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# Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 1

• Last Updated : 28 Mar, 2021

### (i)

Solution:

Here, A =

Cofactors of A are:

C11 = 4      C12 = -2

C21 = -5    C22 = -3

|A|I =  =

Hence Proved

### (ii)

Solution:

Here, A =

Cofactors of A are:

C11 = d      C12 = -c

C21 = -b    C22 = a

|A|I =

Hence Proved

### (iii)

Solution:

Here, A =

Cofactors of A are:

C11 = cos Î±     C12 = -sin Î±

C21 = -sin Î±    C22 = cos Î±

=

=

=

|A|I =

=

=

=

=

=

Hence Proved

### (iv)

Solution:

Here, A =

Cofactors of A are:

C11 = 1    C12 = -(-tan Î±/2) = tan Î±/2

C21 = -tan Î±/2    C22 = 1

=

|A| =

= 1 + tan2 Î±/2

= sec2Î±/2

=

=

|A|I = (sec2Î±/2)

=

=

=

Hence Proved

### (i)

Solution:

Here, A =

Cofactors of A are

C11  = -3

C21  = 2

C31  = 2

C12  = 2

C22 =-3

C32  = 2

C13  = 2

C23  = 2

C33  = -3

=

=

|A| = -3 + 4 + 4 = 5

|A|I= (5) =

Hence Proved

### (ii)

Solution:

Here, A =

Cofactors of A are

C11  = 2

C12  = -3

C13  = 5

C21  = 3

C22  = 6

C23  = -3

C31  = -13

C32  = 9

C33  = -1

=

|A| = 1(3 – 1) – 2(2 + 1) + 5(2 + 3)

= 2 – 6 + 25 = 21

|A|I = (21)

Hence Proved

### (iii)

Solution:

Here, A =

Cofactors of A are

C11  = -22

C12 = – = 4

C13  = 16

C21 = – = 11

C22  = -2

C23 = – = -8

C31  = -11

C32 = – = 2

C33  = 8

|A| = 2(-2 – 20) + 1(-4 – 0) + 3(16 – 0)

= -44 – 4 + 48 = 0

|A|I =

Hence Proved

### (iv)

Solution:

Here, A =

Cofactors of the A are

C11  = 3

C12 = – = -15

C13  = 4

C21  = -1

C22  = 7

C23  = -2

C31  = 1

C32  = -5

C33 = = 2

=

|A| = 2(3 – 0) – 0(15 – 0) – 1(5 – 1)

= 6 – 4 = 2

|A|I = (2)

Hence Proved

### Question 3. For the matrix A =, show that A(adj A) = O.

Solution:

Cofactor of A are,

C11 = 30    C12 = -20    C13 = -50

C21 = 12    C22 = -8     C23 = -20

C31 = -3    C32 = 2       C33 = 5

=

=

= 0

Hence Proved

### Question 4. If A =, show that adj A = A.

Solution:

Here, A =

Cofactor of A are,

C11 = -4    C12 = 1     C13 = 4

C21 = -3    C22 = 0    C23 = 4

C31 = 4    C32 = 4     C33 = 3

=

### Question 5. If A = , show that adj A = 3AT.

Solution:

Here, A =

Cofactor of A are,

C11 = -3    C12 = -6    C13 = -6

C21 = 6    C22 = 3      C23 = -6

C31 = 6    C32 = -6    C33 = 3

=

AT=

Now, 3AT = 3 =

Hence Proved

### Question 6. Find A(adj A) for the matrix A =.

Solution:

Here, A =

Cofactor of A are,

C11 = 9    C12 = 4    C13 = 8

C21 = 19    C22 = 14    C23 = 3

C31 = -4    C32 = 1    C33 = 2

=

=

= 25

= 25I3

### (i)

Solution:

Here, A =

|A| = cos2Î¸ + sin2Î¸ = 1

Hence, inverse of A exist

Cofactors of A are,

Cofactor of A are,

C11 = cos Î¸     C12 = sin Î¸

C21 = -sin Î¸    C22 = cos Î¸

=1/1.

