 # Find Inverse of Matrix

To find the inverse of a matrix, firstly we should know what a matrix is. A matrix is a function which includes an ordered or organised rectangular array of numbers. The values in the array are known as the elements of the matrix. In a matrix, the horizontal arrays are known as rows and the vertical arrays are known as columns. If the number of rows and columns in a matrix is a and b respectively, then the order of the matrix will be a x b, where a and b denote the counting numbers.

Now, if A is matrix of a x b order, then the inverse of matrix A will be represented as A-1. Now the question arises, how to find that inverse of matrix A is A-1. Let us find out here.

## Inverse of a Matrix Definition

If A is a non-singular square matrix, then there exists an inverse matrix A-1, which satisfies the following condition:

AA-1 = A-1A = I, where I is the Identity matrix

### How to find the inverse of 3×3 matrix?

To calculate the inverse of a matrix, we have to follow these steps:

• First, we need to find the matrix of minors
• Now change that matrix into a matrix of cofactors
• Now find the adjoint of the matrix
• At the end, multiply by 1/determinant

Let us solve an example of 3×3 matrix to understand the steps better.

Example: Find the inverse of matrix $$A = \begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & -2\\ 0 & 1 & 1 \end{bmatrix}$$

Solution: To find the inverse of matrix A, we need to find the matrix of minors first;

$$Matrix \ of \ Minors = \begin{bmatrix} 3 & 2 & 2 \\ -1 & 3 & 3\\ -4 & -10 & 1 \end{bmatrix}$$

The next step is to find the Cofactors of minors of the above matrix.

$$Matrix \ of \ Cofactors = \begin{bmatrix} 3 & 2 & 2 \\ -1 & 3 & 3\\ -4 & -10 & 1 \end{bmatrix} \times \begin{bmatrix} + & – & + \\ – & + & – \\ + & – & + \end{bmatrix} \\ = \begin{bmatrix} 3 & -2 & 2\\ 1 & 3 & -3\\ -4 & 10 & 1 \end{bmatrix}$$

After this, find the adjoint or adjugate of the above-generated matrix by swapping the positions of the elements diagonally, such that;

$$Adjoint = \begin{bmatrix} 3 & 1 & -4\\ -2 & 3 & 10\\ 2 & -3 & 1 \end{bmatrix}$$

Now we need to find the determinant of the original or given matrix A. Since we have already calculated the determinants while calculating the matrix of minors. Hence, if we just multiply the elements of the top row of the above adjoint matrix with the cofactors top row, we will get the determinant of the complete matrix.

Hence, the determinant = 3×3 + 1x(-2) + 2×2

D = 9-2+4 = 11

Multiply the adjoint by 1/Determinant, to get the inverse of original matrix A.

Therefore,

$$A^{-1} = 1/11 \begin{bmatrix} 3 & 1 & -4\\ -2 & 3 & 10\\ 2 & -3 & 1 \end{bmatrix} \\ A^{-1} = \begin{bmatrix} 3/11 & 1/11 & -4/11\\ -2/11 & 3/11 & 10/11\\ 2/11 & -3/11 & 1/11 \end{bmatrix}$$