find the invers matrix

[1 2 1

2 -1 -1

3 1 1]

if given equation is  AX = B form then for finding X

we write it like  [ A : B] = X

so for finding inverse

AA^-1 = I

[A : I ] = A^-1

I think u may know about row reduction

Applying the method of Gauss-Jordon
elimination, we get inverse of matrix

?? : ??₃ =

apply row reduction

R₂ - 2R₁ , R₃ - 3R₁ ,

- R₂ / 5

R₁ - 2R₂

R₃ + 5R₂

R₁ + R₃/5

R₂ - 3R₃/5

This is combination of two 3X3 matrices first matrix is Identity matrix  and second matrix is A^-1 matrix

for finding A^-1 we always reduce the first matrix into indentity matrix and second is A^-1

So A^-1 =

like this you can find A^-1

Another process is there to find A^-1 is

A^-1 = adjoint(A) / det( A)

above process is very easy than this

ok thank u , if u like support my answer

invers matrix is that [1 2 3

                                     2 -1 1

                                       1 -1 1]