The Inverse of a Matrix

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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.

How to Find the Inverse of a Square Matrix using a System of Linear Equations:

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:

Matrix A

2×2

C1

C2

R1:

1

4

R2:

−1

−3

Matrix X

2×2

C1

C2

R1:

X11

X12

R2:

X21

X22

Identity Matrix

2×2

C1

C2

R1:

1

0

R2:

0

1

Matrix A×X

=

Identity Matrix

1X11 + 4X21     1X12 + 4X22

−1X11 − 3X21   −1X12 − 3X22

1

0

0

1

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:

Matrix X (A-1)

2×2

C1

C2

R1:

−3

−4

R2:

1

1

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.

Using the Matrix Formula to Find the Inverse:

The matrix formula works for 2×2 matrices. The following shows how the formula works.

Matrix A

a

b

c

d

If matrix A is invertible then adbc ≠ 0. Should adbc ≠ 0 the inverse of matrix A is given by:

A−1 = 1 / (ad − bc) × A

The denominator adbc is the determinant of the 2×2 matrix.

Matrix A

3

−1

−2

2

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.

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