Matrix Inversion by Gauss-Jordan Elimination

GE 120 Lecture overheads
INVERSE OF A MATRIX:
The inverse of a matrix [A] is defined as that
matrix which, when multiplied with the
original matrix [A], produces a unit matrix.
The inverse of [A] is denoted as [A] -1 and is
only defined for a square matrix.
−1
−1
[ A][ A] = [ A] [ A] = [] I
Let us consider the following set of linear
equations.
2x 1 – 4x 2 + 5x 3 = 36 … … (1)
- 3x 1 + 5x 2 + 7x 3 = 7 … … (2)
5x 1 + 3x 2 – 8x 3 = - 31 … … (3)
The three equations are linearly independent,
and therefore, a unique solution exists. The
set of equations can be written in matrix form
as:
Overhead 1 of 16

GE 120 Lecture overheads
⎡
−
⎢⎢⎢
⎣
2
3
5
−
4
5
3
−
5
7
8
⎤
⎥⎥⎥
⎦
⎡ x
x
⎢⎢⎢
⎣x
1
2
3
⎤
⎥⎥⎥
⎦
=
⎡
⎢⎢⎢
⎣−
36⎤
7
31⎥⎥⎥
⎦
[A][x] = [B]
[x] = [A] -1 [B]
⎡ x
x
⎢⎢⎢
⎣x
1
2
3
⎤
⎥⎥⎥
⎦
=
⎡
−
⎢⎢⎢
⎣
2
3
5
− 4
5
3
5
7
− 8
⎤
⎥⎥⎥
⎦
−1
⎡ 36⎤
7
⎢⎢⎢
⎣−
31⎥⎥⎥
⎦
The division of two matrices is not defined in
linear algebra, however, matrix inversion can
be used much the same way division is used to
solve a matrix equation.
A matrix can be inverted if it is a nonsingular
matrix. Meaning:
D [ A] ≠ 0
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GE 120 Lecture overheads
INVERSION OF MATRICES:
By GAUSS-JORDAN ELIMINATION
[ A]
=
⎡
−
⎢⎢⎢
⎣
2
3
5
− 4
5
3
5
7
− 8
Gauss elimination suggests doing elementary
row operations from top to bottom. A
slightly modified form, known as Gauss-
Jordan elimination suggests doing elementary
row operations from bottom to top as well.
⎤
⎥⎥⎥
⎦
Why?
In order to transform the matrix [A] into a
unit matrix. Note down all the row
operations that would be necessary for this.
Purpose:
If we perform the same set of row operations
on a unit matrix, the unit matrix will be
transformed into the inverse of [A].
Overhead 3 of 16

GE 120 Lecture overheads
We can actually perform the row operations
on both [A] and [I] at the same time.
⎡
−
⎢⎢⎢
⎣
2
3
5
−
4
5
3
5
7
− 8
⎤
⎥⎥⎥
⎦
⎡1
0
⎢⎢⎢
⎣0
0
1
0
0⎤
0
1⎥⎥⎥
⎦
R′
2
R′
3
→
3
2
→ −
R
5
2
1
R
+
1
R
+
2
R
3
⎡2
0
⎢⎢⎢
⎣0
− 4
− 1
13
29
− 41
5
2
2
⎤
⎥⎥⎥
⎦
⎡
⎢⎢⎢
⎣−
3
5
1
2
2
0
1
0
0⎤
0
1⎥⎥⎥
⎦
Overhead 4 of 16

GE 120 Lecture overheads
Overhead 5 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
1
0
2
5
0
1
2
3
0
0
1
2
41
13
0
2
29
1
0
5
4
2
3
3
13
1 R
R →
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
13
1
0
26
5
0
1
2
3
0
0
1
26
41
1
0
2
29
1
0
5
4
2

GE 120 Lecture overheads
Overhead 6 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
13
1
0
26
5
0
1
2
3
0
0
1
26
41
1
0
2
29
1
0
5
4
2
3
2
3 R
R
R +
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
13
1
1
13
17
0
1
2
3
0
0
1
13
168
0
0
2
29
1
0
5
4
2

GE 120 Lecture overheads
Overhead 7 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
13
1
1
13
17
0
1
2
3
0
0
1
13
168
0
0
2
29
1
0
5
4
2
3
3
168
13 R
R →
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
168
1
168
13
168
17
0
1
2
3
0
0
1
1
0
0
2
29
1
0
5
4
2

GE 120 Lecture overheads
Overhead 8 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
168
1
168
13
168
17
0
1
2
3
0
0
1
1
0
0
2
29
1
0
5
4
2
3
2
2
3
1
1
2
29
5
R
R
R
R
R
R
−
→
′
−
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
168
1
168
13
168
17
336
29
336
41
336
11
168
5
168
65
168
83
1
0
0
0
1
0
0
4
2

GE 120 Lecture overheads
Overhead 9 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
168
1
168
13
168
17
336
29
336
41
336
11
168
5
168
65
168
83
1
0
0
0
1
0
0
4
2
2
1
1 4R
R
R
−
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
168
1
168
13
168
17
336
29
336
41
336
11
168
53
168
17
168
61
1
0
0
0
1
0
0
0
2

GE 120 Lecture overheads
Overhead 10 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
168
1
168
13
168
17
336
29
336
41
336
11
168
53
168
17
168
61
1
0
0
0
1
0
0
0
2
2
2
1
1
2
1
R
R
R
R
→ −
′
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
168
1
168
13
168
17
336
29
336
41
336
11
336
53
336
17
336
61
1
0
0
0
1
0
0
0
1
Therefore
[ ]
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
=
−
168
1
168
13
168
17
336
29
336
41
336
11
336
53
336
17
336
61
A 1

