# find the invers matrix[1  2  1 2 -1   -1 3  1   1]

mahesh dasari
31 Points
11 years ago

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 =

 1 2 1 1 0 0 2 -1 -1 0 1 0 3 1 1 0 0 1

apply row reduction

R2 - 2R1 , R3 - 3R1 ,

 1 2 1 1 0 0 0 -5 -3 -2 1 0 0 -5 -2 -3 0 1

- R2 / 5

 1 2 1 1 0 0 0 1 3/5 2/5 -1/5 0 0 -5 -2 -3 0 1

R1 - 2R2

R3 + 5R2

 1 0 -1/5 1/5 2/5 0 0 1 3/5 2/5 -1/5 0 0 0 1 -1 -1 1

R1 + R3/5

R2 - 3R3/5

 1 0 0 0 1/5 1/5 0 1 0 1 2/5 -3/5 0 0 1 -1 -1 1

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 =

 0 1/5 1/5 1 2/5 -3/5 -1 -1 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

srinu kummari
32 Points
11 years ago

invers matrix is that [1 2 3

2 -1 1

1 -1 1]