Rd Sharma XII Vol 1 2018 Solutions for Class 12 Science Math Chapter 5 Algebra Of Matrices are provided here with simple step-by-step explanations. These solutions for Algebra Of Matrices are extremely popular among Class 12 Science students for Math Algebra Of Matrices Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2018 Book of Class 12 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2018 Solutions. All Rd Sharma XII Vol 1 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 5.18:

Question 1:

Compute the following sums:
(i) 3-21  4+-24   13

(ii)    213   035-125+1-232   610-31

Answer:

i  3-214+-24133-2-2+41+14+31227

ii 213035-125+1-232610-31 2+11-23+3 0+23+65+1-1+02-35+13-16296-1-16

Page No 5.18:

Question 2:

Let A = 2432, B =    13-25 and C = -25   34. Find each of the following:
(i) 2A − 3B
(ii) B − 4C
(iii) 3AC
(iv) 3A − 2B + 3C

Answer:

i 2A-3B 2A-3B=22432-313-25 2A-3B=4864-39-615 2A-3B=4-38-96+64-15 2A-3B=1-112-11ii B-4CB-4C=13-25-4-2534B-4C=13-25--8201216B-4C=1+83-20-2-125-16B-4C=9-17-14-11iii 3A-C3A-C=32432--25343A-C=61296--25343A-C=6+212-59-36-43A-C=8762iv 3A-2B+3C 3A-2B+3C=32432-213-25+3-2534 3A-2B+3C=61296-26-410+-615912 3A-2B+3C=6-2-612-6+159+4+96-10+12 3A-2B+3C=-221228

Page No 5.18:

Question 3:

If A = 2357, B = -102   341, C = -123   210, find
(i) A + B and B + C
(ii) 2B + 3A and 3C − 4B.

Answer:

i A+B=2357+-102341

It is not possible to add these matrices because the number of elements in A are not equal to the
number of elements in B. So, A + B does not exist.

 B+C=-102341+-123210 B+C=-1-10+22+33+24+11+0 B+C= -225551

ii 2B+3A=2-102341+32357

It is not possible to add these matrices because the number of elements in B are not equal to the
number of elements in A. So, 2B + 3A does not exist.

3C-4B=3-123210-4-1023413C-4B=-369630--408121643C-4B=-3+46-09-86-123-160-43C-4B=161-6-13-4

Page No 5.18:

Question 4:

Let A = -102   314, B = 0-251-31 and C = 1-5   26  0-4. Compute 2A − 3B + 4C.

Answer:

Here,2A-3B+4C=2-102314-30-251-31+41-5260-42A-3B+4C=-204628-0-6153-93+4-208240-162A-3B+4C=-2-0+40+6-204-15+86-3+242+9+08-3-162A-3B+4C=2-14-32711-11

Page No 5.18:

Question 5:

If A = diag (2 − 59), B = diag (11 − 4) and C = diag (−6 3 4), find
(i) A − 2B
(ii) B + C − 2A
(iii) 2A + 3B − 5C

Answer:

 Here,A=2000-50009 , B=10001000-4 and C=-600030004i A-2BA-2B=2000-50009-210001000-4 A-2B=2000-50009-20002000-8A-2B=0000-700017=diag0 -7 17iiB+C-2AB+C-2A=10001000-4+-600030004-22000-50009B+C-2A=10001000-4+-600030004-4000-1000018B+C-2A=1-6-40+0-00+0-00+0-01+3+100+0-00+0-00+0-0-4+4-18B+C-2A=-900014000-18=diag-9 14 -18iii 2A+3B-5C2A+3B-5C=22000-50009+310001000-4-5-6000300042A+3B-5C=4000-1000018+30003000-12--3000015000202A+3B-5C=4+3+300+0-00+0-00+0-0-10+3-150+0-00+0-00+0-018-12-202A+3B-5C=37000-22000-14=diag37 -22 -14

Page No 5.18:

Question 6:

Given the matrices
A = 2   113-100   24, B = 97-135   421   6 and C = 2-431-109   45
Verify that (A + B) + C = A + (B + C).

Answer:

Here, LHS=A+B+C        =2113-10024+97-1354216+2-431-10945        =2+91+71-13+3-1+50+40+22+14+6+2-431-10945        =11806442310+2-431-10945        =11+28-40+36+14-14+02+93+410+5        =134373411715RHS=A+B+C         =2113-10024+97-1354216+2-431-10945         =2113-10024+9+27-4-1+33+15-14+02+91+46+5         =2113-10024+113244411511         =2+111+31+23+4-1+40+40+112+54+11         =134373411715 LHS=RHS

Hence proved.

Page No 5.18:

Question 7:

Find matrices X and Y, if X + Y = 5209 and XY = 3   60-1

Answer:

Given: X+Y+X-Y=5209+360-12X=5+32+60+09-12X=8808X=128808X=4404Now,X+Y-X-Y=5209-360-1X+Y-X+Y=5-32-60-09+12Y=2-4010Y=122-4010Y=1-205 X=4404 and Y=1-205

Page No 5.18:

Question 8:

Find X if Y = 3214 and 2X + Y =    10-32

Answer:


Given:2X+Y=10-322X+3214=10-322X=10-32-32142X=1-30-2-3-12-42X=-2-2-4-2X=12-2-2-4-2X=-1-1-2-1

Page No 5.18:

Question 9:

Find matrices X and Y, if 2XY =     6-60-4   21 and X + 2Y =    32   5-21-7

Answer:

Given:2X-Y=6-60-421                        ...1            X+2Y=325-21-7                        ...2Multiplying eq. 1 by eq. 2, we get22X-Y=26-60-4214X-2Y=12-120-842                 ...3From eq. 3 and eq. 4, we get 4X-2Y+X+2Y=12-120-842+325-21-75X=12+3-12+20+5-8-24+12-75X=15-105-105-5X=1515-105-105-5X=3-21-21-1Putting the value of X in eq. 2, we getX+2Y=325-21-73-21-21-1+2Y=325-21-72Y=325-21-7-3-21-21-12Y=3-32+25-1-2+21-1-7+1Y=02200-3

Page No 5.18:

Question 10:

If XY = 111110100 and X + Y =    351-114 1180, find X and Y.

Answer:


Here,X-Y+X+Y=111110100+351-11411802X=1+31+51+11-11+10+41+110+80+02X=4620241280X=124620241280X=231012640Now,X-Y-X+Y=111110100-351-1141180X-Y-X-Y=1-31-51-11+11-10-41-110-80-0-2Y=-2-4020-4-10-80Y=-12-2-4020-4-10-80Y=120-102540 X=231012640 and Y=120-102540

Page No 5.18:

Question 11:

Find matrix A, if 12-104   9 + A =    9-14-2   13

Answer:


Here,A=9-14-213-12-1049A=9-1-1-24+1-2-01-43-9A=8-35-2-3-6

Page No 5.18:

Question 12:

If A = 9178, B = 15712, find matrix C such that 5A + 3B + 2C is a null matrix.

Answer:

Given: 5A+3B+2C=000059178+315712+2C=00004553540+3152136+2C=000045+35+1535+2140+36+2C=000048205676+2C=00002C=0000-48205676C=12-48-20-56-76C=-24-10-28-38

Page No 5.18:

Question 13:

If A =    2-2   4   2-5   1, B = 8   04-23   6, find matrix X such that 2A + 3X = 5B.

Answer:

Given: 2A+3X=5B22-242-51+3X=5804-2364-484-102+3X=40020-1015303X=40020-101530-4-484-1023X=40-40+420-8-10-415+1030-23X=36412-142528X=1336412-142528X=12434-143253283

Page No 5.18:

Question 14:

If A = 1-322  02 and, B = 2-1-11  0-1, find the matrix C such that A + B + C is zero matrix.

Answer:


Given:A+B+C=0000001-32202+2-1-110-1+C=0000001+2-3-12-12+10+02-1+C=0000003-41301+C=000000C=000000-3-41301C=0-30+40-10-30-00-1C=-34-1-30-1

Page No 5.18:

Question 15:

Find x, y satisfying the matrix equations

(i) x-y2-24x   6+3-2   21   0-1=60052x+y5

(ii) xy+2z-3+y 45=4912

(iii) x21+y35+-8-11=0

Answer:

i Given:x-y2-24x6+3-2210-1=60052x+y5x-y+32-2-2+24+1x+06-1=60052x+y5x-y+3005x5=60052x+y5x-y+3=6x-y=6-3x-y=3                ...1Also,x=2x+y-x=y                   ...2Putting the value of y in eq. 1, we getx--x=32x=3x=32Putting the value of x in eq. 2, we get-32=yy=-32

ii xy+2z-3+y 45=4912x+yy+2+4z-3+5=4912x+yy+6z+2=4912 x+y=4           ...1Also, y+6=9y=3z+2=12z=10Putting the value of y in eq. 1, we getx+3=4                                     x=4-3 x=1                    x=1, y=3 and z=10                     


iii Given:x21+y35+-8-11=002x+3y-8x+5y-11=002x+3y-8=02x+3y=8               ...1Also,x+5y-11=0x+5y=11x=11-5y               ...2Putting the value of x in 1, we get211-5y+3y=822-10y+3y=8-7y=8-22-7y=-14y=2Putting the value of y in 2, we getx=11-52x=11-10x=1 x=1 and y=2



Page No 5.19:

Question 16:

If 2345x+1y01=  70105, find x and y.

Answer:

Given:2345x+1y01=7010568102x+1y01=701056+18+y10+02x+1=7010578+y102x+1=70105 8+y=0y=-8Also,2x+1=52x=4x=42=2 x=2 and y=-8

Page No 5.19:

Question 17:

Find the value of λ, a non-zero scalar, if λ 102345+2   1   23-1-32=44104214

Answer:


 Given: λ102345+2123-1-32=44104214λ02λ3λ4λ5λ+246-2-64=44104214λ+20+42λ+63λ-24λ-65λ+4=44104214λ+2=4     λ=4-2       λ=2         

Page No 5.19:

Question 18:

(i) Find a matrix X such that 2A + B + X = O, where
A = -12   34, B = 3-21   5
(ii) If A = 8   04-23   6 and B =     2-2   4   2-5   1, then find the matrix X of order 3 × 2 such that 2A + 3X = 5B.

Answer:

i 2A+B+X=02-1234+3-215+X=0000-2468+3-215+X=0000-2+34-26+18+5+X=000012713+X=0000 X=-1-2-7-13

ii 2A+3X=5B2804-236+3X=52-242-511608-4612+3X=10-102010-2553X=10-102010-255-1608-46123X=10-16-10-020-810+4-25-65-123X=-6-101214-31-7X=13-6-101214-31-7 X=-2-1034143-313-73

Page No 5.19:

Question 19:

Find x, y, z and t, if
(i) 3xyzt=   x6-1 2t+4x+yz+t3

(ii) 2x57y-3+3412=  7141514

Answer:

i 3xyzt=x6-12t+4x+yz+t33x3y3z3t=x+46+x+y-1+z+t2t+3 3x=x+4     3x-x=4    2x=4           x=2              Also,3y=6+x+y3y-y=6+x2y=6+x             ...1Putting the value of x in eq. 1, we get 2y=6+22y=8 y=4Now,3t=2t+33t-2t=3t=33z=-1+z+t                        3z-z=-1+t                        2z=-1+t             ...2        Putting the value of t in eq. 2, we get2z=-1+32z=2 z=1 x=2, y=4, z=1 and t=3


ii 2x57y-3+3412=71415142x10142y-6+3412=71415142x+310+414+12y-6+2=71415142x+314152y-4=7141514 2x+3=7   2x=4    x=2              Also,2y-4=142y=18y=9          

Page No 5.19:

Question 20:

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

2X+3Y=2340, 3X+2Y=-221-5

Answer:

We have,
32X+3Y-23X+2Y=32340-2-221-56X+9Y-6X-4Y=69120+4-4-2105Y=6+49-412-20+10Y=151051010Y=2122            ...(1)


Also,
22X+3Y-33X+2Y=22340-3-221-54X+6Y-9X-6Y=4680+6-6-315-5X=6+46-68-30+15X=1-5100515X=-20-1-3            ...(2)

From (1) and (2), we get

X=-20-1-3 and Y=2122.

Page No 5.19:

Question 21:

In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.

Answer:

Number of different types of posts in any college is given by

X  = 15611

Total number of posts of each kind in all the colleges = 30X

                                                                                    = 3015611

                                                                                      =  4501803030

Page No 5.19:

Question 22:

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer:

Let the monthly incomes of Aryan and Babban be 3x and 4x, respectively.

