How do you find the inverse of A=((3, 2, 7), (0, 1, 2)) ?

Nov 12, 2016

The additive inverse of $A = \left(\begin{matrix}3 & 2 & 7 \\ 0 & 1 & 2\end{matrix}\right)$ is $- A = \left(\begin{matrix}- 3 & - 2 & - 7 \\ 0 & - 1 & - 2\end{matrix}\right)$

It has no multiplicative inverse. Only square matrices have those.

Explanation:

$\left(\begin{matrix}3 & 2 & 7 \\ 0 & 1 & 2\end{matrix}\right) + \left(\begin{matrix}- 3 & - 2 & - 7 \\ 0 & - 1 & - 2\end{matrix}\right) = \left(\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right)$

So $A = \left(\begin{matrix}3 & 2 & 7 \\ 0 & 1 & 2\end{matrix}\right)$ has an additive inverse $- A = \left(\begin{matrix}- 3 & - 2 & - 7 \\ 0 & - 1 & - 2\end{matrix}\right)$

A multiplicative inverse would have to satisfy:

$A {A}^{- 1} = {A}^{- 1} A = I$

for some identity matrix $I$. Would $I$ be a $2 \times 2$ matrix or a $3 \times 3$ one?

By considering how we multiply matrices, then we notice that $A B$ and $B A$ can only be square matrices if $B$ is a $3 \times 2$ matrix. If $B$ is a $3 \times 2$ matrix then $A B$ is $2 \times 2$ and $B A$ is $3 \times 3$.