GE 120 Lecture overheads
[x] = [A] -1 [B]
⎡ x
x
⎢⎢⎢
⎣x
1
2
3
⎤
⎥⎥⎥
⎦
=
⎡
−
⎢⎢⎢
⎣
61
11
17
336
336
168
17
41
13
336
336
168
53
29
1
336⎤⎡
36 ⎤
336 7
168⎥⎥⎥
⎦⎢⎢⎢
⎣−
31⎥⎥⎥
⎦
x
x
1
1
=
=
{( 61 336) × 36 + ( 17 336) × 7 + ( 53 336) × ( − 31)
}
672 336 = 2
x
x
x
x
2
2
3
3
=
{( − 11 336) × 36 + ( 41 336) × 7 + ( 29 336) × ( − 31)
}
= − 1008 336 = −3
=
{( 17 168) × 36 + ( 13 168) × 7 + ( 1 168) × ( − 31)
}
= 672 168 = 4
Overhead 11 of 16

GE 120 Lecture overheads
Instead of carrying the matrix elements as
fractions during the row transformations we
may carry the elements as decimal numbers.
⎡
−
⎢⎢⎢
⎣
2
3
5
−
4
5
3
−
5
7
8
⎤
⎥⎥⎥
⎦
⎡1
0
⎢⎢⎢
⎣0
0
1
0
0⎤
0
1⎥⎥⎥
⎦
R′
2
R′
3
3
→ R1
+ R2
2
5
→ − R1
+ R
2
3
⎡2
0
⎢⎢⎢
⎣0
− 4
− 1
13
−
5
14.5
20.5
⎤
⎥⎥⎥
⎦
⎡ 1
1.5
⎢⎢⎢
⎣−
2.5
0
1
0
0⎤
0
1⎥⎥⎥
⎦
R ′ →
3
1 R
13
3
⎡2
0
⎢⎢⎢
⎣0
− 4
− 1
1
5
14.5
− 1.577
⎤
⎥⎥⎥
⎦
⎡
⎢⎢⎢
⎣−
1
1.5
0.192
0
1
0
0⎤
0
0.077⎥⎥⎥
⎦
Overhead 12 of 16

GE 120 Lecture overheads
Overhead 13 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
0.077
0
0.192
0
1
1.5
0
0
1
1.577
1
0
14.5
1
0
5
4
2
3
2
3 R
R
R +
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
0.077
1
1.308
0
1
1.5
0
0
1
12.923
0
0
14.5
1
0
5
4
2
3
3
12.923
1
R
R →
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
0.006
0.077
0.101
0
1
1.5
0
0
1
1
0
0
14.5
1
0
5
4
2

GE 120 Lecture overheads
Overhead 14 of 16
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
0.006
0.077
0.101
0
1
1.5
0
0
1
1
0
0
14.5
1
0
5
4
2
3
2
2
3
1
1
14.5
5
R
R
R
R
R
R
−
→
′
−
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
0.006
0.077
0.101
0.087
0.117
0.036
0.030
0.385
0.495
1
0
0
0
1
0
0
4
2
2
1
1 4R
R
R
−
→
′
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
0.006
0.077
0.101
0.087
0.117
0.036
0.318
0.083
0.351
1
0
0
0
1
0
0
0
2

GE 120 Lecture overheads
⎡2
0
⎢⎢⎢
⎣0
0
− 1
0
0
0
1
⎤
⎥⎥⎥
⎦
⎡0.351
0.036
⎢⎢⎢
⎣0.101
−
0.083
0.117
0.077
−
0.318⎤
0.087
0.006⎥⎥⎥
⎦
R′
1
R′
2
1
→ R
2
→ −R
1
2
⎡1
0
⎢⎢⎢
⎣0
0
1
0
0
0
1
⎤
⎥⎥⎥
⎦
⎡
−
⎢⎢⎢
⎣
0.176
0.036
0.101
0.042
0.117
0.077
0.159⎤
0.087
0.006⎥⎥⎥
⎦
[ ]
−
A 1
⎡
= −
⎢⎢⎢
⎣
0.176
0.036
0.101
0.042
0.117
0.077
0.159⎤
0.087
0.006⎥⎥⎥
⎦
[x] = [A] -1 [B]
Overhead 15 of 16

GE 120 Lecture overheads
⎡ x
x
⎢⎢⎢
⎣x
1
2
3
⎤
⎥⎥⎥
⎦
⎡
= −
⎢⎢⎢
⎣
0.176
0.036
0.101
0.042
0.117
0.077
0.159⎤⎡
36 ⎤
0.087 7
0.006⎥⎥⎥
⎦⎢⎢⎢
⎣−
31⎥⎥⎥
⎦
x1 = 0.176 × 36 + 0.042 × 7 + 0.159 × ( −31)
=
1.701
x2 = −0.036
× 36 + 0.117 × 7 + 0.087 × ( −31)
= −3.174
x3 = 0.101 × 36 + 0.077 × 7 + 0.006 × ( −31)
=
3.989
Compare the two solutions:
Elements as fractions
x 1 = 2
x 2 = -3
x 3 = 4
Elements as decimals
x 1 = 1.701
x 2 = -3.174
x 3 = 3.989
Why are the two solutions different?
Overhead 16 of 16