Question

Find the matrix $A$ such that $\left[\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right]A=\left[\begin{array}{ccc}-1& -8& -10\\ 1& -2& -5\\ 9& 22& 15\end{array}\right]$

Answer 1

$\begin{array}{c}\Rightarrow \left[\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right]A=\left[\begin{array}{ccc}-1& -3& -10\\ 1& -2& -15\\ 9& 22& 15\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right]{\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]}_{2\times 3}=\left[\begin{array}{ccc}-1& -3& -10\\ 1& -2& -15\\ 9& 22& 15\end{array}\right]\\ \Rightarrow \left[\begin{array}{ccc}2a_{11}-a_{21}& 2a_{12}-a_{22}& 2a_{13}-a_{23}\\ a_{11}& a_{12}& a_{13}\\ -3a_{11}+4a_{21}& -3a_{12}+4a_{22}& -3a_{13}+4a_{23}\end{array}\right]=\left[\begin{array}{ccc}-1& -3& -10\\ 1& -2& -15\\ 9& 22& 15\end{array}\right]\\ \Rightarrow 2a_{11}-a_{21}=-1\\ \Rightarrow 2a_{12}-a_{22}=-8\\ \Rightarrow 2a_{13}-a_{23}=-10\\ \Rightarrow a_{11}=1\\ \Rightarrow a_{12}=-2\\ \Rightarrow a_{13}=-5\\ \Rightarrow -3a_{11}+4a_{21}=9\\ \Rightarrow -3a_{12}+4a_{22}=22\\ \Rightarrow -3a_{13}+4a_{23}=15\\ So,\\ a_{11}=1a_{12}=-2a_{13}=-5\\ a_{21}=3a_{22}=4a_{23}=0\\ Hence,\\ matrixA=\left[\begin{array}{ccc}1& -2& -5\\ 3& 4& 0\end{array}\right]\end{array}$