Suppose their monthly expenditures are 5y and 7y, respectively.

Since each saves Rs 15,000 per month, 

Monthly saving of Aryan: 3x-5y=15,000Monthly saving of Babban: 4x-7y=15,000

The above system of equations can be written in the matrix form as follows:

3-54-7xy=1500015000

or,
AX = B, where A=3-54-7, X=xy and B=1500015000

Now,

A=3-54-7=-21--20=-1

Adj A=-7-453T=-75-43

So, A-1=1AadjA=-1-75-43=7-54-3

X=A-1Bxy=7-54-31500015000xy=105000-7500060000-45000xy=3000015000x=30,000 and y=15,000

Therefore,

Monthly income of Aryan = 3×Rs 30,000=Rs 90,000

Monthly income of Babban = 4×Rs 30,000= Rs 1,20,000 

From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.



Page No 5.41:

Question 1:

Compute the indicated products:
(i)    ab-baa-bb   a

(ii) 1-22   3   12   3-32-1

(iii) 2343454561-350   243  05

Answer:

i ab-baa-bbaa×a+b×ba×-b+b×a-b×a+a×b  -b×-b+a×aa2+b2-ab+ab-ab+abb2+a2a2+b200a2+b2

ii 1-223123-32-11×1+-2×-31×2+-2×21×3+-2×-12×1+3×-32×2+3×22×3+3×-11+62-43+22-94+66-37-25-7103

iii 2343454561-350243052×1+3×0+4×32×-3+3×2+4×02×5+3×4+4×53×1+4×0+5×33×-3+4×2+5×03×5+4×4+5×54×1+5×0+6×34×-3+5×2+6×04×5+5×4+6×52+0+12-6+6+010+12+203+0+15-9+8+015+16+254+0+18-12+10+020+20+301404218-15622-270

Page No 5.41:

Question 2:

Show that ABBA in each of the following cases:
(i) A=5-16   7 and B=2134

(ii) A=-1    10   0-11   2   34 and B=123010110

(iii) A=130110410 and B=010100051

Answer:

i 
AB=5-1672134AB=10-35-412+216+28AB=713334            ...1Also,BA=21345-167BA=10+6-2+715+24-3+28BA=1653925            ...2 AB ≠ BA             From eqs. (1) and (2)


 ii AB= -1100-11234123010110AB=-1+0+0-2+1+0-3+0+00+0+10-1+10+0+02+0+44+3+46+0+0AB=-1-1-31006116            ...1Also,BA=123010110-1100-11234BA=-1+0+61-2+90+2+120+0+00-1+00+1+0-1+0+01-1+00+1+0BA=58140-11-101            ...2 AB ≠ BA             From eqs. (1) and (2)


iii AB= 130110410010100051AB=0+3+01+0+00+0+00+1+01+0+00+0+00+1+04+0+00+0+0AB=310110140            ...1Also,BA=010100051130110410BA=0+1+00+1+00+0+01+0+03+0+00+0+00+5+40+5+10+0+0BA=110130960            ...2 AB ≠ BA                  From eqs. (1) and (2)
 

Page No 5.41:

Question 3:

Compute the products AB and BA whichever exists in each of the following cases:
(i) A=1-22   3 and B=123231

(ii) A=   32-10-11 and B=456012
(iii) A = [1 −1 2 3] and B=0132

(iv) [a, b]cd + [a, b, c, d]abcd

Answer:

i AB=1-223123231AB=1-42-63-22+64+96+3AB=-3-418139

Since the number of columns in B is greater then the number of rows in A, BA does not exists.

ii AB=32-10-11456012AB=12+015+218+4-4+0-5+0-6+0-4+0-5+1-6+2AB=121722-4-5-6-4-4-4Also, BA=45601232-10-11BA=12-5-68+0+60-1-20+0+2BA=114-32

iii AB=1-1230132AB=0+-1+6+6AB=11Also,BA=01321-123BA=00001-1233-3692-246

iv abcd+abcdabcdac+bd+a2+b2+c2+d2     a2+b2+c2+d2+ac+bd

Page No 5.41:

Question 4:

Show that ABBA in each of the following cases:
(i) A=1   3-12-1-13   0-1 and B=-23-1-12-1-69-4

(ii) A=   10-4-1-11   5   0     9-5   1 and B=121342132

Answer:

i AB=13-12-1-130-1-23-1-12-1-69-4AB=-2-3+63+6-9-1-3+4-4+1+66-2-9-2+1+4-6-0+69+0-9-3-0+4AB=1003-53001                   ...1Also,BA=-23-1-12-1-69-413-12-1-130-1BA=-2+6-3-6-3+02-3+1-1+4-3-3-2+01-2+1-6+18-12-18-9+06-9+4BA=1-900-500-271                   ...2 AB ≠ BA          From eqs. (1) and (2)
 

Page No 5.41:

Question 5:

Evaluate the following:
(i)    1   3-1-4+   3-2-1   1135246

(ii) [1 2 3]102201012246

(iii) 1-10   22   3102201-012102

Answer:

i  13-1-4+3-2-11135246 1+33-2-1-1-4+113524641-2-31352464+212+420+6-2-6-6-12-10-1861626-8-18-28


ii 1231022010122461+4+00+0+32+2+6246531024610+12+6082

iii 1-10223102201-0121021-102231-00-12-22-10-01-21-102231-1010-11-1-1+00+10+20+00-22+3-2+00-30-1120-25-2-3

Page No 5.41:

Question 6:

If A = 1001, B = 1   00-1 and C = 0110, then show that A2 = B2 = C2 = I2.

Answer:

Here,A2=AAA2=10011001A2=1+00+00+00+1A2=1001         ...1B2=BBB2=100-1100-1B2=1+00-00-00+1B2=1001         ...2C2=CCB2=01100110B2=0+10+00+01+0B2=1001         ...3We know, I2=1001                ...4

A2=B2=C2=I2                From eqs. (1), (2), (3) and (4)



Page No 5.42:

Question 7:

If A = 2-13   2 and B =    04-17, find 3A2 − 2B + I

Answer:

Given: A=2-132Now,A2=AAA2=2-1322-132A2=4-3-2-26+6-3+4A2=1-41213A2-2B+I3A2-2B+I=31-4121-204-17+10013A2-2B+I=3-12363-08-214+10013A2-2B+I=3-0+1-12-8+036+2+03-14+13A2-2B+I=4-2038-10

Page No 5.42:

Question 8:

If A =    42-11, prove that (A − 2I) (A − 3I) = O

Answer:

Given: A-2IA-3I A-2IA-3I=42-11-21001 42-11-31001 A-2IA-3I=42-11-2002 42-11-3003 A-2IA-3I=4-22-0-1-01-24-32-0-1-01-3 A-2IA-3I=22-1-112-1-2 A-2IA-3I=2-24-4-1+1-2+2 A-2IA-3I=0000 A-2IA-3I=0Hence proved.

Page No 5.42:

Question 9:

If A = 1101, show that A2 = 1201 and A3 = 1301.

Answer:

Given: A=1101Now,A2=AAA2=11011101A2=1+01+10+00+1A2=1201A3=A2AA3=12011101A3=1+01+20+00+1A3=1301

Hence proved.

Page No 5.42:

Question 10:

If A = ab    b2-a2-ab, show that A2 = O

Answer:

Given: A=abb2-a2-abNow,A2=AAA2=abb2-a2-ababb2-a2-abA2=a2b2-a2b2ab3-ab3-a3b+a3b-a2b2+a2b2A2=0000A2=0Hence proved.

Page No 5.42:

Question 11:

If A =     cos2θsin2θ -sin2θcos2θ, find A2.

Answer:

Given: A=cos 2θsin 2θ-sin 2θcos 2θNow,A2=AAA2=cos 2θsin 2θ-sin 2θcos 2θcos 2θsin 2θ-sin 2θcos 2θA2=cos22θ-sin22θcos2θsin2θ+cos2θsin2θ-cos2θsin2θ-sin2θcos2θ-sin22θ+cos22θA2=cos2×2θ2sin2θcos2θ-2sin2θcos2θcos2×2θ                  cos2θ-sin2θ=cos2θA2=cos 4θsin2×2θ-sin2×2θcos 4θ                               sin2θ=2sinθcosθA2=cos 4θsin 4θ-sin 4θcos 4θ

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Question 12:

If A =    2-3-5-1   4   5   1-3-4 and B = -1   3   5   1-3-5-1   3   5, show that AB = BA = O3×3.

Answer:

Here,AB=2-3-5-1451-3-4-1351-3-5-135AB=-2-3+56+9-1510+15-251+4-5-3-12+15-5-20+25-1-3+43+9-125+15-20AB=000000000AB=03×3       ...1BA=-1351-3-5-1352-3-5-1451-3-4BA=-2-3+53+12-155+15-202+3-5-3-12+15-5-15+20-2-3+53+12-155+15-20BA=000000000  BA=03×3       ...2AB=BA=03×3              From eqs. (1) and (2)

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Question 13:

If A =    0   c-b-c   0   a   b-a  0 and B = a2abacabb2bcacbcc2, show that AB = BA = O3×3.

Answer:

Here,AB=0c-b-c0ab-a0a2abacabb2bcacbcc2AB=0+abc-abc0+b2c-b2c0+bc2-bc2-a2c+0+a2c-abc+0+abc-ac2+0+ac2a2b-a2b+0ab2-ab2+0abc-abc+0AB=000000000AB=O3×3       ...1BA=a2abacabb2bcacbcc20c-b-c0ab-a0BA=0-abc+abca2c+0-a2c-a2b+a2b+00-b2c+b2cabc+0-abc-ab2+ab2+00-bc2+bc2ac2+0-ac2-abc+abc+0BA=000000000BA=O3×3       ...2 AB=BA=O3×3           From eqs. (1) and (2)

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Question 14:

If A =    2-3-5-1   4   5   1-3-4 and B =    2-2-4-1   3   4   1-2-3, show that AB = A and BA = B.

Answer:

Given: AB=2-3-5-1451-3-42-2-4-1341-2-3AB=4+3-5-4-9+10-8-12+15-2-4+52+12-104+16-152+3-4-2-9+8-4-12+12AB=2-3-5-1451-3-4AB=ABA=2-2-4-1341-2-32-3-5-1451-3-4BA=4+2-4-6-8+12-10-10+16-2-3+43+12-125+15-162+2-3-3-8+9-5-10+12BA=2-2-4-1341-2-3BA=B

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Question 15:

Let A = -1   1-1   3-3   3   5   5   5 and B =    0   4   3   1-3-3-1   4   4, compute A2B2.

Answer:

Given: A=-11-13-33555Now,A2=AAA2=-11-13-33555-11-13-33555A2=1+3-5-1-3-51+3-5-3-9+153+9+15-3-9+15-5+15+255-15+25-5+15+25A2=-1-9-13273351535B2=BBB2=0431-3-3-1440431-3-3-144B2=0+4-30-12+120-12+120-3+34+9-123+9-120+4-4-4-12+16-3-12+16B2=100010001A2-B2A2-B2=-1-9-13273351535-100010001A2-B2=-1-1-9-0-1-03-027-13-035-015-035-1A2-B2=-2-9-13263351534

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Question 16:

For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC):
(i) A=   120-101, B=   10-12   03 and C=   1-1

(ii) A=423112301, B=1-110   122-11 and C=12-130   100   1.

Answer:

i ABC=ABC 120-10110-1203 1-1 =120-101 10-12031-1 1-2+00+4+0-1-0+00+0+3 1-1 =120-1011-0-1-20-3 -14-131-1 =120-1011-3-3 -1-4-1-3 =1-6-0-1-0-3 -5-4=-5-4 LHS=RHSHence proved.

ii  ABC=ABC4231123011-110122-11  12-1301001 =423112301 1-110122-1112-13010014+0+6-4+2-34+4+31+0+4-1+1-21+2+23+0+2-3+0-13+0+1  12-1301001=4231123011-3+02-0+0-1-1+10+3+00+0+00+1+22-3+04-0+0-2-1+110-5115-255-4412-1301001=423112301-22-1303-14-210-15+020-0+0-10-5+115-6+010-0+0-5-2+55-12+010-0+0-5-4+4=-8+6-38+0+12-4+6-6-2+3-22+0+8-1+3-4-6+0-16+0+4-3+0-2-520-4-110-2-710-5=-520-4-110-2-710-5 LHS=RHSHence proved.

