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.

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