QJEEM24Ma9b10: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9b10 Solution: Let and . If is the matrix of cofactors of the elements of ,…

$A=\left[\begin{array}{ccc}\beta & \alpha & 3 \\\alpha & \alpha & \beta \\-\beta & \alpha & 2 \alpha\end{array}\right], B=\left[\begin{array}{ccc}3 \alpha & -9 & 3 \alpha \\-\alpha & 7 & -2 \alpha \\-2 \alpha & 5 & -2 \beta\end{array}\right]$

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

$\begin{aligned}& \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\& \Rightarrow \alpha \beta-\alpha=0 \\& \Rightarrow \alpha(\beta-1)=0 \\& \Rightarrow \alpha=0 \text { or } \beta=1[\because \alpha \text { cannot be } 0] \\& \Rightarrow \beta=1 \\& \text { and }-\beta^2+3 \alpha=5 \\& \Rightarrow 3 \alpha=6 \\& \Rightarrow \alpha=2 \\& A=\left[\begin{array}{ccc}1 & 2 & 3 \\2 & 2 & 1 \\-1 & 2 & 4\end{array}\right] \\& \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\& =|A| \cdot|A|^2 \\& =|A|^3 \\& =(6-18+18)^3 \\& =6^3 \\& =216 \\\end{aligned}$