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-
→