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Page No 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.10:

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 6.11:

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 6.11:

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 6.57:

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 6.57:

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 6.58:

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 6.58:

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 6.58:

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 6.58:

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)

Page No 6.58:

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 6.58:

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 6.58:

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 6.58:

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 6.58:

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 6.58:

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 6.58:

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 6.58:

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



Page No 6.59:

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 6.59:

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

Page No 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.59:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.60:

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 6.61:

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 6.61:

Question 41:

1+a1111+aa111+a=a3+3a2

Answer:

=1+a1111+a1111+a=1+a 1+a111+a - 11111+a +111+a11        Expanding=(1+a)(1+a)2 -1-1(1 + a - 1) + (1 - 1 - a)=(1+a)[1 + a2 + 2a - 1] -a -a=1 + a + a2 + a3 + 2a + 2a2 - 2a= a3 + 3a2

Page No 6.61:

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 6.61:

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 6.61:

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 6.61:

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 6.61:

Question 46:

Without expanding, prove that

abcxyzpqr = xyzpqrabc = ybqxapzcr

Answer:

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

Hence proved.

Page No 6.61:

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 6.61:

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 6.61:

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 6.61:

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 6.61:

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 6.61:

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

Answer:

(i)

Let =x+abcax+bcabx+c=x+a+b+cbcx+a+b+cx+bcx+a+b+cbx+c       Applying C1C1+C2+C3=x+a+b+c1bc1x+bc1bx+c =x+a+b+c1bc0 x01bx+c       Applying R2R2-R1=x+a+b+c1bc0 x000x             Applying R3R3-R1=x+a+b+cx2-0=0        Givenx2=0 or x+a+b+c=0x=0 or x=-a+b+c

(ii)
Let =x+axxxx+axxxx+a=3x+axx3x+ax+ax3x+axx+a        Applying C1C1+C2+C3=3x+a1xx1x+ax1xx+a=3x+a1xx0a01xx+a         Applying R2R2-R1=3x+a1xx0a000a              Applying R3R3-R1=3x+aa2-0=0x=-a3

(iii)

Let =3x-83333x-83333x-8=3x-2333x-23x-833x-233x-8              Applying C1=C1+C2+C3=3x-213313x-83133x-8 =3x-213303x-110133x-8             Applying R2R2-R1=3x-213303x-110003x-11           Applying R3R3-R1=3x-23x-112=0x=23,113,113

(iv)
Let =1xx21aa21bb2=1xx20x-ax2-a21bb2         Applying R2R1-R2=1xx20x-ax2-a20x-bx2-b2         Applying R3R1-R3=x-ax-b1xx201x+a01x+b =x-ax-bx+b-x-a=0x=a,b

(v)

Let =x+1352x+2523x+4=x+935x+9x+25x+93x+4              Applying C1=C1+C2+C3=x+91351x+2513x+4 =x+91350x-1013x+4              Applying R2R2-R1=x+91350x-1000x-1             Applying R3R3-R1=x+9x-12=0x=-9, 1, 1

(vi)
Let =1xx31bb31cc3=1xx30b-xb3-x31cc3        Applying R2R2-R1=1xx30b-xb3-x30c-xc3-x3          Applying R3R3-R1=1xx30x-bx3-b30x-cx3-c3=x-bx-c1xx201x2+xb+b201x2+xc+c2 =x-bx-cxc-b-b2+c2=0x=b, c, -b+c


(vii)

Let Δ=15-2x   11-3x   7-x   11           17      14   10          16       13=015-2x-14+2x      11-3x     7-x   11-28                    17        14   10-26                   16         13=0       Applying C1C1-2C3     1         11-3x      7-x-17           17           14 -16            16          13   =0 12-3x        4-2x      7-x      0              3           14       0              3           13    =0        Applying C1C1+C2 and C2C2 -C3 12-3x 3×13-3×14   =012-3x-3=012-3x  =03x =12 x=4

(viii)
Let =11xp+1p+1p+x3x+1x+2=11xppp3x+1x+2           Applying R2R2-R1=p11x1113x+1x+2 =p11x1112x2                        Applying R3R3-R1=p01x0112-xx2                  Applying C1C1-C2=p2-x×1x11          Expanding along C1=p2-x1-x=0x=1, 2


(ix)

Let Δ=3-2sin3θ-78cos2θ-11142=01-2sin3θ18cos2θ3142=0       Applying C1C1+C2 1-2sin3θ010cos2θ-sin3θ0202-3sin3θ=0       Applying R2R2-R1 and R3R3-3R1102-3sin3θ-20cos2θ-sin3θ=020-10sin3θ-20cos2θ=0sin3θ+2cos2θ-2=03sinθ-4sin3θ+2-4sin2θ-2=0-sinθ4sin2θ+4sinθ-3=0sinθ=0     or     4sin2θ+4sinθ-3=0θ=nπ     or     2sinθ+32sinθ-1=0θ=nπ     or     sinθ=-32     or     sinθ=12θ=nπ     or     θ=nπ+-1nπ6 , n



Page No 6.62:

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 6.62:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.71:

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 6.72:

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 6.72:

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 6.72:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.84:

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 6.85:

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 6.85:

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 6.85:

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 6.85:

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 6.85:

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 6.85:

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 6.85:

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 6.89:

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 6.89:

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 6.89:

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 6.89:

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 6.89:

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 6.90:

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 6.90:

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 6.90:

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 6.90:

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 6.90:

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 6.90:

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

Page No 6.90:

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

Page No 6.90:

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

Page No 6.90:

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

Page No 6.90:

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

Page No 6.90:

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

Page No 6.90:

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 = -