Question

Solve the matrix equations:
(i) $[x1]\left[\begin{array}{cc}1& 0\\ -2& -3\end{array}\right]\left[\begin{array}{c}x\\ 5\end{array}\right]=0$

(ii) $[121]\left[\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=0$

(iii) $[x-5-1]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$

(iv) $\left[\begin{array}{cc}2x& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right]\left[\begin{array}{c}x\\ 8\end{array}\right]=0$

Answer 1

​$$\begin{gathered} \left(i\right)\left[\begin{array}{cc}x& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -2& -3\end{array}\right]\left[\begin{array}{c}x\\ 5\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}x-2& 0-3\end{array}\right]\left[\begin{array}{c}x\\ 5\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}x-2& -3\end{array}\right]\left[\begin{array}{c}x\\ 5\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{c}{x}^2-2x-15\end{array}\right]=0 \\ \Rightarrow x^2-2x-15=0 \\ \Rightarrow x^2-5x+3x-15=0 \\ \Rightarrow x\left(x-5\right)+3\left(x-5\right)=0 \\ \Rightarrow \left(x-5\right)\left(x+3\right)=0 \\ \Rightarrow x-5=0\mathrm{or}x+3=0 \\ \Rightarrow x=5\mathrm{or}x=-3 \end{gathered}$$

$$\begin{gathered} \left(ii\right)\left[\begin{array}{ccc}1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}1+4+1& 2+0+0& 0+2+2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}6& 2& 4\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{c}0+4+4x\end{array}\right]=0 \\ \Rightarrow 4+4x=0 \\ \Rightarrow 4x=-4 \\ \therefore x=\frac{-4}{4}=-1 \end{gathered}$$

$$\begin{gathered} \left(iii\right)\left[\begin{array}{ccc}x& -5& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}x-0-2& 0-10-0& 2x-5-3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}x-2& -10& 2x-8\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{c}{x}^2-2x-40+2x-8\end{array}\right]=0 \\ \Rightarrow x^2-48=0 \\ \Rightarrow x^2=48 \\ \Rightarrow x=\pm \sqrt{48} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\left[\begin{array}{cc}2x& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right]\left[\begin{array}{c}x\\ 8\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}2x-9& 4x\end{array}\right]\left[\begin{array}{c}x\\ 8\end{array}\right]=0 \\ \Rightarrow \left[x\left(2x-9\right)+32x\right]=0 \\ \Rightarrow \left[2x^2-9x+32x\right]=0 \\ \Rightarrow \left[2x^2+23x\right]=0 \\ \Rightarrow 2x^2+23x=0 \\ \Rightarrow x\left(2x+23\right)=0 \\ \Rightarrow x=0\mathrm{or}x=-\frac{23}{2} \\ \therefore x=0\mathrm{or}x=-\frac{23}{2} \end{gathered}$$