JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 15)
Explanation
$$A=\left[\begin{array}{ccc} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha \end{array}\right], B=\left[\begin{array}{ccc} 3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta \end{array}\right]$$
Cofactor of $$A$$-matrix is
$$=\left[\begin{array}{ccc} 2 \alpha^2-\alpha \beta & -\left(2 \alpha^2+\beta^2\right) & \alpha^2+\alpha \beta \\ -\left(2 \alpha^2-3 \alpha\right) & (2 \alpha \beta+3 \beta) & -(2 \alpha \beta) \\ \alpha \beta-3 \alpha & -\left(\beta^2-3 \alpha\right) & \beta \alpha-\alpha^2 \end{array}\right]$$
which is equal to matrix $$B$$
So, by comparing elements of two matrix
$$\begin{aligned} & \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\ & \Rightarrow \alpha \beta-\alpha=0 \\ & \Rightarrow \alpha(\beta-1)=0 \\ & \Rightarrow \alpha=0 \text { or } \beta=1[\because \alpha \text { cannot be } 0] \\ & \Rightarrow \beta=1 \\ & \text { and }-\beta^2+3 \alpha=5 \\ & \Rightarrow 3 \alpha=6 \\ & \Rightarrow \alpha=2 \\ & A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right] \\ & \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\ & =|A| \cdot|A|^2 \\ & =|A|^3 \\ & =(6-18+18)^3 \\ & =6^3 \\ & =216 \\ \end{aligned}$$
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