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Question 17:

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC:
(i) A=1-10   2, B=-10   21 and C=0   11-1
(ii) A=   2-1   1   1-1   2, B=0111 and C=1-10   1.

Answer:

i  AB+C = AB+AC1-102-1021+011-1=1-102-1021+1-102011-11-102-1+00+12+11-1=-1-20-10+40+2+0-11+10+20-21-102-1130=-3-142+-122-2-1-31-00+60+0=-3-1-1+24+22-2-4160=-4160 LHS=RHSHence proved.

ii AB+C=AB+AC2-111-120111+1-101=2-111-120111+2-111-121-1012-111-120+11-11+01+1=0-12-10+11+10+2-1+2+2-0-2-11+0-1+1-1+01+22-111-121012=-111221+2-310-132-10-21+10+2-1+20+4=-1+21-31+12+02-11+31-22214=1-22214 LHS=RHSHence proved.

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Question 18:

If A=   1   0-2   3-1   0-2   1   1, B=   05-4-21   3-10   2 and C=   1   52-1   10   0-11, verify that A (BC) = ABAC.

Answer:

Given: AB-C=AB-AC10-23-10-21105-4-213-102-152-1100-11=10-23-10-21105-4-213-102-10-23-10-211152-1100-1110-23-10-2110-15-5-4-2-2+11-13-0-1-00+12-1=0-0+25+0-0-4+0-40+2-015-1+0-12-3+00-2-1-10+1+08+3+2- 1-0-05+0+22+0-23+1+015-1-06-0+0-2-1+0-10+1-1-4+0+110-23-10-211-10-6-103-111=25-8214-15-3-913-1704146-3-10-3-1-0+20+0-2-6+0-2-3+1-00-0+0-18-3+02-1-10+0+112+3+1=2-15-7-8-02-414-14-15-6-3+3-9+1013+31-2-8-20-210116=1-2-8-20-210116 LHS=RHSHence proved.



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Question 19:

Compute the elements a43 and a22 of the matrix:
A=010202032404   2-1-3   2   4  30   1-1   2-23-3   4-4   0

Answer:

We have,

Given: A=0102020324042-1-324301-12-23-34-40A=0102020324040-32+3-2-44+4-4-00+6-3-63+8-6-86+00+94-9-4+128-12-8+0A=010202032404-35-68-46-911-1469-58-4-8A=0+6+00-9-00+11+00-14-00+6-0-6+0+1810-0-10-12+0+1616-0-8-8+0-160+18+180-27-100+33+160-42-80+18-16-12+0+3620-0-20-24+0+3232-0-16-16+0-32A=6-911-14612048-2436-3749-502240816-48 a43=8 and a22=0

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Question 20:

If A=010001pqr, and I is the identity matrix of order 3, show that A3 = pI + qA +rA2.

Answer:

Given: A=010001pqrNow,A2=AAA2=010001pqr010001pqrA2=0+0+00+0+00+1+00+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r2A2=001pqrrpp+rqq+r2A3=A2AA3=001pqrrpp+rqq+r2010001pqrA3=0+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r20+0+pq+r2prp+0+q2+r2q0+p+rq+rq+r3A3=pqrrpp+rqq+r2pq+r2prp+q2+r2qp+2rq+r3         ...1pI+qA+rA2pI+qA+rA2=p100010001+q010001pqr+r001pqrrpp+rqq+r2pI+qA+rA2=p000p000p+0q000qpqq2qr+00rrprqr2r2prp+r2qrq+r3pI+qA+rA2=p+0+00+q+00+0+r0+0+rpp+0+rq0+q+r20+pq+r2p0+q2+rp+r2qp+qr+qr+r3pI+qA+rA2=pqrrpp+rqq+r2pq+r2pq2+r2q+rpp+2qr+r3         ...2 A3=pI+qA+rA2           From eqs. (1) and (2)

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Question 21:

If w is a complex cube root of unity, show that
1w w2w w21 w21w+w  w21 w21ww  w211w  w2=000

Answer:

Here,LHS= 1ww2ww21w21w+ww21w21www211ww2         =1+ww+w2w2+1w+w2w2+11+ww2+w1+w2w+11ww2         =-w2-1-w-1-w-w2-1-w-w21ww2                            1+w+w2=0 and w3=1         =-w2-w-w3-1-w2-w4-1-w2-w4         =-w1+w+w2-1-w2-w3w-1-w2-w3w         =-w×0-1-w2-w-1-w2-w                                        1+w+w2=0 and w3=1          =0-0-0         =000  1ww2ww21w21w+ww21w21www211ww2=000

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Question 22:

If A=   2-3-5-1   4   5   1-3-4, show that A2 = A.

Answer:

Here,A2=AAA2=2-3-5-1451-3-42-3-5-1451-3-4A2=4+3-5-6-12+15-10-15+20-2-4+53+16-155+20-202+3-4-3-12+12-5-15+16A2=2-3-5-1451-3-4 A2=A

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Question 23:

If A=4-1-43   0-43-1-3, show that A2 = I3.

Answer:

Here,A2=AAA2=4-1-430-43-1-34-1-430-43-1-3A2=16-3-12-4+0+4-16+4+1212+0-12-3+0+4-12+0+1212-3-9-3+0+3-12+4+9A2=100010001 A2=I3

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Question 25:

If [x 4 1]21   210   202-4   x   4-1 = 0, find x.

Answer:

Given: x4121210202-4x4-1=02x+4+0x+0+22x+8-4x4-1=02x+4x+22x+4x4-1=02x2+4x+4x+8-2x-4=02x2+6x+4=0x2+3x+2=0x2+2x+x+2=0xx+2+1x+2=0x+2x+1=0x+2=0 or  x+1=0x=-2  or  x=-1 x=-2,-1

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Question 26:

If [1 −1 x]01-121   311   1011 = 0, find x.

Answer:

Given: 1-1x01-1213111011=00-2+x1-1+x-1-3+x011=0-2+xx-4+x011=00+x-4+x=02x-4=02x=4x=42 x=2

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Question 27:

If A=3-24-2 and I=1001, then prove that A2A + 2I = O.

Answer:

Given:A=3-24-2Now,A2=AAA2=3-24-23-24-2A2=9-8-6+412-8-8+4A2=1-24-4A2-A+2I=1-24-4-3-24-2+21001A2-A+2I=1-3-2+24-4-4+2+2002A2-A+2I=-200-2+2002A2-A+2I=-2+20+00+0-2+2A2-A+2I=0000A2-A+2I=0Hence proved.

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Question 28:

If A=   31-12 and I=1001, then find λ so that A2 = 5A + λI.

Answer:

Given: A=31-12Now,A2=AAA2=31-1231-12A2=9-13+2-3-2-1+4A2=85-53A2=5A+λI85-53=531-12+λ100185-53=155-510+λ00λ85-53=15+λ5+0-5+010+λ85-53=15+λ5-510+λThe corresponding elements of two equal matrices are equal. 8=15+λ    8-15=λ  -7=λ λ=-7

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Question 29:

If A=   31-12, show that A2 − 5A + 7I2 = O

Answer:

Given: A=31-12Now,A2=AAA2=31-1231-12A2=9-13+2-3-2-1+4A2=85-53A2-5A+7I2A2-5A+7I2=85-53-531-12+71001A2-5A+7I2=85-53-155-510+7007A2-5A+7I2=8-15+75-5+0-5+5+03-10+7A2-5A+7I2=0000A2-5A+7I2=0Hence proved. 

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Question 30:

If A=   23-10, show that A2 − 2A + 3I2 = O

Answer:

Given: A=23-10Now,A2=AAA2=23-1023-10A2=4-36+0-2+0-3+0A2=16-2-3A2-2A+3I2 A2-2A+3I2=16-2-3-223-10+31001 A2-2A+3I2=16-2-3-46-20+3003 A2-2A+3I2=1-4+36-6+0-2+2+0-3+0+3 A2-2A+3I2=0000=0Hence proved.

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Question 31:

Show that the matrix A=2312 satisfies the equation A3 − 4A2 + A = O

Answer:

We have,  A=2312A2=AAA2=23122312A2=4+36+62+23+4A2=71247A3=A2AA3=712472312A3=14+1221+248+712+14A3=26451526Now, A3-4A2+A A3-4A2+A=26451526-471247+2312 A3-4A2+A=26451526-28481628+2312 A3-4A2+A=26-28+245-48+315-16+126-28+2 A3-4A2+A=0000=OHence proved.

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Question 32:

Show that the matrix A=  53127 is root of the equation A2 − 12AI = O

Answer:

Given: A=53127Now,A2=AAA2=5312753127A2=25+3615+2160+8436+49A2=613614485A2-12A-IA2-12A-I=613614485-1253127-1001A2-12A-I=613614485-603614484-1001A2-12A-I=61-60-136-36+0144-144+085-84-1A2-12A-I=0000Since A is satisfying the equation A2-12A-I, A is the root of the equation A2-12A-I .

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Question 33:

If A=   3-5-4   2, find A2 − 5A − 14I.

Answer:

Given: A=3-5-42Now,A2=AAA2=3-5-423-5-42A2=9+20-15-10-12-820+4A2=29-25-2024A2-5A-14IA2-5A-14I=29-25-2024-53-5-42-141001A2-5A-14I=29-25-2024-15-25-2010-140014A2-5A-14I=29-15-14-25+25+0-20+20+024-10-14A2-5A-14I=0000

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Question 24:

(i) If [1 1 x]102021210111 = 0, find x.
(ii) If 23571-3-24=-46-9x , find x.

Answer:

(i)
Given: 11x102021210111=01+0+2x0+2+x2+1+0111=01+2x2+x3111=01+2x+2+x+3=06+3x=03x=-6x=-63 x=-2

(ii)

Given: 23571-3-24=-46-9x2-6-6+125-14-15+28=-46-9x-46-913=-46-9xx=13 x=13



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Question 34:

 If A=   31-12, show that A2 − 5A + 7I = O use this to find A4.

Answer:

Given: A= 31-12Now,A2=AAA2=31-1231-12A2=9-13+2-3-2-1+4A2=85-53A2-5A+7IA2-5A+7I=85-53-531-12+71001A2-5A+7I=85-53-155-510+7007A2-5A+7I=8-15+75-5+0-5+5+03-10+7A2-5A+7I=0000=0Hence proved.Given: A2-5A+7I=0A2=5A-7I               ...1A3=A5A-7I         Multilpying by A on both sidesA3=5A2-7AIA3=55A-7I-7A        From eq. 1A3=25A-35I-7AA3=18A-35I         A4=18A-35IA         Multilpying by A on both sidesA4=18A2-35AA4=185A-7I-35A     From eq. 1 A4=90A-126I-35AA4=55A-126IA4=5531-12-1261001A4=16555-55110-12600126A4=165-12655-0-55-0110-126A4=3955-55-16

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Question 35:

If A=3-24-2, find k such that A2 = kA − 2I2

Answer:

Given: A=3-24-2Now,A2=AAA2=3-24-23-24-2A2=9-8-6+412-8-8+4A2=1-24-4A2=kA-2I21-24-4=k3-24-2-210011-24-4=3k-2k4k-2k-20021-24-4=3k-2-2k-04k-0-2k-2The corresponding elements of two equal matrices are equal.1=3k-2 1+2=3k  3=3k  k=1   

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Question 36:

If A=   10-17, find k such that A2 − 8A + kI = 0.

Answer:

Given:A=10-17Now,A2=AAA2=10-1710-17A2=1-00+0-1-70+49A2=10-849A2-8A+kI=010-849-810-17+k1001=010-849-80-856+k00k=01-8+k0-0+0-8+8+049-56+k=0-7+k00-7+k=0000The corresponding elements of two equal matrices are equal.-7+k=0 k=7 

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Question 37:

If A=1221, f (x) = x2 − 2x − 3, show that f (A) = 0

Answer:

Here,fx=x2-2x-3fA=A2-2A-3I2Now, A2=AAA2=12211221A2=1+42+22+24+1A2=5445fA=A2-2A-3I2fA=5445-21221-31001fA=5445-2442-3003fA=5-2-34-4-04-4-05-2-3fA=5-5005-5fA=0

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Question 38:

If A=2312 and I=1001, then find λ, μ so that A2 = λA + μI

Answer:

Given: A=2312Now,A2=AAA2=23122312A2=4+36+62+23+4A2=71247A2=λA+μI71247=λ2312+μ100171247=2λ3λλ2λ+μ00μ71247=2λ+μ3λ+0λ+02λ+μ71247=2λ+μ3λλ2λ+μThe corresponding elements of two equal matrices are equal. 7=2λ+μ                    ...1 12=3λλ=123=4Putting the value of λ in eq. 1, we get               7=24+μ7-8=μ μ=-1

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Question 39:

Find the value of x for which the matrix product
2   070   101-21-x14x   7x   0    1     0   x-4x-2x equal an identity matrix.

