Question 1198415
Find the inverse matrix of :

{{{matrix(3,3,
1, 0 ,-1,
1, 1 ,3,
3, -2, 1)}}}


Determinant is not zero, therefore inverse matrix exists.


Write the augmented matrix:
{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
1,	1,	3,	0,	1,	0,
3,	-2,	1,	0,	0,	1)}}}


Find the pivot in the 1st column in the 1st row



{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
1,	1,	3,	0,	1,	0,
3,	-2,	1,	0,	0,	1)}}}



Subtract the 1st row from the 2nd row


{{{matrix(3,6,
1,0,-1,	1,0,0,
0,1,4,-1,1,0,
3,-2,1,	0,0,1)}}}



Multiply the 1st row by 3


{{{matrix(3,6,

3,	0,	-3,	3,	0,	0,
0,	1,	4,	-1,	1,	0,
3,	-2,	1,	0,	0,	1)}}}


Subtract the 1st row from the 3rd row and restore it


{{{matrix(3,6,

1,	0,	-1,	1,	0,	0,
0,	1,	4,	-1,	1,	0,
0,	-2,	4,	-3,	0,	1)}}}



Find the pivot in the 2nd column in the 2nd row


{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
0,	1,	4,	-1,	1,	0,
0,	-2,	4,	-3,	0,	1)}}}



Multiply the 2nd row by -2

{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
0,	-2,	-8,	2,	-2,	0,
0,	-2,	4,	-3,	0,	1)}}}


Subtract the 2nd row from the 3rd row and restore it


{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
0,	1,	4,	-1,	1,	0,
0,	0,	12,	-5,	2,	1)}}}


Make the pivot in the 3rd column by dividing the 3rd row by 12


{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
0,	1,	4,	-1,	1,	0,
0,	0,	1,	-5/12,	1/6,	1/12)}}}



Multiply the 3rd row by -1

{{{matrix(3,6,
1,	0,	-1,	1,	0,	0,
0,	1,	4,	-1,	1,	0,
0,	0,	-1,	5/12,	-1/6,	-1/12)}}}


Subtract the 3rd row from the 1st row and restore it


{{{matrix(3,6,
1,	0,	0,	7/12,	1/6,	1/12,
0,	1,	4,	-1,	1,	0,
0,	0,	1,	-5/12,	1/6,	1/12)}}}


Multiply the 3rd row by 4


{{{matrix(3,6,
1,	0,	0,	7/12,	1/6,	1/12,
0,	1,	4,	-1,	1,	0,
0,	0,	4,	-5/3,	2/3,	1/3)}}}


Subtract the 3rd row from the 2nd row and restore it


{{{matrix(3,6,

1,	0,	0,	7/12,	1/6,	1/12,
0,	1,	0,	2/3,	1/3,	-1/3,
0,	0,	1,	-5/12,	1/6,	1/12)}}}



There is the inverse matrix on the right


{{{matrix(3,3,

7/12,	1/6,	1/12,
2/3,	1/3,	-1/3,
-5/12,	1/6,	1/12)}}}