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-



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