Let α β ≠ 0 and A=left[begin{array}{ccc}β & α & 3 α & α & β -β & α &a

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3 and 4 .Determinants and Matrices

hard

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 $\operatorname{det}(A B)$ is equal to.

A

$343$

B

$125$

C

$64$

D

$216$

(JEE MAIN-2024)

Solution

Equating co-factor fo $\mathrm{A}_{21}$

$ \left(2 \alpha^2-3 \alpha\right)=\alpha $

$ \alpha=0,2 \text { (accept) }$

Now, $2 \alpha^2-\alpha \beta=3 \alpha$

$ \alpha=2 \quad \beta=1 $

$ |A B|=|A \operatorname{cof}(A)|=|A|^3$

$A=\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4\end{array}\right|=6-2(9)+3(6)=6$

Standard 12

Mathematics

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