Let α ∈ (0, ∞) and A = [{1, 2, α}, {1, 0, 1}, {0, 1, 2}]. If det(adj(2A))

Question

Let \(\alpha \in(0, \infty)\) and \(A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]\).

If \(\operatorname{det}\left(\operatorname{adj}\left(2 A-A^{T}\right) \cdot \operatorname{adj}\left(A-2 A^{T}\right)\right)=2^{8}\), then \((\operatorname{det}(\mathrm{A}))^{2}\) is equal to :

(1) 1

(2) 49

(3) 16

(4) 36

Answer 1

Correct option is (3) 16

\(\left|\operatorname{adj}\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=28\)

\(\left|\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=24\)

\(\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|\left|2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right|= \pm 16\)

\(\left(A-2 A^{T}\right)^{T}=A^{T}-2 A\)

\(\left|A-2 A^{T}\right|=\left|A^{T}-2 A\right|\)

\(\Rightarrow\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|^{2}=16\)

\(\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|= \pm 4\)

\(\left[\begin{array}{ccc}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]-\left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2 \alpha & 2 & 4\end{array}\right]\)

\(\left|\begin{array}{ccc}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2\end{array}\right|\)

\(1+3 \alpha=4\)

\(3 \alpha=3\)

\(\alpha=1\)

\(|A|=\left|\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right|=-1-3=-4\)

\(|A|^{2}=16\)