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Question:

(i) If $\left[\begin{array}{lll}1 & 1 & x]\end{array}\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0\right.$, find $x$.

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

Solution:

(i)

Given : $\left[\begin{array}{lll}1 & 1 & x\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}1+0+2 x & 0+2+x & 2+1+0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}1+2 x & 2+x & 3\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0$

$\Rightarrow[1+2 x+2+x+3]=0$

$\Rightarrow 6+3 x=0$

$\Rightarrow 3 x=-6$

$\Rightarrow x=\frac{-6}{3}$

$\therefore x=-2$

(ii)

Given : $\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ -2 & 4\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ -9 & x\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2-6 & -6+12 \\ 5-14 & -15+28\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ -9 & x\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-4 & 6 \\ -9 & 13\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ -9 & x\end{array}\right]$

$\Rightarrow x=13$

$\therefore x=13$