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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.5

### Question 1.

Solution:

A =

A11 = 4; A12 = -3; A21 = -2; A22 = 1

adj A =

adj A =

### Question 2.

Solution:

A =

A11

A11 = 3 – 0 = 3

A12

A12 = -(2 + 10) = -12

A13

A13 = 0 + 6 = 6

A21

A21 = -(-1 – 0) = 1

A22

A22 = 1 + 4 = 5

A23

A23 = -(0 – 2) = 2

A31

A31 = -5 – 6 = -11

A32

A32 = -(5 – 4) = -1

A33

A33 = 3 + 2 = 5

adj A =

adj A =

### Question 3.

Solution:

|A| = -12 -(-12)

|A| = -12 + 12 = 0

so, |A|*I = 0 *

|A|*I = ……… (1)

Now, for adjoint of A

A11 = -6

A12 = 4

A21 = -3

A22 = 2

adj A =

adj A =

Now, A(adj A) =

A(adj A) = [Tex]\begin{bmatrix} 11& 0  &0 \\ 0 & 11 &0 \\ 0 & 0 & 11 \end{bmatrix}[/Tex]

A(adj A) = …………(2)

Now, (adj A)A =

(adj A)A =

(adj A)A =  …………….(3)

From eq(1), (2), and (3), you can see that A(adj A) = (adj A)A = |A|I

### Question 4.

Solution:

A =

|A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0)

|A| = 11

|A| * I =

|A| * I =

Now, for adjoint of A

A11 = 0

A12 = -(9 + 2) = -11

A13 = 0

A21 = -(-3 – 0) = 3

A22 = 3 – 2 = 1

A23 = -(0 + 1) = -1

A31 = 2 – 0 = 2

A32 = -(-2 – 6) = 8

A33 = 0 + 3 = 3

adj A =

Now,

A(adjA) =

A(adj A) =

A(adj A) =

Also,

(adj A).A =

(adj A).A =

(adj A).A =

From above, you can see,

A(adj A) = (adj A)A = I

### Question 5.

Solution:

|A| = 6 – (-8) = 14

|A| â‰  0, So inverse exists.

A11 = 3

A12 = 2

A21 = -4

A22 = 2

adj A =

A-1 = (adj A)/|A|

A-1

### Question 6.

Solution:

A =

|A| = -2 + 15 = 13 â‰  0

Hence, inverse exists.

A11 = 2

A12 = 3

A21 = -5

A22 = -1

adj A =

A-1 = (adj A)/|A|

A-1

### Question 7.

Solution:

A =

|A| = 1(10 – 0) – 2(0 – 0) + 3(0 – 0) = 10

For adj A

A11 = 10 – 0 = 0

A12 = -(0 – 0) = 0

A13 = 0 – 0 = 0

A21 = -(10 – 0) = -10

A22 = 5 – 0 = 5

A23 = -(0 – 0) = 0

A31 = 8 – 6 = 2

A32 = -(4 – 0) = -4

A33 = 2 – 0 = 2

adj A =

A-1 = (adj A)/|A|

A-1

### Question 8.

Solution:

A =

|A| = 1(-3 – 0) – 0 + 0 = -3 â‰  0

Hence, inverse exists.

For adj A

A11 = -3 – 0 = -3

A12 = -(-3 – 0) = 3

A13 = 6 – 15 = -9

A21 = -(0 – 0) = 0

A22 = -1 – 0 = -1

A23 = -(2 – 0) = -2

A31 = 0 – 0 = 0

A32 = -(0 – 0) = 0

A33 = 3 – 0 = 3

adj A =

A-1 = (adj A)/|A|

A-1

### Question 9.

Solution:

A =

|A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -3 â‰  0

Hence, inverse exists.

For adj A

A11 = -1 – 0 = -1

A12 = -(4 – 0) = -4

A13 = 8 – 7 = 1

A21 = -(1 – 6) = 5

A22 = 2 + 21 = 23

A23 = -(4 + 7) = -11

A31 = 0 + 3 = 3

A32 = -(0 – 12) = 12

A33 = -2 – 4 = -6

adj A =

A-1 = (adj A)/|A|

A-1

### Question 10.

Solution:

A =

|A| = 1(8 – 6) – 0 + 3(3 – 4) = -1

Now for adj A

A11 = 8 – 6 =2

A12 = -(0 + 9) = -9

A13 = 0 – 6 = -6

A21 = -(-4 + 4) =0

A22 = 4 – 6 = -2

A23 = -(-2 + 3) = -1

A31 = 3 – 4 = -1

A32 = -(-3 – 0) = 3

A33 = 2 – 0 = 2

adj A =

A-1 = (adj A)/|A|

A-1

A-1

### Question 11.

Solution:

A =

|A| = 1(-cos2Î± – sin2Î±) = -1

Now,

A11 = -cos2Î± – sin2Î± = -1

A12 = 0

A13 = 0

A21 = 0

A22 = -cosÎ±

A23 = -sinÎ±

A31 = 0

A32 = -sinÎ±

A33 = cosÎ±

adj A =

A-1 = (adj A)/|A|

A-1

A-1

### Question 12. Let A =  and B = , verify that (AB) – 1 = B – 1A – 1

Solution:

A =

|A| = 15 – 14 = 1

A11 = 5

A12 = -2

A21 = -7

A22 = 3

A-1 = (adj A)/|A|

A-1

B =

|B| = 54 – 56 = -2

adj B =

B-1 = (adj B)/|B|

B-1

B-1

Now,

B-1A-1

B-1A-1

B-1A-1

Now, AB =

AB =

AB =

|AB| = 67 * 61 – 87 * 47 = -2

adj (AB) =

(AB)-1 = (adj AB)/|AB|

(AB)-1

(AB)-1

From above, you can see that (AB)-1 = B-1A-1.

Hence, it is proved.

### Question 13. A = , show that A2 – 5A + 7I = O. Hence find A-1.

Solution:

A =

A2

A2

A2

So, A2 – 5A + 7I

– 5+ 7

–

= O

Hence, A2 – 5A + 7I = O

It can be written as

A.A – 5A = -7I

Multiplying by A-1 in both sides

A.A(A-1) – 5AA-1 = 7IA-1

A(AA-1) – 5I = -7A-1

AI – 5I = -7A-1

A-1 = -(A – 5I)/7

A-1 =1/7( – )

A-1

### Question 14. For the matrix A =, find the numbers a and b such that A2 + aA + bI = O.

Solution:

A =

A2

A2

A2 =

Now,

A2 – aA + bI = O

Multiplying by A-1 in both sides

(AA)A-1 + aAA-1 + bIA-1 = O

A(AA-1) + aI + b(IA-1) = O

AI + aI + bA-1 = O

A + aI = -bA-1

A-1 = -(A + aI)/b

Now,

A-1 = (adj A)/|A|

A-1

Now,

= -1/b

On comparing elements you will get

-1/b = -1

b = 1

(-3 – a)/b = 1

-3 – a = 1

a = -4

Hence, a = -4 and b = 1

### Question 15. A = , show that A3 – 6A2 + 5A + 11I = O. Hence find A-1

Solution:

A =

A2

A2

A2

A3 = A2.A

A3

A3

A3

A3 – 6A2 + 5A + 11I

– 6+ 5 + 11

–  +  +

= O

Hence, A3 – 6A2 + 5A + 11I = O

Now,

A3 – 6A2 + 5A + 11I = O

(AAA)A-1 – 6(AA)A-1 + 5(AA-1) + 11IA-1 = O

AA(AA-1) – 6A(AA-1) + 5(AA-1) = -11(IA-1)

A2 – 6A + 5I = -11 A-1

A-1 = -1/11(A2 – 6A + 5I) ………….(1)

Now, A2 – 6A + 5I

= – 6 + 5

= –  +

From eq(1) you have

A-1 = -1/11

### Question 16. A = , verify that A3 – 6A2 + 9A – 4I = O and hence fin A-1.

Solution:

A =

A2

A2

A2

A3 = A2.A

A3

A3

A3

Now,

A3 – 6A2 + 9A – 4I

– 6 + 9– 4

= –  + –

= –

= O

So, A3 – 6A2 + 9A – 4I = O

Now,

A3 – 6A2 + 9A – 4I = O

Multiplying by A-1 in both sides

(AAA)A-1 – 6(AA)A-1 + 9AA-1 – 4IA-1 = O

AA(AA-1) – 6A(AA-1) + 9(AA-1) = 4(IA-1)

AAI – 6AI +9I = 4A-1

A2 – 6A + 9I = 4A-1

A-1 = 1/4(A2 – 6A + 9I) ……….(1)

A2 – 6A + 9I

– 6 + 9

From eq(1), you have

A-1

### (A) |A|      (B) |A|2      (C) |A|3      (D) 3|A|

Solution:

You know,

(adj A)A = |A|I

(adj A)A =

|(adj A)A| =

|(adj A)A| = |A|3

|(adj A)A| = |A|3 I

|adj A| = |A|3

Hence, option B is correct.

### (A) det(A)      (B) 1/(det A)      (C) 1      (D) 0

Solution:

Since A is an invertible matrix then A-1 exists.

And A-1 = (adj A)/|A|

Suppose a 2 order matrix is A =

Then |A| = ad – bc

and adj A =

A-1 = (adj A)/|A|

A-1

A-1

|A-1| =

|A-1| =

|A-1| = (ad – bc)/|A|2

|A-1| = |A|/|A|2

|A-1| = 1/|A|

det(A-1) = 1/(det A)

Hence, option B is correct.