Answer:

Here, 2070101-21-x14x7x010x-4x-2x=100010001-2x+0+7x28x+0-28x14x+0-14x0+0+00+1-00+0-0-x-0+x14x-2-4x7x-0-2x=1000100015x00010010x-25x=100010001The corresponding elements of two equal matrices are equal. 5x=1 x=15 

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Question 40:

Solve the matrix equations:
(i) [x 1]   1   0-2-3x5=0

(ii) [1 2 1]12020110202x=0

(iii) [x-5-1]102021203x41=0

(iv) 2x312-30x8=0

Answer:

i  x110-2-3x5=0x-20-3x5=0x-2-3x5=0x2-2x-15=0x2-2x-15=0x2-5x+3x-15=0xx-5+3x-5=0x-5x+3=0x-5=0 or x+3=0x=5 or x=-3

ii 12112020110202x=01+4+12+0+00+2+202x=062402x=00+4+4x=04+4x=04x=-4x=-44=-1

iii x-5-1102021203x41=0x-0-20-10-02x-5-3x41=0x-2-102x-8x41=0x2-2x-40+2x-8=0x2-48=0x2=48x=±48

(iv) 2x312-30x8=02x-94xx8=0x2x-9+32x=02x2-9x+32x=02x2+23x=02x2+23x=0x2x+23=0x=0 or x=-232x=0 or x=-232

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Question 41:

If A=1   203-450-13, compute A2 − 4A + 3I3.

Answer:

Given: A=1203-450-13Now,A2=AAA2=1203-450-131203-450-13A2=1+6+02-8-00+10+03-12+06+16-50-20+150-3+00+4-30-5+9A2=7-610-917-5-314A2-4A+3I3 A2-4A+3I3=7-610-917-5-314-41203-450-13+3100010001A2-4A+3I3=7-610-917-5-314-48012-16200-412+300030003A2-4A+3I3=7-4+3-6-8+010-0+0-9-12+017+16+3-5-20+0-3-0+01+4+04-12+3A2-4A+3I3=6-1410-2136-25-35-5

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Question 42:

If f (x) = x2 − 2x, find f (A), where A=012450023

Answer:

Given: fx=x2-2xfA=A2-2ANow, A2=AAA2=012450023012450023A2=0+4+00+5+40+0+60+20+04+25+08+0+00+8+00+10+60+0+9A2=496202988169fA=A2-2AfA=496202988169-2012450023fA=496202988169-0248100046fA=4-09-26-420-829-108-08-016-49-6fA=472121988123

Page No 5.44:

Question 43:

If f (x) = x3 + 4x2x, find f (A), where A=0   122-301-10

Answer:

Given:  fx=x3+4x2-xfA=A3+4A2-ANow, A2=AAA2=0122-301-100122-301-10A2=0+2+20-3-20+0+00-6+02+9-04-0+00-2+01+3-02-0+0A2=4-50-6114-242 A3=A2AA3=4-50-6114-2420122-301-10A3=0-10+04+15-08-0+00+22+4-6-33-4-12+0+00+8+2-2-12-2-4+0+0A3=-1019826-43-1210-16-4fA=A3+4A2-AfA=-1019826-43-1210-16-4+44-50-6114-242-0122-301-10fA=-1019826-43-1210-16-4+16-200-244416-8168-0122-301-10fA=-10+16-019-20-18+0-226-24-2-43+44+3-12+16-010-8-1-16+16+1-4+8+0fA=6-26044114

Page No 5.44:

Question 44:

If A=102021203, then show that A is a root of the polynomial f (x) = x3 − 6x2 + 7x + 2.

Answer:

Given: fx=x3-6x2+7x+2fA=A3-6A2+7A+2I3Now,  A2=AAA2=102021203102021203A2=1+0+40+0+02+0+60+0+20+4+00+2+32+0+60+0+04+0+9A2=5082458013A3=A2AA3=5082458013102021203A3=5+0+160+0+010+0+242+0+100+8+04+4+158+0+260+0+016+0+39A3=210341282334055A3-6A2+7A+2I3A3-6A2+7A+2I3=210341282334055-65082458013+7102021203+2100010001A3-6A2+7A+2I3=210341282334055-3004812243048078+7014014714021+200020002A3-6A2+7A+2I3=21-30+7+20-0+0+034-48+14+012-12+0+08-24+14+223-30+7+034-48+14+00-0+0+055-78+21+2A3-6A2+7A+2I3=000000000=0Since fA=0, A is the root of fx=x3-6x2+7x+2.

Page No 5.44:

Question 45:

If A=122212221, then prove that A2 − 4A − 5I = O.

Answer:

Given: A=122212221Now,A2=AAA2=122212221122212221A2=1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1A2=988898889A2-4A-5I A2-4A-5I=988898889-4122212221-5100010001 A2-4A-5I=988898889-488848884-500050005 A2-4A-5I=9-4-58-8-08-8-08-8-09-4-58-8-08-8-08-8-09-4-5 A2-4A-5I=000000000=0

Hence proved.

Page No 5.44:

Question 46:

If A=320140005, show that A2 − 7A + 10I3 = O

Answer:

Given:A=320140005Here,A2=AAA2=320140005320140005A2=9+2+06+8+00+0+03+4+02+16+00+0+00+0+00+0+00+0+25A2=1114071800025Now, A2-7A+10I3A2-7A+10I3=1114071800025-7320140005+10100010001A2-7A+10I3=1114071800025-2114072800035+100001000010A2-7A+10I3=11-21+1014-14+00-0+07-7+018-28+100-0+00-0+00-0+025-35+10A2-7A+10I3=000000000=0

Page No 5.44:

Question 47:

Without using the concept of inverse of a matrix, find the matrix xyzu such that
   5-7-2   3xyzu=-16-6     7   2

Answer:

Given: 5-7-23xyzu=-16-6725x-7z5y-7u-2x+3z-2y+3u=-16-672The corresponding elements of two equal matrices are equal.5x-7z=-16                 ...1     5y-7u=-6                       ...2     -2y+3u=2   3u=2+2yu=2+2y3                  ...3     -2x+3z=7 3z=7+2x                     z=7+2x3                ...4             Putting the value of z in eq. 1, we get5x-77+2x3=-165x-49+14x3=-1615x-49-14x3=-16x-49=-48x=-48+49 x=1Putting the value of x in eq. 4, we getz=7+213z=93=3Putting the value of u in eq. 2, we get5y-72+2y3=-65y-14+14y3=-615y-14-14y3=-6y-14=-18y=-18+14 y=-4Putting the value of y in eq. 3, we getu=2+2-43u=-2  xyzu=1-43-2



Page No 5.45:

Question 48:

Find the matrix A such that
(i) 1101 A=335101

(ii) A 123456=-7-8-9   2   4   6

(iii) 413 A=-484-121-363

(iv) 213-10-1-11001110-1=A

(v)     2-1   10-34 A=-1-8-10   1-2  -5   922  15 


(vi) A=123456=-7-8-9   2   4   6 11 10   9
 

Answer:

i Let A=xyzabc1101xyzabc=335101x+ay+bz+c0+a0+b0+c=335101x+ay+bz+cabc=335101The corresponding elements of two equal matrices are equal.x+a=3              ...1y+b=3                  ...2 z+c=5                  ...3 a=1, b=0 and c=1Putting the value of a in eq. 1, we get x+1=3x=3-1 x=2Putting the value of b in eq. 2, we get y+b=3y+0=3 y=3Putting the value of c in eq. 3, we get z+1=5z=5-1 z=4 A=234101

ii  Let A=wxyzA123456=-7-8-9246wxyz123456=-7-8-9246w+4x2w+5x3w+6xy+4z2y+5z3y+6z=-7-8-9246The correspnding elements of two equal matrices are equal. 3w+6x=-9        ...1  y+4z=2  y=2-4z                   ...2 w+4x=-7 w=-7-4x          ...3  2y+5z=4                 ...4                                                                          Putting the value of w in eq. 1, we get3-7-4x+6x=-9-21-12x+6x=-9-6x=12x=-2Putting the value of x in eq. 3, we getw=-7-4-2  w=-7+8w=1Putting the value of y in eq. 4, we get22-4z+5z=44-8z+5z=44-3z=4-3z=0z=0Putting the value of z in eq. 2, we get  y=2-40   y=2 A=1-220


iii Let A=xyz413xyz=-484-121-3634x4y4zxyz3x3y3z=-484-121-363The corresponding elements of two equal matrices are equal.4x=-4              ...14y=8                      ...2 4z=4                      ...3 x=-1, y=2 and z=1 A=-121


iv  Let A=x213-10-1-11001110-1=A213-10-1-11001110-1=x-2-1+00+1+3-2+0+310-1=x-34110-1=x-3+0-1=x-4=xThe corresponding elements of two equal matrices are equal.x=-4  A=-4


v    2-1   10-34A=-1-8-10   1-2  -5   922  15Let A=xyzabc   2-1   10-34xyzabc=-1-8-10   1-2  -5   922  152x-a2y-b2z-cxyz-3x+4a-3y+4b-3z+4c=-1-8-10   1-2  -5   922  15By comparing the elements of second row, we getx=1, y=-2, z=-5By comparing the elements of first row, we get2x-a=-12-a=-1a=3Also,2y-b=-8-4-b=-8b=4And,2z-c=-10-10-c=-10c=0 A=1-2-5340

vi A123456=-7-8-9   2   4   6 11 10   9Let A=xaybzcxaybzc123456=-7-8-9   2   4   6 11 10   9x+4a2x+5a3x+6ay+4b2y+5b3y+6bz+4c2z+5c3z+6c=-7-8-9   2   4   6 11 10   9By comparing the corresponding elements, we getx+4a=-7 and 2x+5a=-8a=-2 and x=1Also,y+4b=2 and 2y+5b=4b=0 and y=2And,z+4c=11 and 2z+5c=10c=4 and z=-5 A=1-220-54
 

Page No 5.45:

Question 49:

Find a 2 × 2 matrix A such that
A 1-21  4=6I2

Answer:


Let A = wxyz

Now,

wxyz1-214=6I2w+x-2w+4xy+z-2y+4z=61001w+x-2w+4xy+z-2y+4z=6006The corresponding elements of two equal matrices are equal. w+x=6 w=6-x            ...1 -2w+4x=0         ...2 Putting the value of w in eq. 2, we get-26-x+4x=0-12+2x+4x=0-12+6x=06x=12x=2Putting the value of x in eq. 1, we getw=6-2w=4Now, y+z=0y=-z              ...3   -2y+4z=6       ...4 Putting the value of y in eq. 4, we get -2-z+4z=62z+4z=66z=6z=1Putting the value of z in eq. 3, we get y=-1 A=42-11

Page No 5.45:

Question 50:

If A=0040, find A16.

Answer:

Given: A=0040Here,A2=AAA2=00400040A2=0+00+00+00+0A2=0000A4=A2A2A4=00000000A4=0000A8=A4A4A8=00000000A8=0000A16=A8A8A16=00000000 A16=0000Thus, A16 is a null matrix.

Page No 5.45:

Question 51:

If A=0-xx0, B=0110 and x2 = −1, then show that (A + B)2 = A2 + B2.

Answer:

Given: A=0-xx0, B=0110 and x2 = −1

To show: (A + B)2 = A2 + B2

LHS:
 A+B=0-xx0+0110        =0+0-x+1x+10+0        =0-x+1x+10A+B2=0-x+1x+100-x+1x+10            =0+1-x1+x0+00+0x+11-x+0            =1-x2001-x2                  ...(1)

RHS:
A=0-xx0A2=0-xx00-xx0    =0-x20+00+0-x2+0    =-x200-x2               ...(2) B=0110 B2=01100110     =0+10+00+01+0     =1001                     ...(3)Adding (2) and (3), we getA2+B2=-x200-x2+1001           =1-x2001-x2             ...(4)

Comparing (1) and (4), we get

(A + B)2 = A2 + B2

Page No 5.45:

Question 52:

If A=10-3213011, then verify that A2 + A = A(A + I), where I is the identity matrix.

