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 = [ (1) {0- 5 × (-5)} – (3) {(-3) × 0 – (-5) × 2} + (-2) {5 × (-3) – 2 × 0}]


= [1 {25} - 3 {0 + 10} - 2 {-15}]


= [1(25) - 3(10) – 2 (-15)]


= [25-30+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- R2 R2 + 3R1



Apply row operation- R3 R3 -2R1



Apply row operation- R2 R2



Apply row operation- R1 R1 – 3R2



Apply row operation- R3 R3 + R2



Apply row operation- R1 R1 - R3



Apply row operation- R2 R2 + R3



Apply row operation- R3 R3



The matrix so obtained is of the form –


[I : A-1]


Hence inverse of the given matrix-



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