# Inverse matrix

### Inverse matrix

The transpose of the matrix obtained by replacing the element of the square matrix A by the corresponding cofactor is called adjoint of the matrix of A and it denote by adj(A)

A square matrix A is said to be an invertible matrix if there exist a square matrix B such that AB = BA = I the matrix B is called invers of A matrix and it denoted by A⁻¹.

If A is an inversible then  $${{A}^{-1}}=\frac{adj(A)}{\det (A)}$$ .

Example: find the adjoint and inverses of the following matrices

1) $$A=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]$$.

Solution: Given that $$A=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]$$,

$$A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right]=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]$$,

$$adj(A)=\left[ \begin{matrix} d & -b \\ -c & d \\\end{matrix} \right]=\left[ \begin{matrix}6 & 3 \\ -4 & 2 \\\end{matrix} \right]$$,

|A| = 12 – (- 12) = 24

$${{A}^{-1}}=\frac{adj(A)}{\det (A)}=\frac{1}{24}\left[ \begin{matrix} 6 & 3 \\ -4 & 2 \\\end{matrix} \right]$$.

2) $$A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$.

Solution: Given that $$A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$,

$$A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$,

$$adj(A)=\left[ \begin{matrix} d & -b \\ -c & d \\\end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$,

|A| = 1 – 0 = 1

$${{A}^{-1}}=\frac{adj(A)}{\det (A)}=\frac{1}{1}\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$,

$${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]$$.