ML Aggarwal Solution Class 10 Chapter 8 Matrices Exercise Test

 Test 

Question 1

Find the values of a and below

[a+3b2+206]=[2a+13b0b25b]

Sol :

[a+3b2+206]=[2a+13b0b25b]

comparing the corresponding elements

a + 3 = 2a + 1

⇒ 2a – a =3 – 1

⇒ a = 2

b² + 2 = 3b

⇒ b² – 3b + 2 = 0

⇒ b² – b – 2b + 2 = 0

⇒ b (b – 1) – 2 (b – 1) = 0

⇒ (b – 1) (b – 2) = 0.

Either b – 1 = 0, then b = 1

or b – 2 = 0, then b = 2

Hence a = 2, b = 2 or 1


Question 2

Find a, b, c and d if 3[abcd]=[4a+bc+d3]+[a612d]

Sol :

Given :

3[abcd]=[4a+bc+d3]+[a612d]

[3a3b3c3d]=[4+aa+b+6c+d13+2d]

Comparing the corresponding elements:

3a=4+a 

3aa=4

2a=4

a=2

3b=a+b+63bb=2+6

2b=8

b=4

3d=3+2d 

3d2d=3

d=3

3c=c+d-1 

3cc=31

2c=2 

c=1

Hence a=2, b=4, c=1, d=3


Question 3

Find X if Y=[3214] and 2X+Y=[1032]

Sol :

Given :

2X+Y=[1032]

2X=2X+Y=[1032]Y

2X=[1032][3214]

=[13023124]=[2242]

X=12[2242]=[1121]


Question 4

Determine the matrices A and B when

A+2B=[1263] and 2AB=[2121]
Sol :

A+2B=[1263]...(i)

2AB=[2121]...(ii)

Multiplying (i) by 1 and (ii) by 2

A+2B=[1263]

4A2B=2[2121]=[4242]

Adding we get 

5 A=[1263]+[4242]=[50105]

A=15[50105]=[1021]

From
(i) A+2B=[1263]

=[1021]+2 B=[1263]

2 B=[1263][1021]=[0242]

B=12[0242]=[0121]

Hence A=[1021] and B=[0121]

Question 5

(i) Find the matrix B if A = [4123] and A2=A+2B

(ii) If A=[1234],B=[0125]
and C=[2011] find A(4B3C)

Sol :

A=[4123]
let B=[abcd]

A2=A×A=[4123][4123]
=[16+24+38+62+9]
=[1871411]

A+2B=[4123]+2[abcd]

=[4123]+[2a2b2c2d]=[4+2a1+2b2+2c3+2d]

A2=A+2B
[1871411]=[4+2a1+2b2+2c3+2d]

Comparing the corresponding elements

4+2a=182a=184=14
a=7
1+2b=72b=71=6
b=3
2+2c=142c=142=12
c=6
3+2d=112d=113=8
d=4

Hence a=7, b=3, c=6, d=4
B=[7364]

(ii)
A=[1234],B=[0125],C=[2011]

4B3C=4[0125]3[2011]

=[04820][6033]

=[0(6)408(3)203]=[0+6408+3203]

=[64517]

Now A(4B3C)=[1234][64517]

=[1×6+2(5)1×4+2×173×6+4×(5)3×4+4×17]

=[6104+34182012+68]=[4383856]


Question 6

If A=[1410],B=[2131] and C=[2305] compute (AB)C = (CB)A ?

Sol :
Given :
A=[1410]
B=[2131]
C=[2305]

(AB) C=[[1410]×[2131]][2305]

=[2+12142+01+0][2305]

=[14321][2305]

=[28+042154+06+5]=[2827411]


(CB) A=[[2305][2131]][1410]

=[4+9230+1505][1410]

=[131155][1410]=[13152+015560+0]

=[12521060]

It is clear from above that

(AB)≠(CB)A


Question 7

If A=[3205] and B=[1012] find the each of the following and state it

they are equal:

(i) (A+B)(AB)

(ii) A2B2

Sol :

Given :

A=[3205]

B=[1012]

(i) (A+B)(A-B)

={[3205]+[1012]}×{[3205][1012]}

=[3+12+00+15+2]×[31200152]

=[4217]×[2213]

=[828+6272+21]=[614523]


(ii) A2B2

=[3205]×[3205][1012]×[1012]

=[9+06+100+00+25][1+00+01+20+4]

=[916025][1034]=[9116003254]

=[816321]

We see that (A+B)(AB)A2B2


Question 8

If A=[3542] find A25A141

Where I is unit matrix of order 2 x 2

Sol :

Given

A=[3542]

A2=A×A=[3542][3542]

=[9+20151012820+4]=[29252024]

5A=5[3542]=[15.252010]

A25 A14I=[29252024]

[15252010]14[1001]

=[29252024][15202010][140014]

=[29151425+25020+20+0241014]=[0000]


Question 9

If A=[33pq] and A2=0 find p and q

Sol :

Given :

A=[33pq]

A2=A×A=[33pq][33pq]

=[9+3p9+3q3p+pq3p+q2]

But A2=0

[9+3p9+3q3p+pq3p+q2]=[0000]

Comparing the corresponding elements

9+3p=0

3p=9p=3

9+3q=0 

3q=9q=3

Hence p=-3, q=-3


Question 10

If A=[3525xy] and A2=1, find x, y

Sol :

Given :

A=[3525xy]

A2=A×A=[3525xy][3525xy]

=[925+25x625+25y35x+xy25x+y2]

But A2=I=[1001]

[925+25x625+25y35x+xy25x+y2]=[1001]

Comparing the corresponding elements,

925+25x=1

25x=1925=1625

x=1625×52=85

625+25y=0

25y=625

y=625×52=35

Hence x=85,y=35


Question 11

If [1001][abcd]=[1001] find a, b, c and d

Sol :

[1001][abcd]=[1001]

[a+0b+00+c0+d]=[1001]

[abcd]=[1001]

Comparing the corresponding elements

a=1a=1

b=0b=0

c=0 and d=1

Hence a=-1, b=0, c=0, d=-1


Question 12

Find a and b if

[abb4b+4a2][2002]=[22140]

Sol :

Given :

[abb4b+4a2][2002]=[22140]

[2a2b+00+2a2b2b+8+00+2a4]=[22140]

[2a2b2a2b2b+82a4]=[22140]

Comparing the corresponding elements

$2 a-4=0

2a=4a=2

2 a-2 b=-2 

2×22b=2

42b=2

2b=24=6

b=3

Hence a=2, b=3


Question 13

If A=[sec60cos903tanA5sin90] and B=[0cos4523sin90]

Find (i) 2A – 3B (ii) A² (iii) BA

Sol :

Given

A=[sec60cos903tanA5sin90] and
B=[0cos4523sin90]

A=[sec60cos903tan45sin90]=[2031]

(sec60)=2,cos90=0,tan45=1sin90=1)

B=[0cot4523sin90]=[0123]

(cot45=1)

(i) 2A-3B

=2[2031]3[0123]

=[4062][0369]

=[40036+629]=[4307]


(ii) A2=A×A=[2031][2031]

=[4+00+0630+1]=[4091]


(iii) BA=[0123][2031]

=[030+1490+3]=[31133]

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