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 [latex]A = \begin{pmatrix}–3 & 5 & 6\\–1 & 2 & 2\\1 & – 1 & – 1 \end{pmatrix}[/latex] and [latex]B = \begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}[/latex].
The matrix A is invertible, since :
[latex]\begin{pmatrix}–3 & 5 & 6\\–1 & 2 & 2\\1 & – 1 & – 1\end{pmatrix}[/latex] × [latex]\begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}[/latex] =
[latex]\begin{pmatrix}0 & 1 & 2\\–1 & 3 & 0\\1 & –2 & 1\end{pmatrix}[/latex] × [latex]\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}[/latex]
