# Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 3

### Question 25. Show that the matrix A =  satisfies the equation A3 – A2 – 3A – I3 = 0. Hence, find A-1.

Solution:

Here, A =

A2

A3

Now A3 – A2 – 3A – I3

=

So, A3 – A2 – 3A – I3 = 0

â‡’ A-1(AAA) – A-1(AA) – 3A-1A – A-1I = 0

â‡’ A2 – A – 3I – A-1 = 0

â‡’ A-1 = A2 – A – 3I

Therefore, A-1 =

### Question 26. If A = . Verify that A3 – 6A2 + 9A – 4I = O and hence find A-1.

Solution:

Here, A =

A2

=

A3 = A2A =

Now, A3 – 6A2 + 9A – 4I

So, A3 – 6A2 + 9A – 4I = O

Multiplying both side by A-1

â‡’ A-1(AAA) – 6A-1(AA) 9A-1A – 4A-1I = O

â‡’ AAI – 6AI + 9I = 4A-1

â‡’ 4A-1 = A2I – 6AI + 9I

=

=

=

â‡’ A-1

### Question 27. If A =, prove that A-1 = AT.

Solution:

Here, A =

AT

Now, Finding A-1

|A| = 1/9[-8(16 + 56) – 1(16 – 7) + 4(-32 – 4)]

= -81

Therefore, inverse of A exists

Cofactors of A are:

C11 = 72      C12 = -9       C13 = -36

C21 = -36    C22 = -36    C23 = -63

C31 = -9      C32 = 72      C33 = -36

=

=

Hence, A-1

= AT

Hence Proved

### Question 28. If A = , show that A-1 = A3.

Solution:

Here, A =

|A| = 3(-3 + 4) + 3(2 – 0) + 4(-2 – 0)

= 3 + 6 – 8

= 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = 1      C12 = -2     C13 = -2

C21 = -1    C22 = 3      C23 = 3

C31 = 0     C32 = -4     C33 = -3

Now, A2

=

A3

=

= A-1

Hence Proved

### Question 29. If A =, show that A2 = A-1.

Solution:

Here, A =

LHS = A2

|A| = -1(1 – 0) – 2(-1 – 0) + 0

= 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 0     C13 = -1

C21 = 0      C22 = 0     C23 = 1

C31 = 21    C32 = 1      C33 = 1

=

Hence, A-1

=

= A

Hence Proved

### Question 30. Solve the matrix equation, where X is a 2Ã—2 matrix.

Solution:

We have,

Let A =  and B =

So. AX = B

â‡’ X = A-1B

Now, |A| = 5 – 4 = 1

Co factors of A are:

C11 = 1         C12 = -1

C21 = -4      C22 = 5

=

A-1

Therefore, X =

X =

### Question 31. Find the matrix X satisfying the matrix equation: X.

Solution:

We have,

Let A =and B =

So, XA = B

XAA-1 = BA-1

XI = BA-1 ………..(i)

Now, |A| = -7

Co factors of A are:

C11 = -2       C12 = 1

C21 = -3      C22 = 5

=

=

Therefore, X =

=

X =

### Question 32. Find the matrix X for which: X

Solution:

Let, A =

B =

C =

Then the given equation becomes

A Ã— B = C

â‡’ X = A-1CB-1

Now |A| = 35 -14 = 21

|B| = -1 + 2 = 1

X = A-1 CB-1

### Question 33. Find the matrix X satisfying the equation:

Solution:

Let A =  B =

AXB = I

X = A-1B-1

|A| = 6 – 5 = 1

|B| = 10 – 9 = 1

A-1 = adj A /|A| =

X =

### Question 34. If A = , Find A-1 and prove that A2 – 4A – 5I = O.

Solution:

Here, A =

A2

Now, A2 + 4A – 5I

Now, A2 – 4A – 5I = O

â‡’ A-1AA – 4A-1A – 5A-1I = O

â‡’ 5A-1 = [A – 4I]

A-1

### Question 35. If A is a square matrix of order n, prove that |A adj A| = |A|n.

Solution:

Given, |A adj A| = |A|n

Taking LHS = |A Adj A|

= |A| |A|n-1

= |A|n-1+1

= |A|n = RHS

Hence Proved

### Question 36. If A-1 =  and B = , find (AB)-1.

Solution:

Here, B =

|B| = 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = 3       C12 = 1      C13 = 2

C21 = 2      C22 = 1      C23 = 2

C31 = 6     C32 = 2      C33 = 5

Therefore,

B-1

Hence, (AB)-1 = B-1A-1

### Question 37. If A = , find (AT)-1.

Solution:

Assuming B = AT

|B| = 1(-1 – 8) – 0 – 2(-8 + 3)

= -9 + 10 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = -9      C12 = 8      C13 = -5

C21 = -8     C22 = 7      C23 = -4

C31 = -2     C32 = 2      C33 = -1

=

B-1

or (AT)-1

### Question 38. Find the adjoint of the matrix A = and hence show that A (adj A) = |A|I3.

Solution:

Here, A =

|A| = -1(1 – 4) – 2(2 + 4) – 2(-4 – 2)

= 3 + 12 + 12 = 27

Therefore, inverse of A exists

Cofactors of A are:

C11 = -3     C12 = -6      C13 = -6

C21 = 6      C22 = 3       C23 = -6

C31 = 6      C32 = -6      C33 = 3

or A (adj A) = 27

Hence, A (adj A) = |A|I

Hence Proved

### Question 39. If A = , A-1 and show that A-1 = 1/2(A2 – 3I).

Solution:

Here, A =

|A| = 0 – 1(0 – 1) + 1(1 – 0)

= 1 + 1 = 2

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 1      C13 = 1

C21 = 1      C22 = -1    C23 = 1

C31 = 1      C32 = 1      C33 = -1

Hence, A-1

Now, A2 – 3I =

=

=

Hence, A-1 = 1/2(A2 – 3I)

Hence Proved

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