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Let $A=\left(\begin{array}{cc}x+2 & 3 x \\ 3 & x+2\end{array}\right), B=\left(\begin{array}{cc}x & 0 \\ 5 & x+2\end{array}\right)$. Then
all solutions of the equation det $(A B)=0$ is
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all solutions of the equation det $(A B)=0$ is
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Verified Answer
The correct answer is:
1,4,0,-2
We know that, $\operatorname{det}(A B)=$ det $(A)$ det $(B)$
$\therefore$ $\operatorname{det}(A B)=0$
$\Rightarrow$ det$(A) \cdot$ det $(B)=0$ $\Rightarrow\left|\begin{array}{cc}x+2 & 3 x \\ 3 & x+2\end{array} \| \begin{array}{cc}x & 0 \\ 5 & x+2\end{array}\right|=0$
$\Rightarrow\left\{(x+2)^{2}-9 x\right\}\{x(x+2)-0\}=0$
$\Rightarrow \quad\left(x^{2}+4 x+4-9 x\right) \times(x+2)=0$
$\Rightarrow \quad x(x+2)\left(x^{2}-5 x+4\right)=0$
$\Rightarrow \quad x(x+2)(x-1)(x-4)=0$
$\Rightarrow \quad x=0,-2,1,4$
$\therefore$ $\operatorname{det}(A B)=0$
$\Rightarrow$ det$(A) \cdot$ det $(B)=0$ $\Rightarrow\left|\begin{array}{cc}x+2 & 3 x \\ 3 & x+2\end{array} \| \begin{array}{cc}x & 0 \\ 5 & x+2\end{array}\right|=0$
$\Rightarrow\left\{(x+2)^{2}-9 x\right\}\{x(x+2)-0\}=0$
$\Rightarrow \quad\left(x^{2}+4 x+4-9 x\right) \times(x+2)=0$
$\Rightarrow \quad x(x+2)\left(x^{2}-5 x+4\right)=0$
$\Rightarrow \quad x(x+2)(x-1)(x-4)=0$
$\Rightarrow \quad x=0,-2,1,4$
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