Question

(i) If [1 1 x]$\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ = 0, find x.
(ii) If $\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left[\begin{array}{cc}1& -3\\ -2& 4\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ -9& x\end{array}\right]$ , find x.

Answer 1

(i)
$$\begin{gathered} \mathrm{Given}:\left[\begin{array}{ccc}1& 1& x\end{array}\right]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}1+0+2x& 0+2+x& 2+1+0\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}1+2x& 2+x& 3\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{c}1+2x+2+x+3\end{array}\right]=0 \\ \Rightarrow 6+3x=0 \\ \Rightarrow 3x=-6 \\ \Rightarrow x=\frac{-6}{3} \\ \therefore x=-2 \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Given}:\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left[\begin{array}{cc}1& -3\\ -2& 4\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ -9& x\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2-6& -6+12\\ 5-14& -15+28\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ -9& x\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}-4& 6\\ -9& 13\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ -9& x\end{array}\right] \\ \Rightarrow x=13 \\ \therefore x=13 \end{gathered}$$