A square matrix A of order n is invertible if there exists a square matrix B of order n such that AB = BA = I, an identity matrix.
Example
Consider the matrices \(A = \begin{pmatrix}–3 & 5 & 6\\–1 & 2 & 2\\1 & – 1 & – 1 \end{pmatrix}\) and \(B = \begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}\).
The matrix A is invertible, since :
\(\begin{pmatrix}–3 & 5 & 6\\–1 & 2 & 2\\1 & – 1 & – 1\end{pmatrix}\) × \(\begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}\) =
\(\begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}\) × \(\begin{pmatrix}–3 & 5 & 6\\–1 & 2 & 2\\1 & – 1 & – 1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\)