Question

If $\left[\begin{array}{ccc}1& x& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 5& 1\\ 0& 3& 2\end{array}\right]\left[\begin{array}{c}x\\ 1\\ -2\end{array}\right]=0$, then the value of $x$ is

• A. $0$
• B. $\frac{2}{3}$
• C. $\frac{5}{4}$
• D. $-\frac{4}{5}$

Answer 1

The correct option is C

$\frac{5}{4}$

Explanation for the correct option

It is given that $\left[\begin{array}{ccc}1& x& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 5& 1\\ 0& 3& 2\end{array}\right]\left[\begin{array}{c}x\\ 1\\ -2\end{array}\right]=0$

$$\begin{gathered} \Rightarrow \left[\begin{array}{ccc}\left(1\times 1\right)+\left(x\times 0\right)+\left(1\times 0\right)& \left(1\times 2\right)+\left(x\times 5\right)+\left(1\times 3\right)& \left(1\times 3\right)+\left(x\times 1\right)+\left(1\times 2\right)\end{array}\right]\left[\begin{array}{c}x\\ 1\\ -2\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{ccc}1& 5x+5& x+5\end{array}\right]\left[\begin{array}{c}x\\ 1\\ -2\end{array}\right]=0 \\ \Rightarrow \left[\left(1\times x\right)+\left\{\left(5x+5\right)\times 1\right\}+\left\{\left(x+5\right)\times \left(-2\right)\right\}\right]=0 \\ \Rightarrow \left[x+5x+5-2x-10\right]=0 \\ \Rightarrow \left[4x-5\right]=0 \\ \Rightarrow 4x-5=0 \\ \Rightarrow x=\frac{5}{4} \end{gathered}$$

Therefore, the value of $x$ is $\frac{5}{4}$.

Hence, option(C) is the correct option.