A square matrix that has an inverse is invertible (non-singular). Not every square matrix has an inverse.
Non-square matrices do not have an inverse and are singular. To show this we examine two matrices; A and B. If A is of order m × n and B is of order n × m where m ≠ n, the products of AB and BA cannot be equal, and by matrix multiplication definition AB and BA cannot be multiplied.
The inverse of a square matrix is easiest to understand if we begin with the equation ax = b where a ≠ 0. To solve this equation for x we multiply both sides of the equation by a-1:
ax = b
a-1 ax = a-1 b
x = a-1 b
a-1 is the multiplicative inverse of a because a-1a = 1. This is similar to the definition of the multiplicative inverse of a matrix:
If we let A be an n × n matrix and let In be the n × n identity matrix then,
AA-1 = In = A-1 A
This identifies A-1 as the multiplicative inverse of A, the A inverse.
By applying matrix multiplication to a square matrix of which we want to find the inverse and using the matrix equation AX = I to solve for X, when operations have been completed the square matrix X is the inverse matrix A-1, X = A-1, and we will have solved AA-1 = In. We show how to find the inverse of a 2 × 2 matrix; however this scheme applies to any square matrix:
2×2 | C1 | C2 |
R1: | 1 | 4 |
R2: | -1 | -3 |
Matrix A
2×2 | C1 | C2 |
R1: | X11 | X12 |
R2: | X21 | X22 |
Matrix X
2×2 | C1 | C2 |
R1: | 1 | 0 |
R2: | 0 | 1 |
Identity Matrix
1 | 0 |
0 | 1 |
Identity Matrix
Matrix A × X
1X11 + 4X21 1X12 + 4X22
-1X11 – 3X21 -1X12 – 3X22
=
Next, equate corresponding entries to obtain two systems of linear equations…
Linear Equations System 1
X11 + 4X21 = 1
-X11 – 3X21 = 0
Linear Equations System 2
X12 + 4X22 = 0
-X12 – 3X22 = 1
From the first system we determine that X11 = -3 and X21 = 1.
From the second system we determine -X12 = -4 and X22 = 1.
We now write the inverse of A as:
It is recommended that you also understand the Gauss-Jordan elimination method, especially if you are working with 3 × 3 matrices or larger. By solving both systems of linear equations simultaneously it is more efficient than solving for the inverse of a matrix.
2×2 | C1 | C2 |
R1: | -3 | -4 |
R2: | 1 | 1 |
Matrix X (A-1)
The matrix formula works for 2 × 2 matrices. The following shows how the formula works.
a | b |
c | d |
Matrix A
If matrix A is invertible then ad – bc ≠ 0. Should ad – bc ≠ 0 the inverse of matrix A is given by:
A-1 = 1 / (ad – bc) × A
The denominator ad – bc is the determinant of the 2 × 2 matrix.
3 | -1 |
-2 | 2 |
Matrix A
ad – bc = (3)(2) – (-2)(-1) = 4
And A-1 = 1 / (ad – bc) × A = ¼ A
The inverse is a scalar multiplication of ¼ by the array elements of matrix A.
Copyright © DigitMath.com
All Rights Reserved.