MCQ
Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{ccc}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$. If $B=\left[\begin{array}{ccc}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}( AB )$ is equal to :
  • A
    343
  • B
    125
  • C
    64
  • D
    216

Answer

Equating co $-$ factor fo $A _{21}$$\left(2 \alpha^2-3 \alpha\right)=\alpha$
$\alpha=0,2 \  ($accept$)$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$
$ \alpha=2 \quad \beta=1 \\ |AB|=|A \operatorname{cof}(A)|=|A|^3 $
$A=\left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right|$
$=6-2(9)+3(6)=6$

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