Linear Algebra Toolkit

Calculating the inverse using row operations

v. 1.25

PROBLEM

Find (if possible) the inverse of the matrix

2 7	-6 7	3 7	1
6 7	3 7	2 7	2
-3 7	2 7	6 7	3
0	0	0	1

SOLUTION

Step 1: Adjoin the identity matrix to the given matrix

Adjoining I₄ to the given matrix, we obtain the 4x8 matrix:

2 7	-6 7	3 7	1	1	0	0	0
6 7	3 7	2 7	2	0	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

Step 2: Transform the matrix to the reduced row echelon form (Hide details)

Row
Operation
1:

2 7	-6 7	3 7	1	1	0	0	0
6 7	3 7	2 7	2	0	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

multiply the 1st row by 7/2

1	-3	3 2	7 2	7 2	0	0	0
6 7	3 7	2 7	2	0	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

Row
Operation
2:

1	-3	3 2	7 2	7 2	0	0	0
6 7	3 7	2 7	2	0	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

add -6/7 times the 1st row to the 2nd row

1	-3	3 2	7 2	7 2	0	0	0
0	3	-1	-1	-3	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

Row
Operation
3:

1	-3	3 2	7 2	7 2	0	0	0
0	3	-1	-1	-3	1	0	0
-3 7	2 7	6 7	3	0	0	1	0
0	0	0	1	0	0	0	1

add 3/7 times the 1st row to the 3rd row

1	-3	3 2	7 2	7 2	0	0	0
0	3	-1	-1	-3	1	0	0
0	-1	3 2	9 2	3 2	0	1	0
0	0	0	1	0	0	0	1

Row
Operation
4:

1	-3	3 2	7 2	7 2	0	0	0
0	3	-1	-1	-3	1	0	0
0	-1	3 2	9 2	3 2	0	1	0
0	0	0	1	0	0	0	1

multiply the 2nd row by 1/3

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	-1	3 2	9 2	3 2	0	1	0
0	0	0	1	0	0	0	1

Row
Operation
5:

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	-1	3 2	9 2	3 2	0	1	0
0	0	0	1	0	0	0	1

add 1 times the 2nd row to the 3rd row

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	7 6	25 6	1 2	1 3	1	0
0	0	0	1	0	0	0	1

Row
Operation
6:

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	7 6	25 6	1 2	1 3	1	0
0	0	0	1	0	0	0	1

multiply the 3rd row by 6/7

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	1	25 7	3 7	2 7	6 7	0
0	0	0	1	0	0	0	1

Row
Operation
7:

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	1	25 7	3 7	2 7	6 7	0
0	0	0	1	0	0	0	1

add -25/7 times the 4th row to the 3rd row

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Row
Operation
8:

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	-1 3	-1	1 3	0	0
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

add 1/3 times the 4th row to the 2nd row

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	0	-1	1 3	0	1 3
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Row
Operation
9:

1	-3	3 2	7 2	7 2	0	0	0
0	1	-1 3	0	-1	1 3	0	1 3
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

add -7/2 times the 4th row to the 1st row

1	-3	3 2	0	7 2	0	0	-7 2
0	1	-1 3	0	-1	1 3	0	1 3
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Row
Operation
10:

1	-3	3 2	0	7 2	0	0	-7 2
0	1	-1 3	0	-1	1 3	0	1 3
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

add 1/3 times the 3rd row to the 2nd row

1	-3	3 2	0	7 2	0	0	-7 2
0	1	0	0	-6 7	3 7	2 7	-6 7
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Row
Operation
11:

1	-3	3 2	0	7 2	0	0	-7 2
0	1	0	0	-6 7	3 7	2 7	-6 7
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

add -3/2 times the 3rd row to the 1st row

1	-3	0	0	20 7	-3 7	-9 7	13 7
0	1	0	0	-6 7	3 7	2 7	-6 7
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Row
Operation
12:

1	-3	0	0	20 7	-3 7	-9 7	13 7
0	1	0	0	-6 7	3 7	2 7	-6 7
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

add 3 times the 2nd row to the 1st row

1	0	0	0	2 7	6 7	-3 7	-5 7
0	1	0	0	-6 7	3 7	2 7	-6 7
0	0	1	0	3 7	2 7	6 7	-25 7
0	0	0	1	0	0	0	1

Step 3: Interpret the reduced row echelon form

The matrix in reduced row echelon form obtained above can be interpreted as [ C | D ], where both C and D are 4x4 matrices.

Since C =

1	0	0	0
0	1	0	0
0	0	1	0
0	0	0	1

equals I₄

then D =

2 7	6 7	-3 7	-5 7
-6 7	3 7	2 7	-6 7
3 7	2 7	6 7	-25 7
0	0	0	1

is the inverse of the original matrix.

Comments

Shortcut: To decide whether the matrix is invertible (without actually determining its inverse), stop the row operations as soon as the pattern of leading entries is established. If every column of the left half has a leading entry, the matrix is invertible; otherwise, it is singular.
Reference: Bogacki Linear Algebra: Concepts and Applications, p. 89 - Procedure for finding the inverse.

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