Question 317501
{{{A = (matrix(3,3,2, 1,  0,
                   0, 2,  1,
                   1, 0, -1))}}}
<pre><b>
Augment the matrix with the identity matrix:

{{{A = (matrix(3,7,2, 1,  0, "|", 1, 0, 0, 
                   0, 2,  1, "|", 0, 1, 0,
                   1, 0, -1, "|", 0, 0, 1))}}}

Use row operations to get the identity on the left:

R1-2R3->R3

{{{A = (matrix(3,7,2, 1,  0, "|", 1, 0, 0, 
                   0, 2,  1, "|", 0, 1, 0,
                   0, 1,  2, "|", 1, 0, -2))}}}

-1R3+R1->R1

{{{A = (matrix(3,7,2, 0, -2, "|", 0, 0, 2, 
                   0, 2,  1, "|", 0, 1, 0,
                   0, 1,  2, "|", 1, 0, -2))}}}

-R2+2R3->R3

{{{A = (matrix(3,7,2, 0, -2, "|", 0,  0, 2, 
                   0, 2,  1, "|", 0,  1, 0,
                   0, 0,  3, "|", 2, -1, -4))}}}

2*R3+3*R1->R1

{{{A = (matrix(3,7,6, 0,  0, "|", 4, -2, -2, 
                   0, 2,  1, "|", 0,  1,  0,
                   0, 0,  3, "|", 2, -1, -4))}}}

R3-3R2->R2

{{{A = (matrix(3,7,6,  0,  0, "|", 4, -2, -2, 
                   0, -6,  0, "|", 2, -4, -4,
                   0,  0,  3, "|", 2, -1, -4))}}}

{{{1/6}}}R1->R1
{{{-1/6}}}R2->R2
{{{1/3}}}R3->R3

{{{A = (matrix(3,7,1,  0,  0, "|", 2/3, -1/3, -1/3, 
                   0,  1,  0, "|", -1/3, 2/3, 2/3,
                   0,  0,  1, "|", 2/3, -1/3, -4/3))}}}

We end up with the identity matrix augmented with the inverse.

That is, {{{A^(-1)}}} is the 3x3 matrix on the right:

{{{A^(-1) = (matrix(3,3,2/3, -1/3, -1/3, 
                  -1/3, 2/3, 2/3,
                   2/3, -1/3, -4/3))}}}