Solution comments
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a
b
c
d
x
y
z
AC
i
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Number format
3131313131351515151515≈52188552198585858586
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10
How to find the inverse by Gaussian elimination with back-substitution
Augment the matrix with the identity to form [A|I]. Apply forward elimination to reduce A to upper-triangular form, propagating each row operation through the identity side. Then use back-substitution to clear the upper triangle. When A reaches the identity, the right side holds the inverse.
Gaussian elimination inverse — worked example (4×4)
Write the initial matrix
A
:
A
=
2
1
0
3
1
4
2
0
3
0
5
1
0
2
1
4
To find the inverse matrix of matrix
A
, we can add to the right of it the identity matrix of the same size;
After that, using the
Gaussian elimination
method, we transform the matrix so that the left part becomes a identity matrix, then in the right part we get the inverse matrix of the matrix
A
;
Write the expanded matrix (adding the identity matrix to the right of matrix
A
):
2
1
0
3
1
4
2
0
3
0
5
1
0
2
1
4
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
Gaussian Run Forwards
3
Iteration 1Divide
1
th row by
2
;
1
1
0
3
1
2
4
2
0
1
1
2
0
5
1
0
2
1
4
1
2
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
From
2
th row we subtract
1
th row;
From
4
th row we subtract
1
th row, multiplied by
3
;
1
0
0
0
1
2
3
1
2
2
-1
1
2
1
1
2
-1
1
2
5
-3
1
2
0
2
1
4
1
2
-
1
2
0
-1
1
2
0
1
0
0
0
0
1
0
0
0
0
1
4
Iteration 2Divide
2
th row by
3
1
2
;
1
0
0
0
1
2
1
2
-1
1
2
1
1
2
-
3
7
5
-3
1
2
0
4
7
1
4
1
2
-
1
7
0
-1
1
2
0
2
7
0
0
0
0
1
0
0
0
0
1
From
3
th row we subtract
2
th row, multiplied by
2
;
From
4
th row we subtract
2
th row, multiplied by
-1
1
2
;
1
0
0
0
1
2
1
0
0
1
1
2
-
3
7
5
6
7
-4
1
7
0
4
7
-
1
7
4
6
7
1
2
-
1
7
2
7
-1
5
7
0
2
7
-
4
7
3
7
0
0
1
0
0
0
0
1
5
Iteration 3Divide
3
th row by
5
6
7
;
1
0
0
0
1
2
1
0
0
1
1
2
-
3
7
1
-4
1
7
0
4
7
-
1
41
4
6
7
1
2
-
1
7
2
41
-1
5
7
0
2
7
-
4
41
3
7
0
0
7
41
0
0
0
0
1
From
4
th row we subtract
3
th row, multiplied by
-4
1
7
;
1
0
0
0
1
2
1
0
0
1
1
2
-
3
7
1
0
0
4
7
-
1
41
4
31
41
1
2
-
1
7
2
41
-1
21
41
0
2
7
-
4
41
1
41
0
0
7
41
29
41
0
0
0
1
6
Iteration 4Divide
4
th row by
4
31
41
;
1
0
0
0
1
2
1
0
0
1
1
2
-
3
7
1
0
0
4
7
-
1
41
1
1
2
-
1
7
2
41
-
62
195
0
2
7
-
4
41
1
195
0
0
7
41
29
195
0
0
0
41
195
7
Iteration 1From
3
th row we subtract
4
th row, multiplied by
-
1
41
;
From
2
th row we subtract
4
th row, multiplied by
4
7
;
1
0
0
0
1
2
1
0
0
1
1
2
-
3
7
1
0
0
0
0
1
1
2
53
1365
8
195
-
62
195
0
386
1365
-
19
195
1
195
0
-
116
1365
34
195
29
195
0
-
164
1365
1
195
41
195
8
Iteration 2From
2
th row we subtract
3
th row, multiplied by
-
3
7
;
From
1
th row we subtract
3
th row, multiplied by
1
1
2
;
1
0
0
0
1
2
1
0
0
0
0
1
0
0
0
0
1
57
130
11
195
8
195
-
62
195
19
130
47
195
-
19
195
1
195
-
17
65
-
2
195
34
195
29
195
-
1
130
-
23
195
1
195
41
195
9
Iteration 3From
1
th row we subtract
2
th row, multiplied by
1
2
;
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
16
39
11
195
8
195
-
62
195
1
39
47
195
-
19
195
1
195
-
10
39
-
2
195
34
195
29
195
2
39
-
23
195
1
195
41
195
Answer
B = A⁻¹16
39
11
195
8
195
-
62
195
1
39
47
195
-
19
195
1
195
-
10
39
-
2
195
34
195
29
195
2
39
-
23
195
1
195
41
195
SIZE4×4METHODGaussian elimination