# What is the condition of a matrix to have an inverse? Find the inverse of the matrix $$A = \begin{bmatrix} 5 & 1 & 2\\ 1 & 0 & 3\\ 4 & -3 & 8 \end{bmatrix}$$  If it exists.

The condition of a matrix to have an inverse is If A is an invertible matrix then there is a matrix C such that

AC = I = CA

In such case, C is called inverse of A and write as C = A-1

Problem Part:

Given

A = $$\begin{bmatrix}1 & -2 & -1\\ -1 & 5 & 6\\ 5 & -4 & 5\end{bmatrix}$$

~ [A:I]

$$~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ -1 & 5 & 6 & : & 0 & 1 & 0\\ 5 & -4 & 5 & : & 0 & 0 & 1\end{bmatrix}$$

Applying Reduced Echelon Form

R2 → R2 + R1,   R3 → R3 – 5R1

$$~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ 0 & 3 & 5 & : & 1 & 1 & 0\\ 0 & 6 & 10 & : & -5 & 0 & 1\end{bmatrix}$$

$$R_2 \to \frac{1}{3} R_2$$

$$~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ 0 & 1 & \frac{5}{3} & : & \frac{1}{3} & \frac{1}{3} & 0\\ 0 & 6 & 10 & : & -5 & 0 & 1\end{bmatrix}$$

R1 → R1 + 2R2 ,   R3 → R3 – 6R2

$$~\begin{bmatrix}1 & 0 & \frac{7}{3} & : & \frac{5}{3} & \frac{2}{3} & 0\\ 0 & 1 & \frac{5}{3} & : & \frac{1}{3} & \frac{1}{3} & 0\\ 0 & 0 & 0 & : & -7 & -2 & 1\end{bmatrix}$$

To have inverse, We should have one pivot element in each row but in above matrix R3 doesn’t have pivot element. So,  Inverse of Matrix doesn’t exists.