Matrix inverse by Gauss elimination

Write the augmented matrix [A | I] and reduce A to I by row operations; the right half then becomes A⁻¹. (Gauss–Jordan form of Gauss elimination.)

[A | I]
[      4       3      -1 |     1       0       0 ]
[      1       1       1 |     0       1       0 ]
[      3       5       3 |     0       0       1 ]

R1 → R1/4 (pivot 4 becomes 1)

[      1     3/4    -1/4 |   1/4       0       0 ]
[      1       1       1 |     0       1       0 ]
[      3       5       3 |     0       0       1 ]

R2 → R2 − 1·R1, R3 → R3 − 3·R1

[      1     3/4    -1/4 |   1/4       0       0 ]
[      0     1/4     5/4 |  -1/4       1       0 ]
[      0    11/4    15/4 |  -3/4       0       1 ]

R2 → 4·R2 (second pivot 1/4 becomes 1)

[      1     3/4    -1/4 |   1/4       0       0 ]
[      0       1       5 |    -1       4       0 ]
[      0    11/4    15/4 |  -3/4       0       1 ]

R1 → R1 − (3/4)·R2, R3 → R3 − (11/4)·R2

[      1       0      -4 |     1      -3       0 ]
[      0       1       5 |    -1       4       0 ]
[      0       0     -10 |     2     -11       1 ]

R3 → R3/(−10) (third pivot becomes 1)

[      1       0      -4 |     1      -3       0 ]
[      0       1       5 |    -1       4       0 ]
[      0       0       1 |  -1/5   11/10   -1/10 ]

R1 → R1 + 4·R3, R2 → R2 − 5·R3

[      1       0       0 |   1/5     7/5    -2/5 ]
[      0       1       0 |     0    -3/2     1/2 ]
[      0       0       1 |  -1/5   11/10   -1/10 ]

Hence

        1  [  2   14  −4 ]
A⁻¹ = ──── [  0  −15   5 ]  =  [  0.2   1.4  −0.4 ]
       10  [ −2   11  −1 ]     [  0.0  −1.5   0.5 ]
                               [ −0.2   1.1  −0.1 ]

Check: first row of A times first column of A⁻¹ = 4(0.2) + 3(0) + (−1)(−0.2) = 1, and row 1 × column 2 = 4(1.4) + 3(−1.5) + (−1)(1.1) = 0, as required (det A = −10).

Answer: A⁻¹ = (1/10)[2 14 −4; 0 −15 5; −2 11 −1]