# How do you solve Ax=B given A^-1=((2, -1), (3, -2)) and B=((2), (3))?

Apr 30, 2016

I found: $x = \left(\begin{matrix}{x}_{1} \\ {x}_{2}\end{matrix}\right) = \left(\begin{matrix}1 \\ 0\end{matrix}\right)$

#### Explanation:

As in a normal equation we can multiply both sides for the same amount:
$\textcolor{red}{{A}^{-} 1} A x = \textcolor{red}{{A}^{-} 1} B$
$I x = {A}^{-} 1 B$
where $I$ is the $4 \times 4$ identity matrix giving;
$x = {A}^{-} 1 B$
multiplying the two matrices on the right we get:
$x = \left(\begin{matrix}4 - 3 \\ 6 - 6\end{matrix}\right) = \left(\begin{matrix}1 \\ 0\end{matrix}\right)$
where we get, for the incognitas' vector $x$, two solutions:
$x = \left(\begin{matrix}{x}_{1} \\ {x}_{2}\end{matrix}\right) = \left(\begin{matrix}1 \\ 0\end{matrix}\right)$