Correct Answer - Option 2 :
\(\frac{1}{49}\)
Concept:
For any square matrix A
det(A-1) = \(\rm \dfrac{1}{det(A)}\)
det(adj(A)) = [det(A)]n-1
Where n × n is the order of matrix A
Calculation:
det(P) = \({\begin{vmatrix} 2&α &3\\ { - α }&2&0\\ 3&{ - 2}&α \end{vmatrix}}\)
det(P) = 2 (2α - 0) - α (-α2 - 0) + 3 (2α - 6)
det(P) = α3 + 10α - 18
As det(P) = cofactor of second diagonal element of P
α3 + 10α - 18 = 2α - 9
α3 + 8α - 9 = 0
(α - 1)(α2 + α + 9) = 0
∵ α is a real number α = 1
det(P) = 1 + 10 - 18 = -7
Now, det(adj(P-1)) = [det(P-1)]3-1
det(adj(P-1)) = \(\rm \left[\dfrac{1}{det(P)}\right]^2\)
det(adj(P-1)) = \(\rm \left[\dfrac{1}{-7}\right]^2\)= \(\boldsymbol{\rm \dfrac{1}{49}}\)