Answer:

To verify: A2 + A = A(A + I),

Given: A=10-3213011
A2=10-321301110-3213011    =1+0+00+0-3-3+0-32+2+00+1+3-6+3+30+2+00+1+10+3+1    =1-3-6440224

LHS:
A2+A=1-3-6440224+10-3213011          =1+1-3+0-6-34+24+10+32+02+14+1          =2-3-9653235

RHS:
A+I=10-3213011+100010001       =1+10+0-3+02+01+13+00+01+01+1       =20-3223012AA+I=10-321301120-3223012            =2+0+00+0-3-3+0-64+2+00+2+3-6+3+60+2+00+2+10+3+2            =2-3-9653235

Therefore, LHS = RHS.

Hence, A2 + A = A(A + I) is verified.

Page No 5.45:

Question 53:

If A=3-5-42, then find A2 − 5A − 14I. Hence, obtain A3.

Answer:

Given: A=3-5-42

A2=3-5-423-5-42    =9+20-15-10-12-820+4    =29-25-2024

A2-5A-14I=29-25-2024-53-5-42-141001                    =29-15-14-25+25-0-20+20-024-10-14                    =0000

Therefore, A2 − 5A − 14I = 0       ...(1)

Premultiplying the (1) by A, we get

A(A2 − 5A − 14I) = A.0
⇒ A3 − 5A2 − 14= 0
⇒ A3 = 5A2 + 14A
A3=529-25-2024+143-5-42       =145+42-125-70-100-56120+28       =187-195-156148

Page No 5.45:

Question 54:

(i) If P(x)=cosxsinx-sinxcosx, then show that P(x) P(y) = P(x + y) = P(y) P(x).

(ii) If P=x000y000z and Q=a000b000c, prove that PQ=xa000yb000zc=QP

Answer:

(i) Given: P(x)=cosxsinx-sinxcosx

then, P(y)=cosysiny-sinycosy

Now,
P(x) P(y)=cosxsinx-sinxcosxcosysiny-sinycosy              =cosxcosy-sinxsinycosxsiny+sinxcosy-sinxcosy-cosxsiny-sinxsiny+cosxcosy              =cosx+ysinx+y-sinx+ycosx+y                 ...(1)

Also,  P(x+y)=cosx+ysinx+y-sinx+ycosx+y          ...(2)

Now,
P(y) P(x)=cosysiny-sinycosycosxsinx-sinxcosx              =cosycosx-sinysinxcosysinx+sinycosx-sinycosx-cosysinx-sinysinx+cosycosx              =cosx+ysinx+y-sinx+ycosx+y                ...(3)

From (1), (2) and (3), we get

P(xP(y) = P(x + y) = P(yP(x)

(ii) Given: P=x000y000z and Q=a000b000c

Now,
PQ=x000y000za000b000c     =xa+0+00+0+00+0+00+0+00+yb+00+0+00+0+00+0+00+0+zc     =xa000yb000zc                 ...(4)

Also,
QP=a000b000cx000y000z     =ax+0+00+0+00+0+00+0+00+by+00+0+00+0+00+0+00+0+cz     =xa000yb000zc                 ...(5)

From (4) and (5), we get
PQ=xa000yb000zc=QP

Page No 5.45:

Question 55:

If A=2012131-10, find A2 − 5A + 4I and hence find a matrix X such that A2 − 5A + 4I + X = 0.

Answer:

Given: A=2012131-10

 A2=2012131-102012131-10    =4+0+10+0-12+0+04+2+30+1-32+3+02-2+00-1-01-3+0    =5-129-250-1-2

Now,
A2-5A+4I=5-129-250-1-2-52012131-10+4100010001                  =5-10+4-1-0+02-5+09-10+0-2-5+45-15+00-5+0-1+5+0-2-0+4                  =-1-1-3-1-3-10-542

Now, A2 − 5A + 4I + = 0
⇒ = −(A2 − 5A + 4I)
X=--1-1-3-1-3-10-542       =11313105-4-2

Page No 5.45:

Question 56:

If A=1101, prove that An=1n01 for all positive integers n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral powers of matrix, we have
A1=1101=A
So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
Am=1m01                   ...(1)

Now, we shall show that the result is true for n=m+1.
Here,
Am+1=1m+101

By definition of integral power of matrix, we have
 Am+1=AmA             =1m011101           From eq. 1             =1+01+m0+00+1             =11+m01

This shows that when the result is true for n = m, it is also true for n = m + 1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 5.45:

Question 57:

If A=ab01, prove that An=  anb(an-1)/a-101 for every positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=a1ba1-1/a-101=ab01=A

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
Am=ambam-1/a-101                       ...(1)

Now, we shall show that the result is true for n=m+1.
Here,
Am+1=am+1bam+1-1/a-101

By definition of integral power of matrix, we have
Am+1=AmAAm+1=ambam-1/a-101ab01       From eq. 1Am+1=ama+0amb+bam-1/a-10+00+1Am+1=am+1am+1b-amb+amb-b/a-101Am+1=am+1bam+1-1/a-101

This shows that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 5.45:

Question 58:

If A=cosθi sinθi sinθ cosθ, then prove by principle of mathematical induction that

An= cos nθ  i sinθi sin nθcos nθ for all n ∈ N.

Answer:


We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

A1=cos 1θi sin1θi sin 1θcos 1θ=cosθi sinθi sinθcosθ=A

Thus, the result is true for n=1.

Step 2: Let the result be true for n = m. Then,

Am=cosmθi sinmθi sinmθcosmθ

Now we shall show that the result is true for n=m+1.
Here,
Am+1=cos m+1θi sinm+1θi sin m+1θcos m+1θ                 ...(1)

By definition of integral power of matrix, we have
Am+1=AmAAm+1=cos mθi sinmθi sin  mθcos mθcosθi sinθi sinθcosθ                             From eq. 1Am+1=cos mθ.cosθ+i sinmθ.i sinθcos mθ.i sinθ+i sinmθ.cosθi sin mθ.cosθ+cos mθ.i sinθi sin mθ.i sinθ+cos mθ.cosθAm+1=cos mθ.cosθ-sinmθ.sinθicos mθ.sinθ+sinmθ.cosθi sin mθ.cosθ+cos mθ sinθ-sin mθ.sinθ+cos mθ.cosθAm+1=cos mθ.cosθ-sinmθ.sinθicos mθ.sinθ+sinmθ.cosθi sin mθ.cosθ+cos mθ sinθcos mθ.cosθ-sin mθ.sinθAm+1=cosmθ+θi sinmθ+θi sinmθ+θcosmθ+θAm+1=cosm+1θi sinm+1θi sinm+1θcosm+1θ

This shows that when the result is true for n = m, it is true for n=m+1.
Hence, by the principle of mathematical induction, the result is valid for all nN.

Disclaimer: n is missing before θ in a12 in An.



Page No 5.46:

Question 59:

If A=cos α+sin α2sin α-2sin αcos α-sin α, prove that

An=cos n α+sin n α2sin n α-2sin n αcos n α-sin n α for all nN.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1:  If n = 1, by definition of integral power of a matrix, we have

A1=cos 1α+sin 1α2sin 1α-2 sin 1αcos 1α-sin 1α=cos α+sin α 2sin α-2sin αcos α-sin α=A

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
Am=cos mα+sin mα 2sin mα-2sin mαcos mα-sin mα                    ...(1)

Now we shall show that the result is true for n=m+1.
Here,
Am+1=cos m+1α+sin m+1α 2sin m+1α-2sin m+1αcos m+1α-sin m+1α

By definition of integral power of matrix, we have
Am+1=Am.AAm+1=cos mα+sin mα 2sin mα-2sin mαcos mα-sin mαcos α+sin α 2sin α-2sin αcos α-sin α                      From eq. 1Am+1=cos mα+sin mαcos α+sin α -2sin mα2sin αcos mα+sin mα 2sin α+2sin mαcos α-sin α-2sin mαcos α+sin α-cos mα-sin mα2sin α-2sin mα2sin α+cos mα-sin mαcos α-sin αAm+1=cos mα cosα+sin mα cosα+cos mα sinα+sin mα sinα-2sin mα sinα2sin α cos mα+2sin α sin mα+2sin mα cosα-2sin ma sinα-2sin ma cosα-2sin ma sinα-2sin α cos mα+2sin α sin mα-2sin α sin mα+cos mα cosα-sin mα cosα-cos mα sinα+sin mα sinαAm+1=cosmα-α+sinmα+α-cosmα-α+cosmα+α2sinmα+α-2sinmα+αcosmα+α-sinmα+αAm+1=cosmα+α+sinmα+α2sinmα+a-2sinmα+αcosmα+α-sinmα+αAm+1=cosm+1α+sinm+1α2sinm+1α-2sinm+1αcosm+1α-sinm+1α

This show that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for all nN.

Page No 5.46:

Question 60:

Let A=111011001. Use the principle of mathematical induction to show that

An=1nn(n+1)/201n001 for every positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

A1=1111+1/2011001=111011001=A

Thus, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

Am=1mmm+1/201m001                      ...(1)

Now, we shall show that the result is true for n=m+1.
Here,

Am+1=1m+1m+1m+1+1/201m+1001Am+1=1m+1m+1m+2/201m+1001          

By definition of integral power of matrix, we have
Am+1=AmAAm+1=1mmm+1/201m001111011001                 From eq. 1Am+1=1+0+01+m+01+m+mm+1/20+0+00+1+00+1+m0+0+00+0+00+0+1Am+1=11+m2+2m+m2+m/2011+m001Am+1=11+mm2+3m+2/2011+m001Am+1=11+mm+1m+2/2011+m001


This shows that when the result is true for n = m, it is also true for n = m + 1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 5.46:

Question 61:

If B, C are n rowed square matrices and if A = B + C, BC = CB, C2 = O, then show that for every nN, An+1 = Bn (B + (n + 1) C).

Answer:

Let Pn be the statement given by Pn : An+1=BnB+n+1C.

For n = 1, we have
P1 : A2=BB+2CHere,LHS =A2         =B+CB+C         =BB+C+CB+C         =B2+BC+CB+C2         =B2+2BC                           BC=CB and C2=O         =BB+2C=RHS

Hence, the statement is true for n = 1.

If the statement is true for n = k, then
 Pk : Ak+1=BkB+k+1C                   ...(1)

For Pk+1 to be true, we must have
Pk+1 : Ak+2=Bk+1B+k+2C

Now,
Ak+2=Ak+1A          =BkB+k+1CB+C        From eqn. 1          =Bk+1+k+1BkCB+C          =Bk+1B+C+k+1BkCB+C          =Bk+2+Bk+1C+k+1BkCB+k+1BkC2          =Bk+2+Bk+1C+k+1BkBC       BC=CB and C2=0          =Bk+2+Bk+1C+k+1Bk+1C          =Bk+2+k+2Bk+1C          =Bk+1B+k+2C

So the statement is true for n = k+1.
Hence, by the principle of mathematical induction, Pn is true for all nN.

Page No 5.46:

Question 62:

If A = diag (a, b, c), show that An = diag (an, bn, cn) for all positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

A1=a1000b1000c1=a000b000c=A

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

Am=am000bm000cm                           ...(1)

Now, we shall check if the result is true for n=m+1.
Here,
Am+1=am+1000bm+1000cm+1

By definition of integral power of matrix, we have

Am+1=AmAAm+1=am000bm000cma000b000c                                    From eq. 1Am+1=aam+0+00+0+00+0+00+0+00+bbm+00+0+00+0+00+0+00+0+ccmAm+1=am+1000bm+1000cm+1

This shows that when the result is true for n = m, it is also true for n=m+1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 5.46:

Question 63:

If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ∈ ℕ.

Answer:

Let the given statement P(n), be given as
P(n): (AT)n = (An)T for all n ∈ ℕ.

We observe that
P(1): (AT)1 = AT = (A1)T
Thus, P(n) is true for n = 1.

Assume that P(n) is true for n = k ∈ ℕ.
i.e., P(k): (AT)k = (Ak)T

To prove that P(k + 1) is true, we have
(AT)k + 1 = (AT)k.(AT)1
               = (Ak)T.(A1)T
               = (A+ 1)T
Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of mathematical induction, P(n) is true for all n ∈ ℕ.

Page No 5.46:

Question 64:

A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal.

