

In an identity matrix, all the diagonal elements are 1 and rest of the elements are 0.
Here,

I3=1  0  00  1  00  0  1  =1 × 1   00   1     Expanding along C1= 1I3= 1

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Question 16:

A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?

Answer:

If A = ai j  is a square matrix of order n and k is a constant, thenkA = kn A   Here, Number of rows = n k is a common factor from each row of k3A  = 33A = 27 × 5 = 135              Given matrix is 3×3 such that A = 5Thus, 3A  = 135

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Question 17:

On expanding by first row, the value of the determinant of 3 × 3 square matrix A = aij is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by second column.

Answer:

If A=ai j is a square matrix of order n, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefore,

i=1nai j Ci j=A and j=1nai j Ci j=AGiven: A = a11C11 + a12C12 + a13C13          Expanding along R1Now,A=a12 C12 + a22C22 + a32C32                Expanding along R2    a12 , a22 and a32 are elements of C2 

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Question 18:

Let A = [aij] be a square matrix of order 3 × 3 and Cij denote cofactor of aij in A. If |A| = 5, write the value of a31C31  +  a32C32a33C33.

Answer:

 If A=ai j is a square matrix of order n and Ci j is a cofactor of ai j, theni=1nai j Ci j=A and j=1nai j Ci j=AGiven: A = 5 and matrix A  is of order 3×3Since a13 C13 + a23C23 + a33C33  represent expansion of A along third column, we geta13 C13 + a23C23 + a33C33 = A = 5a13 C13 + a23C23 + a33C33 = 5

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Question 19:

In question 18, write the value of a11C21 + a12C22 + a13C23.

Answer:

We know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column ) is zero. Therefore,

A=ai j is a square matrix of order n.j=1nai j Ck j=0 and i=1nai j Ci k = 0a11C21  +  a12C22  + a13 C23 = 0    Since the elements are of first row and the cofactors are of elements of second row a11C21  +  a12C22  + a13 C23 = 0  

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Question 20:

Write the value of sin 20°-cos 20°sin 70°cos 70°.

Answer:

Let Δ=sin 20°    -cos 20°sin 70°       cos 70° =sin 20° cos 70° + cos 20°sin 70°=sin (20° + 70°)          trignometric identity=sin 90° =1

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Question 21:

If A is a square matrix satisfying AT A = I, write the value of |A|.

Answer:

Let A=ai j be a square matrix of order n. Here,A=AT              By property of determinantsGiven: ATA = I ATA = 1Then,ATA = AT  A       Since the determinants are of the same order  AT A =1A = 1ATA = 1A             A = ATA2= 1A = ±1

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Question 22:

If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.

Answer:

Since A & B are square matrices of the same order, by the property of determinants we get

AB  = A × BA = 3, AB = I AB = 1A × B = 13 × B = 1B = 13

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Question 23:

A is a skew-symmetric of order 3, write the value of |A|.

Answer:

We know that if a skew symmetric matrix A is of odd order, then A=0

Since the order of the given matrix is 3, A=0.

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Question 24:

If A is a square matrix of order 3 with determinant 4, then write the value of |−A|.

Answer:

  A  = 4    Here, Order of the matrix n = 3Using properties of matrices, we getkA = knA         For a square matrix of order n and constant k-A = -13  A = -1 × 4 = -4

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Question 25:

If A is a square matrix such that |A| = 2, write the value of |A AT|.

Answer:

In a square matrix, A  = AT.Since they are of same order, A AT = AAT .Given: A=2 A AT=2 × 2 = 4

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Question 26:

Find the value of the determinant 2431563008152100-304.

Answer:

 243    156    300  81      52    100 -3       0      4=243 -81 × 3   156 - 52 × 3    300 - 100 × 3   81                            52                         100 -3                             0                           4                   Applying  R1 R1-3 R2=   0            0          0  81       52       100 -3       0            4 =0

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Question 27:

Write the value of the determinant 2-354-6106-915.

Answer:

A=2   -3     54   -6    106    -9   15=2           -3            54 - 4   -6 + 6    10 - 106           -9            15                   Applying R2R2-2R 1=2    -3    50       0     06    -9   15 =0

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Question 28:

If the matrix  5x2-101 is singular, find the value of x.

Answer:

A  matrix is said to be singular if its determinant is zero. Since the given matrix is singular, we get

A= 5x      2 -10    1   A=   5x      2 -10     1 = 05x + 20 = 0         Expandingx = -205 = - 4



Page No 5.98:

Question 29:

If A is a square matrix of order n × n such that |A|=λ, then write the value of |−A|.

