Let \[\alpha \beta \neq 0\] and \[A=\left[\[\begin{gathered} \beta \quad \alpha \quad 3 \\ \alpha \quad \alpha \quad \beta \\ -\beta \quad \alpha \quad 2 \alpha \end{gathered}\]\right]\]. If \[B=\left[\[\begin{gathered} 3 \alpha \quad -9 \quad 3 \alpha \\ -\alpha \quad 7 \quad -2 \alpha \\ -2 \alpha \quad 5 \quad -2 \beta \end{gathered}\]\right]\] is the matrix of cofactors of the elements of A, then \[\operatorname{det}(A B)\] is equal to :
Solutions · 1
Official solution
\[A=\left[\[βα−βααα3β2α\]\right], B=\left[\[3α−α−2α−9753α−2α−2β\]\right]\]
Cofactor of A-matrix is
\[=\left[\[2α2−αβ−(2α2−3α)αβ−3α−(2α2+β2)(2αβ+3β)−(β2−3α)α2+αβ−(2αβ)βα−α2\]\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[\[\begin{array}{|ccccccc|}1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4\end{array}\end{aligned}\]\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 \\
\]