Answer:

Here,X a+b×a+2Y b+1×a+3Since XY exists, the number of columns in X is equal to the number of rows in Y. a+2=b+1                 ...1Similarly, since YXexists, the number of columns in Y is equal to the number of rows in X.a+b=a+3b=3Putting the value of b in 1, we geta+2=3+1a=2

Since the order of the matrices XY and YX is not same, XY and YX are not of the same type and they are unequal.

Page No 5.46:

Question 65:

Give examples of matrices
(i) A and B such that ABBA
(ii) A and B such that AB = O but A ≠ 0, B ≠ 0.
(iii) A and B such that AB = O but BAO.
(iv) A, B and C such that AB = AC but BC, A ≠ 0.

Answer:

i Let A=1-232 and B=23-12AB=1-23223-12AB=2+23-46-29+4AB=4-1413Now,BA=23-121-232BA=2+9-4+6-1+62+4BA=11256

Thus, AB ≠ BA.

ii Let A=0200 and B=1000 AB=0+00+00+00+0 =0000=O

Thus, AB = O while A ≠ 0 and B ≠ 0.

iii Let A=0100 and B=1000 AB=O and  BA=10000100 BA=0+01+00+00+0BA=0100

Thus, AB = O but BAO.

iv Let A=1000 , B=0001 and C=0002AB=10000001AB=0+00+00+00+0=0000and AC=10000002AC=0+00+00+00+0=0000

Thus,
AB = AC
But B ≠ C and A ≠ 0.

Page No 5.46:

Question 66:

Let A and B be square matrices of the same order. Does (A + B)2 = A2 + 2AB + B2 hold? If not, why?

Answer:

LHS=A+B2         =A+BA+B         =AA+B+BA+B         =A2+AB+BA+B2

We know that a matrix does not have commutative property. So,
ABBA
Thus,
A+B2A2+2AB+B2

Page No 5.46:

Question 67:

If A and B are square matrices of the same order, explain, why in general
(i) (A + B)2A2 + 2AB + B2
(ii) (A B)2A2 − 2AB + B2
(iii) (A + B) (AB) ≠ A2B2.

Answer:

i LHS =A+B2              =A+BA+B              =AA+B+BA+B              =A2+AB+BA+B2

We know that a matrix does not have commutative property. So,
ABBA
Thus,
A+B2A2+2AB+B2

ii LHS=A-B2              =A-BA-B              =AA-B-BA-B              =A2-AB-BA+B2

We know that a matrix does not have commutative property. So,
ABBA
Thus,
A-B2A2-2AB+B2

iii LHS=A+BA-B               =AA-B+BA-B                =A2-AB+BA-B2

We know that a matrix does not have commutative property. So,
ABBA
Thus,
A+BA-BA2-B2

Page No 5.46:

Question 68:

Let A and B be square matrices of the order 3 × 3. Is (AB)2 = A2 B2? Give reasons.

Answer:

Yes, (AB)2 = A2 B2 if AB = BA.

If AB = BA, then
 (AB)2 = (AB)(AB)
           = A(BA)B      (associative law)
           = A(AB)B
           = A2 B2
 

Page No 5.46:

Question 69:

If A and B are square matrices of the same order such that AB = BA, then show that (A + B)2 = A2 + 2AB + B2.

Answer:

(A + B)2 = (A + B)(A + B)
               = A2 + AB + BA B2
               = A2 + 2AB + B2          (∵ AB = BA)

Hence, (A + B)2 = A2 + 2AB + B2.

Page No 5.46:

Question 70:

Let A=111333, B=   31   52-24 and C=   42-35   50.
Verify that AB = AC though BC, AO.

Answer:

Here,  A=111333, B=3152-24 and C=42-3550Now, AB=1113333152-24AB=3+5-21+2+49+15-63+6+12AB=671821AC=11133342-3550AC=4-3+52+5+012-9+156+15+0AC=671821

So, AB = AC though B ≠ C , A ≠ O.

Page No 5.46:

Question 71:

Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill.

Answer:

 

Shopkeepers Notebooks
In dozen
Pens
In dozen
Pencils
In dozen
A 12 5 6
B 10 6 7
C 11 3 8

Here,
Cost of notebooks per dozen = 12×40 paise = Rs 4.80
Cost of pens per dozen = Rs. 12×1.25  = Rs 15
Cost ofpPencils per dozen = 12×35 paise = Rs 4.20

 12561067111384.80154.20=12×4.80+5×15+6×4.2010×4.80+6×15+7×4.2011×4.80+13×15+8×4.20                                     =57.60+75+25.2048+90+29.4052.80+195+33.60                                     =157.80167.40281.40

Thus, the bills of A, B and C are Rs 157.80, Rs 167.40 and Rs 281.40, respectively.

Page No 5.46:

Question 72:

The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each respectively. Find the total amount the store will receive from selling all the items.

Answer:

Stock of various types of books in the store is given by

        Physics  Chemistry    MathematicsX=120           96                     60

Selling price of various types of books in the store is given by

Y=8.303.454.50PhysicsChemistryMathematics   

Total amount received by the store from selling all the items is given by

XY=12096608.303.454.50      =1208.30+963.45+604.50      =996+331.20+270      =1597.20

Required amount = Rs 1597.20



Page No 5.47:

Question 73:

In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as
A=Cost per contact           40         100  50TelephoneHouse callLetter
The number of contacts of each type made in two cities X and Y is given in matrix B as
      Telephone                  House call                  LetterB=1000                       500                    5000    3000                          1000                    10000XY
Find the total amount spent by the group in the two cities X and Y.

Answer:

The cost per contact in paise is given by

  A=4010050TelephoneHousecallLetter

The number of contacts of each type made in the two cities X and Y is given by

          Telephone     Housecall      LetterB=1000            500             50003000            1000           10000   XY

Total amount spent by the group in the two cities X and Y is given by

 BA=1000500500030001000100004010050      =40000+50000+250000120000+100000+500000      =340000720000XY

Thus,
Amount spent on X = Rs 3400
Amount spent on Y = Rs 7200

Page No 5.47:

Question 74:

A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of
(i) Rs 1800 (ii) Rs 2000

Answer:

If Rs x are invested in the first type of bond and Rs 30000-x are invested in the second type of bond, then the matrix A=x30000-x represents investment and the matrix B=51007100 represents rate of interest.

i x30000-x51007100=18005x100+730000-x100=18005x+210000-7x100=1800210000-2x=1800002x=30000x=15000

Thus,
Amount invested in the first bond = Rs 15000

Amount invested in the second bond = Rs 30000-15000
                                                           = Rs 15000

ii x30000-x51007100=20005x100+730000-x100=20005x+210000-7x100=2000210000-2x=2000002x=10000x=5000

Thus,
Amount invested in the first bond = Rs 5000

Amount invested in the second bond = Rs 30000-5000
                                                            = Rs 25000

Page No 5.47:

Question 75:

To promote making of toilets for women, an organisation tried to generate awarness through (i) house calls, (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below:
(i) ₹50       (ii) ₹20       (iii) ₹40

The number of attempts made in three villages XY and Z are given below:
          (i)               (ii)              (iii)
X      400              300             100
Y      300              250               75
Z      500              400             150

Find the total cost incurred by the organisation for three villages separately, using matrices.

Answer:

According to the question,

Let A be the matrix showing number of attempts made in three villages XY and Z.
A=40030010030025075500400150

And, B be a matrix showing the cost for each mode per attempt.
B=502040

Now, the total cost per village will be shown by AB.
AB=40030010030025075500400150502040     =20000+6000+400015000+5000+300025000+8000+6000     =300002300039000

Hence, the total cost incurred by the organisation for three villages separately is
X: ₹30,000
Y: ₹23,000
Z: ₹39,000

Page No 5.47:

Question 76:

There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question?

Answer:

According to the question,

Let X be the matrix showing number of family members in family A and B.
X=462224Family AFamily B

And, Y be a matrix showing the recommend daily amount of calories.
Y=240019001800

And, Z be a matrix showing the recommend daily amount of proteins.
Z=455533

Now, the total requirement of calories of the two families will be shown by XY.
XY=462224240019001800     =9600+11400+36004800+3800+7200     =2460015800

Also, the total requirement of proteins of the two families will be shown by XZ.
XZ=462224455533     =180+330+6690+110+132     =576332

Hence, the total requirement of calories and proteins for each of the two families is shown as:

               Calories   ProteinsFamily A:Family B: 2460015800       576332

Page No 5.47:

Question 77:

In a parliament election, a political party hired a public relations firm to promote its candidates in three ways − telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as

A=140200150TelephoneHouse callsLetters

The number of contacts of each type made in two cities X and Y is given in the matrix B as

     Telephone       House calls        LettersB=1000                500               50003000               1000              10000City XCity Y

Find the total amount spent by the party in the two cities.

What should one consider before casting his/her vote − party's promotional activity of their social activities?

Answer:

According to the question,

Let A be the matrix showing the cost per contact (in paisa).
A=140200150TelephoneHouse callsLetters

And, B be a matrix showing the number of contacts of each type made in two cities X and Y.
     Telephone       House calls        LettersB=1000                500               50003000               1000              10000City XCity Y

Now, the total amount spent by the party in the two cities will be shown by BA.
BA=100050050003000100010000140200150     =140000+100000+750000420000+200000+1500000     =9900002120000

Hence, the total amount spent by the party in the two cities is
X: ₹9900
Y: ₹21200

One should consider social activities of a party before casting his/her vote.



Page No 5.48:

Question 78:

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer:

Let the monthly incomes of Aryan and Babban be 3x and 4x, respectively.

Suppose their monthly expenditures are 5y and 7y, respectively.

Since each saves Rs 15,000 per month, 

Monthly saving of Aryan: 3x-5y=15,000Monthly saving of Babban: 4x-7y=15,000

The above system of equations can be written in the matrix form as follows:

3-54-7xy=1500015000

or,
AX = B, where A=3-54-7, X=xy and B=1500015000

Now,

A=3-54-7=-21--20=-1

Adj A=-7-453T=-75-43

So, A-1=1AadjA=-1-75-43=7-54-3

X=A-1Bxy=7-54-31500015000xy=105000-7500060000-45000xy=3000015000x=30,000 and y=15,000

Therefore,

Monthly income of Aryan = 3×Rs 30,000=Rs 90,000

Monthly income of Babban = 4×Rs 30,000= Rs 1,20,000 

From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.

Page No 5.48:

Question 79:

A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interes. Using matrix method, find the amount invested by the trust.

Answer:

Let Rs x be invested in the first bond and Rs y be invested in the second bond.
Let A be the investment matrix and B be the interest per rupee matrix. Then,

A=xy and B=1010012100Total annual interest=AB=xy1010012100=10x100+12y10010x100+12y100=280010x+12y=280000       .....1
If the rates of interest had been interchanged, then the total interest earned is Rs 100 less than the previous interest.
 12x100+10y100=270012x+10y=270000       .....2
The system of equations (1) and (2) can be expressed as
PX = Q, where P=10121210, X=xy, Q=280000270000
P=10121210=100-144=-440
Thus, P is invertible.
 X=P-1QX=adj PPQxy=1-4410-12-1210T280000270000xy=1-4410-12-1210280000270000xy=2800000-3240000-44-3360000+2700000-44=1000015000x=10000 and y= 15000

Therefore, Rs 10,000 be invested in the first bond and Rs 15,000 be invested in the second bond.



Page No 5.54:

Question 1:

Let A=   2-3-7   5 and B=1   02-4, verify that
(i) (2A)T = 2AT
(ii) (A + B)T = AT + BT
(iii) (AB)T = AT BT
(iv) (AB)T = BT AT

Answer:

Given:A=2-3-75AT=2-7-35B=102-4 BT=120-4i 2AT=2AT22-3-75T=22-7-354-6-1410T=4-14-6104-14-610=4-14-610        LHS=RHSii A+BT=AT+BT2-3-75+102-4T=2-7-35+120-42+1-3+0-7+25-4T=2+1-7+2-3+05-4 3-3-51T=3-5-313-5-31=3-5-31 LHS=RHSiii A-BT=AT-BT2-3-75-102-4T=2-7-35-120-42-1-3-0-7-25+4T=2-1-7-2-3-05+41-3-99T=1-9-391-9-39=1-9-39 LHS=RHSiv ABT=BTAT2-3-75102-4T=120-4 2-7-352-60+12-7+100-20T=2-6-7+100+120-20-4123-20T=-4312-20-4312-20=-4312-20 LHS=RHS 

Page No 5.54:

Question 2:

If A=352 and B = [1 0 4], verify that (AB)T = BT AT

Answer:

Given:A=352AT=352B=104BT=104Now, AB=352104AB=30125020208ABT=35200012208          ...1BTAT=104352BTAT=35200012208          ...2 ABT=BTAT          From eqs. (1)and (2)

Page No 5.54:

Question 3:

Let A=1-102   131   21 and B=123213011. Find AT, BT and verify that
(i) (A + B)T = AT + BT
(ii) (AB)T = BT AT
(iii) (2A)T = 2AT.

