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Let \(P = \left[ {\begin{array}{*{20}{c}} 2&α &3\\ { - α }&2&0\\ 3&{ - 2}&α \end{array}} \right]\), where α is real number such that det(P) = cofactor of second diagonal element of P. Then det(adj(p-1)) equal
1. 49
2. \(\frac{1}{49}\)
3. \(-\frac{1}{7 }\)
4. -7

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Best answer
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))\(\rm \left[\dfrac{1}{-7}\right]^2\)\(\boldsymbol{\rm \dfrac{1}{49}}\)

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