Let α ∈ (0, ∞) and If det(adj(2A – AT). adj(A – 2AT)) = 28, then (det(A))2 is equal to

Engineering

Mathematics

Transpose and Adjoint of a Matrix

Question

Let α ∈ (0, ∞) and $A = [\begin{array}{l} 1 & 2 & α \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}]$ If det(adj(2A – A^T). adj(A – 2A^T)) = 2⁸, then (det(A))² is equal to

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∣adj⁡(2A – A^T)adj(⁡A – 2A^T)| = |adj{(⁡A – 2A^T)(2A– A^T)}| = |adj⁡{–(2A – A^T)^T⋅(2A – A^T)}|

= |adj⁡(–B^T⋅B)| where B = 2 A – A^T

= |–B^T⋅BI²

= |–B^T|²|B|²

=∣B⁴ (∵ |B| = |B^T|)

2⁸= |2A – A^T|⁴

⇒ 2A – A^T = ±4

$\Rightarrow |\begin{matrix} 1 & 3 & 2 α \\ 0 & 0 & 1 \\ - α & 1 & 2 \end{matrix}| = \pm 4$

⇒ –1 (1 + 3α) = 4 or –1 (1 + 3α) = – 4

$\Rightarrow α = - \frac{5}{3} (Re j) or α = 1$

Now |A| = (α – 5) = – 4 (det (A))² = 16