Answer:

Given: A=1-10213121 and B=123213011AT=121-112031 and  BT=120211331i A+B=1-10213121+123213011 A+B=1+1-1+20+32+21+13+31+02+11+1A+B=213426132A+BT=241123362               ...1Now, AT+BT=121-112031+120211331AT+BT=1+12+21+0-1+21+12+10+33+31+1AT+BT=241123362                   ...2A+BT=AT+BT           From eqs. 1 and 2 ii AB=1-10213121 123213011 AB=1-2+02-1+03-3+02+2+04+1+36+3+31+4+02+2+13+6+1AB=-11048125510ABT=-14518501210              ...1Now,BTAT=120211331121-112031BTAT=1-2+02+2+01+4+02-1+04+1+32+2+13-3+06+3+33+6+1BTAT=-14518501210              ...2ABT=BTAT              From eqs. (1)and (2)iii 2A=21-10213121 2A=2-204262422AT=242-224062               ...1Now, 2AT=2121-1120312AT=242-224062               ...2 2AT=2AT             From eqs. (1) and (2)



Page No 5.55:

Question 4:

If A=-2   4   5, B = [1 3 −6], verify that (AB)T = BT AT

Answer:

Given: A=-245AT=-245B=13-6 BT=13-6Now, AB=-245 13-6 AB=-2-612412-24515-30ABT=-245-6121512-24-30           ...1 BTAT=13-6-245BTAT=-245-6121512-24-30           ...2 ABT=BTAT         From eqs. (1) and (2)

Page No 5.55:

Question 5:

If A=   24-1-10   2, B=   34-12   21, find (AB)T

Answer:

Here,AB=24-1-10234-1221AB=6-4-28+8-1-3-0+4-4+0+2AB=0151-2ABT=0115-2

Page No 5.55:

Question 6:

(i) For two matrices A and B, A=213410, B=1-10   25   0 verify that
(AB)T = BT AT.

(ii) For the matrices A and B, verify that (AB)T = BT AT, where
A=1324, B=1425

Answer:

i  Given: A=213410 AT=241130B=1-10250BT=105-120Now,AB=2134101-10250 AB =2+0+15-2+2+04+0+0-4+2+0AB =1704-2ABT=1740-2            ...1BTAT=105-120241130BTAT=2+0+154+0+0-2+2+0-4+2+0BTAT=1740-2            ...2 ABT=BTAT          From eqs. (1) and (2)

ii Given:A=1324AT=1234B=1425BT=1245Now,AB=1324 1425AB=1+64+152+88+20AB=7191028ABT=7101928            ...1Also, BTAT=12451234BTAT=1+62+84+158+20BTAT=7101928            ...2 ABT=BTAT          From eqs. (1) and (2)

Page No 5.55:

Question 7:

If AT=34-1201 and B=-121123, find AT − BT.

Answer:

Given: AT=34-1201 and B=-121123

BT=-112213

Now,
AT-BT=34-1201--112213            =3+14-1-1-22-20-11-3            =43-30-1-2

Therefore, AT-BT=43-30-1-2.

Page No 5.55:

Question 8:

If A=  cos αsin α-sin αcos α, then verify that AT A = I2.

Answer:

Given: A=cos αsin α-sin αcos α AT=cos α-sin αsin αcos αNow, ATA=I2Consider: LHS=ATA=cos α-sin αsin αcos αcos αsin α-sin αcos α=cos2α+sin2αcos α sin α-sin α cos αsin α cos α-cos α sin αsin2α+cos2α=1001=RHS

Hence proved.

Page No 5.55:

Question 9:

If A=   sin αcos α-cos αsin α, verify that AT A = I2.

Answer:

Given: A=sinαcosα-cosαsinα AT=sinα-cosαcosαsinαNow, ATA=sinα-cosαcosαsinαsinαcosα-cosαsinα ATA=sinαsinα+-cosα-cosαsinαcosα+-cosαsinαcosαsinα+sinα-cosαcosαcosα+sinαsinαATA=sin2α+cos2αsinα cosα-sinα cosαsinα cosα-sinα cosαcos2α+sin2αATA=1001=I

Page No 5.55:

Question 10:

If li, mi, nii = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AAT = I, where A=l1m1n1l2m2n2l3m3n3.

Answer:

Given,
l1, m1, n1,  l2, m2, n2, l3, m3, n3 are the direction cosines of three mutually perpendicular vectors in space.

l12+m12+n12=1l22+m22+n22=1l32+m32+n32=1                         .....il1l2+m1m2+n1n2=0l2l3+m2m3+n2n3=0l3l1+m3m1+n3n1=0            .....ii
Let A=l1m1n1l2m2n2l3m3n3
AT=l1l2l3m1m2m3n1n2n3
AAT=l1m1n1l2m2n2l3m3n3l1l2l3m1m2m3n1n2n3AAT=l12+m12+n12l1l2+m1m2+n1n2l3l1+m3m1+n3n1l1l2+m1m2+n1n2l22+m22+n22l2l3+m2m3+n2n3l3l1+m3m1+n3n1l2l3+m2m3+n2n3l32+m32+n32
From (i) and (ii), we get
AAT=100010001=I
Hence proved.



Page No 5.6:

Question 1:

If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Answer:

We know that if a matrix is of order m×n, then it has mn elements.

The possible orders of a matrix with 8 elements are given below:
1×8, 2×4, 4×2, 8×

Thus, there are 4 possible orders of the matrix.


The possible orders of a matrix with 5 elements are given below:
1×5, 5×1

Thus, there are 2 possible orders of the matrix.

Page No 5.6:

Question 2:

If A = [aij] = 23-514   907-2 and B = [bij] =    2-1-3  4 1 2
then find (i) a22 + b21 (ii) a11 b11 + a22 b22

Answer:

i

a22+b21

Here,a22=4 and b21=-3 a22+b21=4-3=1        

ii

a11b11+a22b22

Here,a11=2, b11=2, a22=4 and b22=4a11b11+a22b22=2×2+4×4a11b11+a22b22=4+16a11b11+a22b22=20

Page No 5.6:

Question 3:

Let A be a matrix of order 3 × 4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2

Answer:

The order of R1 is 1×4 and the order of C2 is 3×1.



Page No 5.60:

Question 1:

If A=2345, prove that AAT is a skew-symmetric matrix.

Answer:

Given:A=2345 AT=2435Now,A-AT=2345-2435A-AT=2-23-44-35-5A-AT=0-110                ...1A-ATT=0-110TA-ATT=01-10A-ATT=-0-110A-AT=-A-ATT       Using eq. 1Thus, A-AT is a skew-symmetric matrix.



Page No 5.61:

Question 2:

If A=3-41-1, show that AAT is a skewsymmetric matrix.

Answer:

 Given:A=3-41-1AT=31-4-1Now, A-AT=3-41-1-31-4-1A-AT=3-3-4-11+4-1+1A-AT=0-550            ...1A-ATT=0-550TA-ATT=05-50A-ATT=-0-550A-ATT=-A-AT                 From eq. 1Thus, A-AT is a skew-symmetric matrix.

Page No 5.61:

Question 3:

If the matrix A=52   xyz-34t-7 is a symmetric matrix, find x, y, z and t.

Answer:

Given:A=52xyz-34t-7AT=5y42ztx-3-7Since A is a symmetric matrix, AT=A. 5y42ztx-3-7=52xyz-34t-7The corresponding elements of two equal matrices are equal.  x=4       y=2       z=z       t=-3Hence, x=4, y=2, t=-3 and z can have any value.

Page No 5.61:

Question 4:

Let A=   327   143-258. Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew-symmetric matrix.

Answer:

Given: A=327143-258  AT=31-2245738Let X=12A+AT=12327143-258+31-2245738=3325232445248Let Y=12A-AT=12327143-258-31-2245738=01292-120-1-9210XT=3325232445248T=3325232445248T=XYT=01292-120-1-9210T=0-12-92120192-10=-01292-120-1-9210=YThus, X is a symmetric matrix and Y is skew-symmetric matrix.Now,X+Y=3325232445248+01292-120-1-9210=327143-258=A X=3325232445248  and Y=01292-120-1-9210

Page No 5.61:

Question 5:

Express the matrix A=4   2-13   5   71-2   1 as the sum of a symmetric and a skew-symmetric matrix.

Answer:

Given: A=42-13571-21AT=43125-2-171Let  X=12A+AT=1242-13571-21+43125-2-171=4520525520521XT=4520525520521T=4520525520521=XLet  Y=12A-AT=1242-13571-21-43125-2-171 =0-12-1120921-920YT=0-12-1120921-920T=0121-120-92-1920=-0-12-1120921-920=-YThus, X is a symmetric matrix and Y is a skew-symmetric matrix.Now, X+Y=4520525520521+0-12-1120921-920=42-13571-21=A

Page No 5.61:

Question 6:

Define a symmetric matrix. Prove that for A=2456, A + AT is a symmetric matrix where AT is the transpose of A.

Answer:

A square matrix A is called a symmetric matrix, if AT=A.Given: A=2456 AT=2546Now, A+AT=2456+2546      A+AT=49912            ...1A+ATT=49912T                =49912                =A+AT               From eq. 1 A+ATT=A+ATThus, A+AT is a symmetric matrix.

Page No 5.61:

Question 7:

Express the matrix A=3-41-1 as the sum of a symmetric and a skew-symmetric matrix.

Answer:

Given: A=3-41-1AT=31-4-1Let  X=12A+AT=123-41-1+31-4-1=3-32-32-1XT=3-32-32-1T=3-32-32-1=XLet Y=12A-AT=123-41-1-31-4-1  =0-52520YT=0-52520T=052-520=-0-52520=Y X is a symmetric matrix and Y is a skew-symmetric matrix.X+Y=3-32-32-1+0-52520=3-41-1=A

Page No 5.61:

Question 8:

Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result:    3-2-4   3-2-5-1   1   2.

Answer:


Here,  A=3-2-43-2-5-112 AT=33-1-2-21-4-52Let  X=12A+AT=123-2-43-2-5-112+33-1-2-21-4-52=312-5212-2-2-52-22XT=312-5212-2-2-52-22T=312-5212-2-2-52-22=XLet Y=12A-AT=123-2-43-2-5-112-33-1-2-21-4-52=0-52-32520-33230    YT=0-52-32520-33230T=05232-5203-32-30=-0-52-32520-33230=-Y X is a symmetric matrix and Y is a skew-symmetric matrix.Now,         X+Y=312-5212-2-2-52-22+0-52-32520-33230=3-2-43-2-5-112=A                  

Page No 5.61:

Question 1:

If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.

Answer:

Given: Order of A = m×n
            Order of B = n×p

Since  the number of columns in A are equal to the number of rows in B, i.e. n, AB exists.
Order of AB = Number of rows in A× Number of columns in B
                           = m×p

Page No 5.61:

Question 2:

If A=214415and B=3-12   21   3. Write the orders of AB and BA.

Answer:


The order of matrix A is 2×3 and the order of matrix B is 3×2.

Since the number of columns in A is equal to the number of rows in B, AB exists and it is of order 2×2.
Also, since the number of columns in B is equal to the number of rows in A, BA exists and it is of order 3×3.



Page No 5.62:

Question 3:

If A=4312 and B=-4   3, write AB.

Answer:

AB=4312-43AB=-16+9-4+6AB=-72

Page No 5.62:

Question 4:

If A=123, write AAT.

Answer:

Given: A=123 AT=123Now, AAT=123123AAT=123246369

Page No 5.62:

Question 5:

Given an example of two non-zero 2 × 2 matrices A and B such that AB = O.

Answer:

Let the two matrices be A=5090 and B=0012-21, such that AB=50900012-21.AB=0000

Page No 5.62:

Question 6:

If A=2357, find A + AT.

Answer:

Given:A=2357 AT=2537Now, A+AT=2357+2537A+AT=2+23+55+37+7A+AT =48814      

Page No 5.62:

Question 7:

If A=i00i, write A2.