### (ii)

Solution:

Here, A =

|A| = -1

Hence, inverse of A exist

Cofactor of A are,

C11 = 0      C12 = -1

C21 = -1    C22 = 0

### (iii)

Solution:

Here, A =

|A| = a(1 + bc)/a – bc = 1 + bc – bc = 1

Hence, inverse of A exists.

Cofactor of A are,

C11 = (1 + bc)/a     C12 = -c

C21 = -b                 C22 = a

= 1/1

### (iv)

Solution:

Here, A =

|A| = 2 + 15 = 17

Hence, inverse of A exists.

Cofactor of A are,

C11 = 1      C12 = 3

C21 = -5   C22 = 2

### (i)

Solution:

Here, A =

|A| = 1(6 – 1) – 2(4 – 3) + 3(2 – 9)

= 5 – 2 – 21 = -18

Therefore, inverse of A exists

Cofactors of A are:

C11 = 5    C12 = -1      C13 = -7

C21 = -1    C22 = -7    C23 = 5

C31 = -7    C32 = 5     C33 = -1

Hence, A-1

=

### (ii)

Solution:

Here, A =

|A| = 1(1 + 3) – 2(-1 + 2) + 5(3 + 2)

= 4 – 2 – 25 = 27

Therefore, inverse of A exists

Cofactors of A are:

C11 = 4        C12 = -1     C13 = 5

C21 = -17    C22 = -11   C23 = 1

C31 = 3       C32 = 6      C33 = -3

Hence, A-1

### (iii)

Solution:

Here, A =

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

= 6 – 1 – 1 = 4

Therefore, inverse of A exists

Cofactors of A are:

C11 = 3    C12 = 1      C13 = -1

C21 = 1    C22 = 3     C23 = 1

C31 = -1    C32 = 1    C33 = 3

Hence, A-1

### (iv)

Solution:

Here, A =

|A| = 2(3 – 0) – 0 + 1(5)

= 6 – 5 = 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = 3     C12 = -15     C13 = 5

C21 = -1   C22 = 6        C23 = -2

C31 = 1     C32 = -5      C33 = 2

Hence, A-1

### (v)

Solution:

Here, A =

|A| = 0 – 1(16 – 12) – 1(-12 + 9)

= -4 + 3 = -1

Therefore, inverse of A exists

Cofactors of A are:

C11 = 0    C12 = -4    C13 = -3

C21 = -1   C22 = 3     C23 = 3

C31 = 1    C32 = -4    C33 = -4

Hence, A-1

### (vi)

Solution:

Here, A =

|A| = 0 – 0 – 1(-12 + 8)

= -1(-4) = 4

Therefore, inverse of A exists

Cofactors of A are:

C11 = -8    C12 = 11      C13 = -4

C21 = 4     C22 = -2     C23 = 0

C31 = 4    C32 = -3      C33 = 0

Hence, A-1

### (vii)

Solution:

Here, A =

|A| = -cos2Î± – sin2Î±

= -(cos2Î± + sin2Î±) = -1

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 0           C13 = -0

C21 = 0      C22 = -cosÎ±   C23 = -sinÎ±

C31 = 0      C32 = -sinÎ±     C33 = cosÎ±

Hence, A-1

Question 9. (i)

Solution:

Here, A =

|A| = 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3 = 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = 7       C12 = -1   C13 = -1

C21 = -3    C22 = 1     C23 = 0

C31 = -3    C32 = 0    C33 = 1

Hence, A-1 = 1/1

=

To verify A-1A =

### (ii)

Solution:

Here, A =

|A| = 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

= 2 – 3(3) + 1(9) = 2

Therefore, inverse of A exists

Cofactors of A are:

C11 = 1      C12 = -3    C13 = 9

C21 = 1     C22 = 1      C23 = -5

C31 = -1   C32 = 1      C33 = -1