# Definition of an Inverse of a Matrix

Assuming that we have a square matrix *A*, which is non-singular (i.e. det (*A*) does not equal zero), then there exists an *n × n* matrix *A*^{-1} which is called the inverse of *A *such that:

*AA*^{-1} = A^{-1}A = I, where *I* is the identity matrix.

**The inverse of a 2×2 matrix**

Take for example an arbitrary 2×2 Matrix *A* whose determinant (ad − bc) is not equal to zero.

where *a*, *b*, *c* and *d* are numbers.

The inverse is:

The inverse of a general *n × n* matrix *A* can be found by using the following equation.

where the *adj *(*A*) denotes the adjoint of a matrix. It can be calculated by the following method:

Given the *n × n* matrix *A*, define *B** = b*_{ij} to be the matrix whose coefficients are found by taking the determinant of the *(n-1) × (n-1)* matrix obtained by deleting the *i*^{th} row and *j*^{th} column of *A*.

The terms of *B* (i.e. *B = b*_{ij}) are known as the cofactors of *A*.

Define the matrix *C*, where *c*_{ij} = (−1)^{i+j} *b*_{ij}.

The transpose of *C* (i.e. *C*^{T}) is called the adjoint of matrix *A*.

**Example 1: **. Find the adj *A*.

__Solution__**:**

Computation of adj *A*:

Cofactor of 1 = a_{11} = - 4

Cofactor of 3 = a_{12} = -1

Cofactor of 7 = a_{13} = 6

Cofactor of 4 = a_{21} = 11

Cofactor of 2 = a_{22} = -6

Cofactor of 3 = a_{23} = 1

Cofactor of 1 = a_{31} = -5

Cofactor of 2 = a_{32} =-25

Cofactor of 1 = a_{33} = -10

Therefore we have:

**Example 2: **Find the inverse of

__Solution__**: **

The following method to find the inverse is only applicable for 2 *× *2 matrices.

1. Interchange leading diagonal elements:

-7 → 2; 2 → -7

2. Change signs of the other 2 elements:

-3 → 3; 4 → -4

3. Find the determinant |*A*|

4. Multiply result of [2] by 1/ |*A*|

** **

** **

**Example 3: **Find the inverse of

__Solution__**: **

The cofactor matrix for A can be calculated as follows:

Cofactor of 1 = a_{11} = 24

Cofactor of 2 = a_{12} = 5

Cofactor of 3 = a_{13} = -4

Cofactor of 0 = a_{21} = -12

Cofactor of 4 = a_{22} = 3

Cofactor of 5 = a_{23} = 2

Cofactor of 1 = a_{31} = -2

Cofactor of 0 = a_{32} = -5

Cofactor of 6 = a_{33} = 4

So the cofactor of

Therefore, the adjoint of .

And finally, the inverse of *A* is given by,

**Example 4: **Compute the inverse of

__Solution__**: **The cofactor matrix for *A* can be calculated as follows:

Cofactor of 3: a_{11} = 12

Cofactor of 2: a_{12} = 6

Cofactor of -1: a_{13} = -16

Cofactor of 1: a_{21} = 4

Cofactor of 6: a_{22} = 2

Cofactor of 3: a_{23} = 16

Cofactor of 2: a_{31} = 12

Cofactor of -4: a_{32} = -10

Cofactor of 0: a_{33} = 16

So the cofactor of

Therefore the adjoint of .

And finally, the inverse of *A* is given by:

**Example 5:** Find the inverse of

**Solution:** Write

Since

We have:

*a + c = 1*

*-a + 2c = 0*

*b + d = 0*

*-b + 2d = 1*

or

*a = 2/3*

*b= -1/3*

*c=1/3*

*d= 1/3*

The inverse of *A* is therefore:

We know that the inverse matrix is unique when it exists. So if *A* is invertible, then *A*^{-1} is also invertible and (*A*^{-1})^{-1} = *A*.