Question

If $A=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right]$ are two square matrices, find AB and hence solve the system of linear equations:
x − y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Answer 1

Here,

$$\begin{gathered} A=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right] \\ \mathrm{Now}, \\ AB=\left[\begin{array}{ccc}1& -1& 0\\ 2& 3& 4\\ 0& 1& 2\end{array}\right]\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right] \\ \Rightarrow AB=6\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ \Rightarrow AB=6I_3 \\ \Rightarrow \frac{1}{6}AB=I_3 \\ \Rightarrow \left(\frac{1}{6}B\right)A=I_3\left(\because AB=BA\right) \\ \Rightarrow A^{-1}=\frac{1}{6}B \\ \Rightarrow A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right] \\ X=A^{-1}B \\ X=\frac{1}{6}\left[\begin{array}{ccc}2& 2& -4\\ -4& 2& -4\\ 2& -1& 5\end{array}\right]\left[\begin{array}{c}3\\ 17\\ 7\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6+34-28\\ -12+34-28\\ 6-17+35\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12\\ -6\\ 24\end{array}\right] \\ \therefore x=2,y=-1\mathrm{and}z=4 \end{gathered}$$