**Question 1.**

**Question 2.**

By R

_{1}– R

_{2}, we get,

By R

_{1}– R

_{3}and By R

_{2}– R

_{3}, we get

**Question 3.Check whether the following matrices are invertible or not:**

∴ A is a non-singular matrix.

Hence, A^{-1} exist.

= sec^{2}θ – tan^{2}θ = 1 ≠ 0.

∴ A is a non-singular matrix.

Hence, A^{-1} exist.

= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)

= 15 – 20 + 9 = 4 ≠ 0

∴ A is a non-singular matrix.

Hence, A^{-1} exist.

= 1 (-3 -6) – 2 (6 – 3) + 3 (4 + 1)

= -9 – 6 + 15 = 0

∴ A is a singular matrix.

Hence, A^{-1} does not exist.

= 1(32 – 30) – 2(24 – 20) + 3(18 – 16)

= 2 – 8 + 6 = 0

∴ A is a singular matrix.

Hence, A^{-1} does not exist.

**Question 4.Solution:**

∴ A is a non-singular matrix.

Hence, (AB)

^{-1}exist.

**Question 5.Solution:**

Since A is a non-singular matrix, then find A

^{-1}by using elementary row transformations.

We write AA

^{-1}= I

**Question 6.Solution:**

We will reduce the matrix A to the identity matrix by using row transformations. During this pro¬cess, I will be converted to the matrix X.

We have AX = I.

Question 7.

Find the inverse of each of the following matrices (if they exist).

∴ A^{-1} exists.

Consider AA^{-1} = I

∴ A^{-1} exists.

Consider AA^{-1} = I

∴ A^{-1} exists.

Consider AA^{-1} = I

∴ A^{-1} exists.

Consider AA^{-1} = I

∴ A^{-1} exists.

Consider AA^{-1} = I

∴ A^{-1} exists.

Consider AA^{-1} = I

= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)

= 20 – 15 – 30 = -25 ≠ 0

∴ A^{-1} exists.

Consider AA^{-1} = I

= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)

= 25 + 30 -30

= 25 ≠ 0

∴ A^{-1} exists.

Consider AA^{-1} = I

= 2(3 – 0) – 0 – 1(5 – 0)

= 6 – 0 – 5 = 1 ≠ 0

∴ A^{-1} exists.

Consider AA^{-1} = I

|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)

= -3 – 2 + 4

= -1 ≠ 0

∴ A^{-1} exists.

We have

AA^{-1} = I

**Question 8.**

= cosθ (cosθ – 0) + sinθ (sinθ – 0) + 0

= cos^{2}θ + sin^{2}θ = 1 ≠ 0

∴ A^{-1} exists.

(i) Consider AA^{-1} = I

**(ii) elementary column transformationsSolution:**

Consider A

^{-1}A = I

**Question 9.**

From (1) and (2), (AB)

^{-1}= B

^{-1}∙ A

^{-1}.

**Question 10.**

∴ A^{-1} exists.

Consider AA^{-1} = I

**Question 11.Solution:**

AX = B

**Question 12.Solution:**

AX = B

**Question 13.**

First we perform the row transformations.

**Question 14.**

= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)

= -13 + 6 + 6 = -1 ≠ 0

∴ A

^{-1}exists.

First we have to find the cofactor matrix

**Question 15.**

|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)

|A| = -4 – 2

|A| = -6 ≠ 0

∴ A

^{-1}exists.

First we have to find the cofactor matrix

= [A

_{ij}]3×3, where A

_{ij}= (-1)

^{i+j}M

_{ij}

**Question 16.**

= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)

= 0 + 12 – 9 = 3 ≠ 0

∴ A

^{-1}exists.

A

^{-1}by adjoint method :

We have to find the cofactor matrix

= [A

_{ij}]

_{3×3}, where A

_{ij}= (-1)

^{i+j }M

_{ij}

**Question 17.**

= 1 (2 – 6) – 0 + 1 (0 – 2)

= -4 – 2= -6 ≠ 0

∴ A

^{-1}exists.

Consider A

^{-1}A = I

By C

_{3}– C

_{1}, we get,

**Question 18.**

= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)

= -13 + 6 + 6 = -1 ≠ 0

∴ A^{-1} exists.

Consider AA^{-1} = I

**Question 19.Show with usual notations that for any matrix A = [a _{ij}]_{3×3}(i) a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0Solution:**

= -(a

_{12}a

_{33}– a

_{13}a

_{32})

= -a

_{12}a

_{33}+ a

_{13}a

_{32}

= a

_{11}a

_{33}– a

_{13}a

_{31}

= -(a

_{11}a

_{32}– a

_{12}a

_{31})

= -a

_{11}a

_{32}+ a

_{12}a

_{31}

∴ a

_{11}A

_{21}+ a

_{12}A

_{22}+ a

_{13}A

_{23}

= a

_{11}(-a

_{1233}+ a

_{13}a

_{32}) + a

_{12}(a

_{11}a

_{33}– a

_{13}a

_{31}) + a

_{13}(-a

_{11}a

_{32}+ a

_{12}a

_{31})

= -a

_{11}a

_{12}a

_{33}+ a

_{11}a

_{13}a

_{32}+ a

_{11}a

_{12}a

_{33}– a

_{12}a

_{13}a

_{31}– a

_{11}a

_{13}a

_{32}+ a

_{12}a

_{13}a

_{31}

= 0

(ii) a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = |A|

Solution:

**Question 20.Solution:**

Consider XA = B