Question

Let $A=\begin{pmatrix}x+2&3x\\\ 3&x+2\end{pmatrix}, B=\begin{pmatrix}x&0\\\ 5&x+2\end{pmatrix}$. Then all solutions of the equation det $(AB) = 0$ is

• A. $1, -1, 0, 2$
• B. $1, 4, 0, -2$
• C. $1, -1, 4, 3$
• D. $-1, 4, 0, 3$

Answer 1

Answer: (B)

$1, 4, 0, -2$

Answer 2

We know that,
$\text{det}(A B)=\text{det}(A) \text{det}(B)$
$\therefore \text{det}(A B)=0$
$\Rightarrow \text{det}(A) \cdot \text{det}(B)=0$
$\Rightarrow \begin{vmatrix}x+2 & 3 x \\\ 3 & x+2\end{vmatrix} \begin{vmatrix} x & 0 \\\ 5 & x+2\end{vmatrix} =0$
\Rightarrow \left\\{(x+2)^{2}-9 x\right\\}\\{x(x+2)-0\\}=0
$\Rightarrow \left(x^{2}+4 x+4-9 x\right) x(x+2)=0$
$\Rightarrow x(x+2)\left(x^{2}-5 x+4\right)=0$
$\Rightarrow x(x+2)(x-1)(x-4)=0$
$\Rightarrow x=0,-2,1,4$