Question

If $\begin{bmatrix}1&x&1\end{bmatrix} \begin{bmatrix}1&3&2\\\ 2&5&1\\\ 15&3&2\end{bmatrix}\begin{bmatrix}1\\\ 2\\\ x\end{bmatrix} = 0$, then x can be

• A. -2
• B. 2
• C. 14
• D. -14

Answer 1

Answer: (D)

-14

Answer 2

Given that, $\begin{bmatrix}1 & x & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 2 \\\ 2 & 5 & 1 \\\ 15 & 3 & 2\end{bmatrix}\begin{bmatrix}1 \\\ 2 \\\ x\end{bmatrix}=0$
$\Rightarrow \begin{bmatrix}1+2 x+15 \\\ 3+5 x+3 \\\ 2+x+2\end{bmatrix}\begin{bmatrix}1 \\\ 2 \\\ x\end{bmatrix}=0$
$\Rightarrow \begin{bmatrix}2 x+16 \\\ 5 x+6 \\\ x+4\end{bmatrix}\begin{bmatrix}1 \\\ 2 \\\ x\end{bmatrix}=0$
$\Rightarrow 2 x+16+2(5 x+6)+x(x+4)=0$
$\Rightarrow \, 2 x+16+10 x+12+x^{2}+4 x=0$
$\Rightarrow \, x^{2}+16 x+28=0$
$\Rightarrow \, x^{2}+2 x+14 x+28=0$
$\Rightarrow \, x(x+2)+14(x+2)=0$
$\Rightarrow \,(x+2)(x+14)=0$
$\Rightarrow \,x=-2,-14$