Answer:

A=λ             Order of A is n-A=-1n A=-1nλ

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Question 30:

Find the value of the determinant 222324232425242526.

Answer:


22    23   2423   24   2524   25   26 = 22×23 ×24 1     2   221     2   221     2   22                         Taking out common factors from R1, R2 and R3= 22×23 ×24×2 1     1   221     1   221     1   22=0              Two rows being identical22    23   2423   24   2524   25   26=0

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Question 31:

If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.

Answer:

Let A & B be non-singular matrices of order n.

A0 and B 0            By definition Since they are of same order, AB=ABAB=0 iff either A=0 or B=0           But it is not the case here. Thus, AB is non-zero and AB is non-singular matrix.

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Question 32:

A matrix of order 3 × 3 has determinant 2. What is the value of |A (3I)|, where I is the identity matrix of order 3 × 3.

Answer:

Let A be the given matrix. Then, A  = 2             Order=n = 3   I  = 1              I is an identity matrix3I =  3A3I =  3A  = 33A                A being of order 3=27 × 2 = 54A3I = 54

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Question 33:

If A and B are square matrices of order 3 such that |A| = − 1, |B| = 3, then find the value of |3 AB|.

Answer:

K A = KnA                     n is the order of A 3AB  = 33 AB         ...(1)If A and B are square matrices of the same order, then AB=A B. So,3AB=33A B                From eq. (1)         =27 × -1 × 3         =-81

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Question 34:

Write the value of a+ibc+id-c+ida-ib.

Answer:

a+ibc+id-c+ida-ib=a2-iab+iab-i2b2-(-c2-icd+icd+i2d2)=a2-i2b2+c2-i2d2Here, i2=-1a+ibc+id-c+ida-ib=a2+b2+c2+d2

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Question 35:

Write the cofactor of a12 in the following matrix 2-3560415-7.

Answer:

Given: 2 -3    56  0   41 5-7Here, a12=-3Cofactor of a12=-11+26   41-7 = -(-42 - 4) = 46

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Question 36:

If 2x+535x+29=0, find x.

Answer:

2x + 5 35x + 29 = 09(2x + 5) - 3(5x + 2) = 018x + 45 - 15x - 6 = 03x + 39 = 03x = -39x = -393 = -13

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Question 37:

Find the value of x from the following : x422x=0

Answer:

x42   2x = 02x2 - 8 = 02x2 = 8x2 = 82 = 4x = 4 = ±2

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Question 38:

Write the value of the determinant 2345686x9x12x

Answer:

2345686x9x12x= 2 3 45 6 82 3 4         Taking 2x common from R3= 0                         R1 and R3 are identical

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Question 39:

If |A| = 2, where A is 2 × 2 matrix, find |adj A|.

Answer:

For any square matrix A of order n, adjA=An-1.Given:A = 2Here, order is 2.adjA = 22-1 = 2

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Question 40:

What is the value of the determinant 020234456 ?

Answer:

020234456=0(18 - 20) - 2(12 - 16) + 0(10 - 12)=8

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Question 41:

For what value of x is the matrix 6-x43-x1 singular?

Answer:

6 - x 43 - x 1 is singular when its determinant is 0.6 - x  43 - x1 =06 - x - 43 - x = 06 - x - 12 + 4x = 03x - 6 = 03x = 6x = 63 = 2

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Question 42:

A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2 A|.

Answer:

 KA  = Kn A  Here, n is the order of A.Given:  A  = 42A = 23 × 4 = 32

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Question 43:

Evaluate: cos 15°sin 15°sin 75°cos 75°

Answer:

cos 15°    sin 15°sin 75°     cos 75°=cos 15° cos 75° - sin 15° sin 75°= cos (15° + 75°)                      cos A cos B - sin A sin B = cos (A + B)=cos 90 °=0cos 15°    sin 15°sin 75°     cos 75°  = 0

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Question 44:

If A=538201123. Write the cofactor of the element a32.

Answer:

Minor of a32 = M32 = 5821=5-16=-11

Cofactor of a32 = A32 = (−1)3+2 M32 = 11

Hence, the cofactor of the element a32 is 11.

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Question 45:

If x+1x-1x-3x+2=4-113, then write the value of x.

Answer:

x+1x-1x-3x+2=4-113x+1x+2-x-1x-3=12+1x2+3x+2-x2+4x-3=137x-1=137x=14x=2Hence, the value of x is 2.

