# Class 12-science 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

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### Solution 13

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### Solution 15

### Solution 16(i)

### Solution 16(ii)

### Solution 16(iii)

### Solution 17

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### Solution 39

## Adjoint and Inverse of a Matrix Exercise Ex. 7.2

### Solution 1

### Solution 2

### Solution 3

Let

AA^{-1} =
I

Applying R_{2}→ R_{2}
- 2R_{1}

Applying R_{2}→ R_{2}/-5

Applying R_{1}→ R_{1}
- 2R_{2}

### Solution 4

### Solution 5

### Solution 6

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### Solution 13

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### Solution 17

Let

AA^{-1} =
I

Applying R_{2}→ R_{2}
- 2R_{1} and R_{3}→ R_{3} + 2R_{1}

Applying R_{1}→ R_{1} -
2R_{2}

Applying R_{1}→ R_{1}
- R_{3} and R_{2}→ R_{2} - R_{3}

Hence,

### Solution 18

Let

AA^{-1} =
I

Applying R_{3 }↔ R_{1}

Applying R_{2}→ R_{2}
- 3R_{1} and R_{3}→ R_{3} - 2R_{1}

Applying R_{2}→ -R_{2}

Applying R_{1}→ R_{1}
- R_{2} and R_{3}→ R_{3} + 5R_{2}

Applying R_{3}→ -R_{3}

Applying R_{2}→ R_{2}
+ 2R_{3}

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^{-1}A^{-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: A^{2} = A

(I - A)^{3} + A = (I - A)^{2}(I - A) + A

= (I^{2} - 2AI + A^{2})(I - A) + A

= (I- 2A + A)(I - A) + A … (Since A^{2} = A)

= (I - A)(I - A) + A

= I^{2} - 2AI + A^{2} + 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

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### 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 C_{2}→ C_{2}
+ 2C_{1}

### 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 R_{2}→ R_{2}
+ R_{1}

### Solution 31

Given: |A| = 4