Question

Find a matrix A such that $\left[\begin{array}{ccc}1& 3& 2\\ 2& 1& 1\\ 4& 0& 3\end{array}\right]$ A=$\left[\begin{array}{ccc}7& 1& 3\\ 1& 0& 3\\ -1& -3& 7\end{array}\right]$

Short answer

$A=\left[\begin{array}{ccc}-1& 0& 1\\ 2& 1& 0\\ 1& -1& 1\end{array}\right]$

Step-by-step solution

Step: 1

Given

$\left[\begin{array}{ccc}1& 3& 2\\ 2& 1& 1\\ 4& 0& 3\end{array}\right]$A=$\left[\begin{array}{ccc}7& 1& 3\\ 1& 0& 3\\ -1& -3& 7\end{array}\right]$

A=${\left[\begin{array}{ccc}1& 3& 2\\ 2& 1& 1\\ 4& 0& 3\end{array}\right]}^{-1}$$\left[\begin{array}{ccc}7& 1& 3\\ 1& 0& 3\\ -1& -3& 7\end{array}\right]$

Step: 2

Now to find

${\left[\begin{array}{ccc}1& 3& 2\\ 2& 1& 1\\ 4& 0& 3\end{array}\right]}^{-1}=\frac{1}{11}\left[\begin{array}{cccc}-3& 9& -1& \\ 1& 5& -3& \\ 4& -12& 5& \end{array}\right]$

Step: 3

$\begin{array}{r}A=\frac{1}{11}\left[\begin{array}{ccc}-3& 9& -1\\ 2& 5& -3\\ 4& -12& 5\end{array}\right]\left[\begin{array}{ccc}7& 1& 3\\ 1& 0& 3\\ -1& -3& 7\end{array}\right]\\ =\frac{1}{11}\left[\begin{array}{ccc}-21+9+1& -3+0+3& -9+27-7\\ 14+5+3& 2+0+9& 6+15-21\\ 28-12-5& 4+0-15& 12-36+35\end{array}\right]\\ =\frac{1}{11}\left[\begin{array}{ccc}-11& 0& 11\\ 22& 11& 0\\ 11& -11& 11\end{array}\right]\\ =\left[\begin{array}{ccc}-1& 0& 1\\ 2& 1& 0\\ 1& -1& 1\end{array}\right]\\ A=\left[\begin{array}{ccc}-1& 0& 1\\ 2& 1& 0\\ 1& -1& 1\end{array}\right]\end{array}$