Using elementary transformations, find the inverse of each of the matrices.

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = [ (2) {2×2- 3×(-2)} – (-3) {2 × 2 - 3×3} + (3) {2× (-2) – 2×3}]

= [2 {4-(-6)} + 3 {4-9} + 3 { -4-6}]

= [2(10) + 3(-5) + 3(-10)]

= [20-15-30]

= -25

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -2R₁

→

Apply row operation- R₃→ R₃ - 3R₁

→.

Apply row operation- R₂→ R₂

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₃→ R₃ - R₂

→

Apply row operation- R₃→ R₃

→

Apply row operation- R₁→ R₁ R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→