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Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$. If $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}(A B)$ is equal to :
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2521 Upvotes
Verified Answer
The correct answer is:
216
Equating co-factor fo $\mathrm{A}_{21}$
$\begin{aligned}
& \left(2 \alpha^2-3 \alpha\right)=\alpha \\
& \alpha=0,2 \text { (accept) }
\end{aligned}$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$
$\begin{aligned}
& \alpha=2 \quad \beta=1 \\
& |\mathrm{AB}|=|\mathrm{A} \operatorname{cof}(\mathrm{A})|=|\mathrm{A}|^3 \\
& \mathrm{~A}=\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 2 & 1 \\
-1 & 2 & 4
\end{array}\right|=6-2(9)+3(6)=6
\end{aligned}$
$\begin{aligned}
& \left(2 \alpha^2-3 \alpha\right)=\alpha \\
& \alpha=0,2 \text { (accept) }
\end{aligned}$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$
$\begin{aligned}
& \alpha=2 \quad \beta=1 \\
& |\mathrm{AB}|=|\mathrm{A} \operatorname{cof}(\mathrm{A})|=|\mathrm{A}|^3 \\
& \mathrm{~A}=\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 2 & 1 \\
-1 & 2 & 4
\end{array}\right|=6-2(9)+3(6)=6
\end{aligned}$
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