Answer:

Given: A=i00iA2=AAA2=i00ii00iA2=i2+00+00+00+i2A2=i200i2A2=-100-1                      i2=-1  

Page No 5.62:

Question 8:

If A=  cos xsin x-sin xcos x, find x satisfying 0 < x < π2 when A + AT = I

Answer:

Given: A=cos xsin x-sin xcos xAT=cos x-sin xsin xcos xNow, A+AT=Icos xsin x-sin xcos x+cos x-sin xsin xcos x=1001cos x+cos xsin x-sin x-sin x+sin xcos x+cos x=10012cos x002cos x=10012cos x=1 cos x=12x=π3

Page No 5.62:

Question 9:

If A=cos x-sin x sin x cos x, find AAT

Answer:

Given:A=cosx-sinxsinxcosx AT=cosxsinx-sinxcosxAAT=cosx-sinxsinxcosxcosxsinx-sinxcosxAAT=cos2x+sin2xcosxsinx-sinxcosxcosxsinx-sinxcosxsin2x+cos2xAAT=1001

Page No 5.62:

Question 10:

If 10y5+2x   01-2 = I, where I is 2 × 2 unit matrix. Find x and y.

Answer:

Given:10y5+2x01-2=I10y5+2x02-4=10011+2x0+0y+25-4=10011+2x0y+21=1001 1+2x=1 and y+2=02x=1-1 and y=-22x=0x=02=0

Page No 5.62:

Question 11:

If A=   1-1-1   1, satisfies the matrix equation A2 = kA, write the value of k.

Answer:

A2=AAA2=1-1-111-1-11A2=1+1-1-1-1-11+1A2=2-2-22Now, A2=kA2-2-22=k1-1-112-2-22=k-k-kk k=2

Page No 5.62:

Question 12:

If A=1111 satisfies A4 = λA, then write the value of λ.

Answer:

A2=A.AA2=11111111A2=1+11+11+11+1A2=2222Now, A4=A2A2A4=22222222A4=4+44+44+44+4A4=8888Also, A4=λA8888=λ11118888=λλλλ λ=8

Page No 5.62:

Question 13:

If A=-1   0  0   0-1   0  0   0-1, find A2.

Answer:

Here,A2=AAA2=-1000-1000-1-1000-1000-1A2=1+0+00+0+00+0+00+0+00+1+00+0+00+0+00+0+00+0+1A2=100010001

Page No 5.62:

Question 14:

If A=-1   0   0   0-1   0   0   0-1, find A3.

Answer:

Here,A2=AAA2=-1000-1000-1-1000-1000-1A2=1+0+00+0+00+0+00+0+00+1+00+0+00+0+00+0+00+0+1A2=100010001Now, A3=A2AA3=100010001-1000-1000-1A3=-1+0+00+0+00+0+00+0+00-1+00+0+00+0+00+0+00+0-1A3=-1000-1000-1=A

Page No 5.62:

Question 15:

If A=-3   0   0-3, find A4.

Answer:

Here,A2=AAA2=-300-3-300-3A2=9+00+00+00+9A2=9009Now,A4=A2A2A4=90099009A4=81+00+00+00+81A4=810081

Page No 5.62:

Question 16:

If [x  2] 34=2, find x

Answer:

Given:x234=23x+8=23x=2-83x=-6x=-63x=-2

Page No 5.62:

Question 17:

If A = [aij] is a 2 × 2 matrix such that aij = i + 2j, write A.

Answer:

Here, aij=i+2j, 1i2 and 1j2 a11=1+21=3, a12=1+22=1+4=5a21=2+21=4 and a22=2+22=2+4=6 A=a11a12a21a22=3546

Page No 5.62:

Question 18:

Write matrix A satisfying A+   23-14=   3-6-3   8.

Answer:


Given: A+23-14=3-6-38A=3-6-38-23-14A=3-2-6-3-3+18-4A=1-9-24

Page No 5.62:

Question 19:

If A = [aij] is a square matrix such that aij = i2j2, then write whether A is symmetric or skew-symmetric.

Answer:

Here,aij=i2-j2 , 1i2 and 1j2 a11=12-12=1-1=0  , a12=12-22=1-4=-3a21=22-12=4-1=3  and  a22=22-22=4-4=0 A=a11a12a21a22=0-330AT=03-30AT=-0-330AT=-ASince AT=-A, A is skew-symmetric.

Page No 5.62:

Question 20:

For any square matrix write whether AAT is symmetric or skew-symmetric.

Answer:

Here,

AATT=ATTAT        ABT=BTAT AATT=AAT           ATT=A         

Thus, AAT is a symmetric matrix.

Page No 5.62:

Question 21:

If A = [aij] is a skew-symmetric matrix, then write the value of i aij.

Answer:

Given: A=aij is a skew-symmetric matrix.aij=-aij                For all values of i, jaii=-aii               For all values of iaij+aii=02aii=0aii=0                    For all values of iiaii=0+0+...+0      i times           =0Thus,iaii=0

Page No 5.62:

Question 22:

If A = [aij] is a skew-symmetric matrix, then write the value of ij aij.

Answer:

 Given: A=aij is a skew symmetric matrix.aij=-aji       For all values of i,  jaii=-aii       For all values of iaij=0Now, ijaij=a11+a12+a13+...+a21+a22+a23+...+a31+a32+a33+...                =0+a12+a13+...-a12+0+a23+...-a13-a23+0+...                =0Thus, ijaij=0

Page No 5.62:

Question 23:

If A and B are symmetric matrices, then write the condition for which AB is also symmetric.

Answer:

Given: AB is symmetric.

 ABT=ABBTAT=AB            ABT=BTATBA=AB                 A and B are symmetric matrices, so AT=A and BT=B

Thus, AB is also symmetric, if AB = BA.

Page No 5.62:

Question 24:

If B is a skew-symmetric matrix, write whether the matrix AB AT is symmetric or skew-symmetric.

Answer:

If B is a skew-symmetric matrix, then BT=-B.

ABATT=ATTBTAT      ABCT=CTBTATABATT=ABTAT             ATT=AABATT=A-B AT         BT=-BABATT=-ABAT  ABAT is a skew-symmetric matrix.



Page No 5.63:

Question 25:

If B is a symmetric matrix, write whether the matrix AB AT is symmetric or skew-symmetric.

Answer:

If B is a symmetric matrix, then BT=B.

AB ATT= ATT BTAT                ABCT=CTBTAT AB ATT=ABTAT                   ATT=A AB ATT=AB AT                    BT=B

AB AT is a symmetric matrix.

Page No 5.63:

Question 26:

If A is a skew-symmetric and nN such that (An)T = λAn, write the value of λ.

Answer:

Given: A is skew symmetric matrix.

AT=-A

       AnT=λAnATn=λAn-An=λAn-1n An=λAnλ=-1n

Page No 5.63:

Question 27:

If A is a symmetric matrix and nN, write whether An is symmetric or skew-symmetric or neither of these two.

Answer:

If A is a symmetric matrix, then AT=A.Now, AnT=ATn              for all nNAnT=An            AT=A

Hence, An is a symmetric matrix.

Page No 5.63:

Question 28:

If A is a skew-symmetric matrix and n is an even natural number, write whether An is symmetric or skew-symmetric or neither of these two.

Answer:

If A is a skew-symmetric matrix, then AT=-A.AnT=ATn               For all nNAnT=-An         AT=-A AnT=-1n AnAnT=An, if n is even or -An, if n is odd.

Hence, An is symmetric when n is an even natural number.

Page No 5.63:

Question 29:

If A is a skew-symmetric matrix and n is an odd natural number, write whether An is symmetric or skew-symmetric or neither of the two.

Answer:

If A is a skew-symmetric matrix, then AT=-A.AnT=ATn               For all nNAnT=-An          AT=-AAnT=-1n AnAnT=An, if n is even or -An, if n is odd.

Hence, An is skew-symmetric when n is an odd natural number.

Page No 5.63:

Question 30:

If A and B are symmetric matrices of the same order, write whether ABBA is symmetric or skew-symmetric or neither of the two.

Answer:

Since A and B are symmetric matrices, AT=A and BT=B.

Here,
   AB-BAT=ABT-BATAB-BAT=BTAT-ATBT            ABT=BTATAB-BAT=BA-AB                    BT=B and AT=AAB-BAT=-AB-BATherefore, AB-BA is skew-symmetric.

Page No 5.63:

Question 31:

Write a square matrix which is both symmetric as well as skew-symmetric.

Answer:

Let A=0000 AT=0000Since AT=A, A is a symmmetric matrix.Now,  -A=-0000  -A=0000Since AT=-A,A is a skew-symmetric matrix.Thus, A=0000 is an example of a matrix that is both symmetric and skew-symmetric.    

Page No 5.63:

Question 32:

Find the values of x and y, if 2130x+y012=5618

Answer:

Given: 2130x+y012=56182602x+y012=56182+y6+00+12x+2=56182+y612x+2=5618 2+y=5  and 2x+2=8y=5-2 and 2x=8-2y=3 and 2x=6x=62=3

Page No 5.63:

Question 33:

If x+3      4y-4x+y=5439, find x and y

Answer:

The corresponding elements of two equal matrices are equal.

Given:x+34y-4x+y=5439x+3=5 and y-4=3x=5-3 and y=3+4x=2 and y=7 x=2 and y=7

Page No 5.63:

Question 34:

Find the value of x from the following:  2x-y53       y=6   53-2

Answer:

The corresponding elements of two equal matrices are equal.

Given: 2x-y53y=653-22x-y=6           ...1y=-2Putting the value of y in eq. 12x--2=62x+2=62x=6-22x=4x=42=2 x=2

Page No 5.63:

Question 35:

Find the value of y, ifx-y2x      5=2235

Answer:

Given: x-y2x5=2235The corresponding elements of two equal matrices are equal.x-y=2              ...1 and x=3Putting the value of x in eq. 13-y=23-2=y y=1

Page No 5.63:

Question 36:

Find the value of x, if3x+y-y2y-x   3=   12-53

Answer:

The corresponding elements of two equal matrices are equal.

Given:3x+y-y2y-x3=12-533x+y=1               ...1 -y=2y=-2Putting the value of y in eq. 13x+-2=13x-2=13x=1+23x=3x=33=1  x=1

Page No 5.63:

Question 37:

If matrix A = [1 2 3], write AAT.

Answer:

Given: A=123 AT=123AAT=123123AAT=1+4+9AAT=14

Page No 5.63:

Question 38:

If 2x+y3y0        4=6064, then find x.

Answer:

The corresponding elements of two equal matrices are equal.

Given:2x+y3y04=60642x+y=6             ...1 3y=0y=0Putting the value of y in eq. 12x+0=62x=6x=62 x=3

Page No 5.63:

Question 39:

If A=1234, find A + AT.

Answer:

Given: A=1234 AT=1324A+AT=1234 +1324A+AT=1+12+33+24+4A+AT=2558

Page No 5.63:

Question 40:

If a+b25      b=6522, then find a.

Answer:

The corresponding elements of two equal matrices are equal.

 a+b25b=6522a+b=6               ...1 b=2Putting the value of b in eq. 1a+2=6a=6-2 a=4

Page No 5.63:

Question 41:

If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, find the order of the matrix of AB.

Answer:

If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, then the order of matrix AB is given by the number of rows in A and number of columns in B, respectively.

Thus, the order of matrix AB is 3×3.

Page No 5.63:

Question 42:

If A=cos α-sin αsin α  cos α is identity matrix, then write the value of α.

Answer:

 Here,  A=cos α-sin αsin αcos α=Icos α-sin αsin αcos α=1001The corresponding elements of  equal matrices are equal. cos α=1        α=0°

Page No 5.63:

Question 43:

If 12343125=711k23, then write the value of k.

Answer:

Given: 12343125=711k233+41+109+83+20=711k237111723=711k23The corresponding elements of two equal matrices are equal. k=17

Page No 5.63:

Question 44:

If I is the identity matrix and A is a square matrix such that A2 = A, then what is the value of (I + A)2 = 3A?

Answer:

Given: A is a square matrix, such that A2=A.

Here,
              I+A2-3A=I+AI+A-3AI+A2-3A=I×I+I×A+A×I+A×A-3A           using distributive propertyI+A2-3A=I+A+A+A2-3A                               using I×I=I and  IA=AI=AI+A2-3A=I+2A+A-3A                                       A2=AI+