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Question 46:

If 2xx+32x+1x+1=1533, then write the value of x.

Answer:

2xx+32x+1x+1=15332xx+1-2x+1x+3=3-15x+12x-2x-6=-12-6x-6=-12-6x=-6x=1Hence, the value of x is 1.

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Question 47:

If 3x7-24=8764, find the value of x.

Answer:

3x7-24=876412x+14=32-4212x+14=-1012x=-24x=-2 x=-2.



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Question 48:

If 2x58x=6-273, write the value of x.

Answer:

2x58x=6-2732x2-40=18+142x2-40=322x2=72x2=36x=±6Hence, the value of x is ±6.

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Question 49:

If A is a 3 × 3 matrix, A0 and 3A=kA then write the value of k.

Answer:

Let A=a1a2a3b1b2b3c1c2c3.then, 3A=3a13a23a33b13b23b33c13c23c3.3A=3a13a23a33b13b23b33c13c23c3       =33a1a2a3b1b2b3c1c2c3         Taking 3 common from each row       =27AHence, the value of k is 27.

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Question 50:

Write the value of the determinant pp+1p-1p.

Answer:

pp+1p-1p=p2-p+1p-1                   =p2-p2-1                   =p2-p2+1                   =1

Hence, the value of the determinant pp+1p-1p is 1.

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Question 51:

Write the value of the determinant x+yy+zz+xzxy-3-3-3.

Answer:

x+yy+zz+xzxy-3-3-3=x+y+zx+y+zz+x+yzxy-3-3-3       Applying R1R1+R2=x+y+z111zxy-3-3-3       Taking x+y+z common from R1=x+y+z111zxy-3-3-3       Applying R3R3+3R1=x+y+z111zxy000=0       Expanding along the last row


Hence, the value of the determinant x+yy+zz+xzxy-3-3-3 is 0.

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Question 52:

If A=cosθsinθ-sinθcosθ, then for any natural number, find the value of Det(An).

Answer:

Let A=cosθsinθ-sinθcosθ.Then, A2=cosθsinθ-sinθcosθcosθsinθ-sinθcosθ             =cos2θ-sin2θcosθsinθ+sinθcosθ-sinθcosθ-cosθsinθ-sin2θ+cos2θ             =cos2θsin2θ-sin2θcos2θSimilarly, An=cosnθsinnθ-sinnθcosnθTherefore,An=cosnθsinnθ-sinnθcosnθ      =cos2nθ+sin2nθ      =1Hence, Det(An)=1.

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Question 53:

Find the maximum value of  1111  1+sin θ1111+cos θ 

Answer:

Let =1111  1+sinθ1111+cosθ 

Applying R2R2-R1 and R3R3-R1, we get

=1110  sinθ000cosθ 

=sinθcosθ=sin2θ2
We know that −1 ≤ sin2θ ≤ 1.
∴ Maximum value of ∆ = 12×1=12

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Question 54:

If x ∈ N and x+3-2-3x 2x = 8, then find the value of x.

Answer:

x+3-2-3x 2x = 8

x+32x--2-3x=82x2+6x-6x=82x2=8x2-4=0x2=4x=2    x-2  xN

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Question 55:

If xsin θcos θ-sin θ-x1cos θ1x=8, write the value of x.

Answer:

xsin θcos θ-sin θ-x1cos θ1x=8
Expanding along R1, we get
x-x2-1-sinθ-xsinθ-cosθ+cosθ-sinθ+xcosθ=8-x3-x+xsin2θ+sinθcosθ-sinθcosθ+xcos2θ=8-x3-x+xsin2θ+cos2θ=8-x3-x+x=8x3+8=0x+2x2-2x+4=0x+2=0            x2-2x+4>0 xx=-2

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Question 56:

If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k.

Answer:

As we know that

A-1=Adj AA A-1=Adj AA= A3-1A    If A is a non singular matrix of order n, then adjA=An-1
​=A2A=A
As we are given that A-1=Ak 
 k=1

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Question 57:

A and B are square matrices of the same order 3, such that AB = 2I and |A| = 2, write the value of |B|.

Answer:

Given: 
A and B are square matrices of order 3
|A| = 2
AB = 2I


Now,AB=2IAB=2IAB=2I                 AB=AB, if they are square matrices of same order2B=2I                     A=22B=23I                    Order of I is 3×32B=8×1                    I=1B=4

Hence, the value of |B| is 4.



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