Class 12-commerce RD SHARMA Solutions Maths Chapter 7: Adjoint and Inverse of a Matrix
Adjoint and Inverse of a Matrix Exercise Ex. 7.1
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7(i)
Solution 7(ii)
Solution 7(iii)
Solution 7(iv)
Solution 8(i)
Solution 8(ii)
Solution 8(iii)
Solution 8(iv)
Solution 8(v)
Solution 8(vi)
Solution 8(vii)
Solution 9(i)
Solution 9(ii)
Solution 10(i)
Solution 10(ii)
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16(i)
Solution 16(ii)
Solution 16(iii)
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Adjoint and Inverse of a Matrix Exercise Ex. 7.2
Solution 1
Solution 2
Solution 3
Let 
AA-1 = I
Applying R2→ R2 - 2R1
Applying R2→ R2/-5
Applying R1→ R1 - 2R2
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Let 
AA-1 = I
Applying R2→ R2 - 2R1 and R3→ R3 + 2R1
Applying R1→ R1 - 2R2
Applying R1→ R1 - R3 and R2→ R2 - R3
Hence, 
Solution 18
Let 
AA-1 = I
Applying R3 ↔ R1
Applying R2→ R2 - 3R1 and R3→ R3 - 2R1
Applying R2→ -R2
Applying R1→ R1 - R2 and R3→ R3 + 5R2
Applying R3→ -R3
Applying R2→ R2 + 2R3
Hence, 
Adjoint and Inverse of a Matrix Exercise MCQ
Solution 1
Correct option: (a)
|A-1| = |A|-1, (AT)-1 = (A-1)T, |A| ≠ 0 are properties of an invertible matrix.
Solution 2
Correct option: (c)
Solution 3
Correct option:(d)
Solution 4
Correct option: (b)
Solution 5
Correct option:(b)
If A is singular matrix then adjoint of A is also singular.
Solution 6
Correct option: (a)
Solution 7
Correct option: (c)
Solution 8
Correct option:(a)
Solution 9
Correct option: (c)
Solution 10
Correct option: (c)
Solution 11
Correct option: (d)
Solution 12
Correct option: (d)
Solution 13
Correct option: (a)
Solution 14
Correct option: (b)
Solution 15
Correct option: (c)
Solution 16
Correct option: (d)
Solution 17
Correct option: (d)
Solution 18
Correct option: (b)
Solution 19
Correct option: (d)
Solution 20
Correct option: (c)
Solution 21
Correct option: (c)
Adj A = |A| A-1, Det(A-1) = (det A)-1, (AB)-1 = B-1A-1 are all the properties of invertible matrix.
Solution 22
Correct option: (b)
Solution 23
Correct option: (a)
Solution 24
Correct option: (b)
Solution 25
Solution 26
Correct option: (d)
Solution 27
Correct option: (b)
Solution 28
Correct option: (d)
Solution 29
Correct option: (b)
Solution 30
Correct option: (a)
Solution 31
Correct option: (a)
Solution 32
Given: A and B are invertible matrices
Relation between inverse and adjoint of a matrix is given by
We know that, AA-1 = I
As 
But 
Solution 33
A matrix is invertible or its inverse exists if determinant is 0.
Now, 
Therefore, A-1 exists if |A| = 0
i.e. if 
i.e. if 
Solution 34
Given: A2 = A
(I - A)3 + A = (I - A)2(I - A) + A
= (I2 - 2AI + A2)(I - A) + A
= (I- 2A + A)(I - A) + A … (Since A2 = A)
= (I - A)(I - A) + A
= I2 - 2AI + A2 + A
= I - 2A + A + A
= I
Hence, (I - A)3 + A = I.
Solution 35
The matrix is non-invertible if its determinant is 0.
This matrix will not be invertible if 
i.e. if 
Adjoint AND Inverse of a Matrix Exercise Ex. 7VSAQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Given: 
which is in
the form X = AB
Now, the column operation is applicable on X and B simultaneously when the equation is X = AB
Therefore, we'll apply the column operation on the first and third matrices.
Applying C2→ C2 + 2C1
Solution 30
Given: 
which is in
the form AB = X
Now, the row operation is applicable on A and X simultaneously when the equation is AB = X
Therefore, we'll apply the row operation on the first and third matrices.
Applying R2→ R2 + R1
Solution 31
Given: |A| = 4
