Let B be a skew symmetric matrix of order $3 \times 3$ with real entries.
Given I – B and I + B are non–singular matrices.
If $A = (I + B) {(I - B)}^{- 1}$ where det(A) > 0, then the value of det(2A) – det(adjA) is
[Here det(P) denotes determinant of square matrix P and det(adj P) denotes determinant of adjoint of square matrix P respectively]

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answer is 7.

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Detailed Solution

$A = (I + B) {(I - B)}^{- 1} A A^{T} = (I + B) {(I - B)}^{- 1} {({(I - B)}^{T})}^{- 1} {(I + B)}^{T} = (I + B) {(I - B)}^{- 1} {(I + B)}^{- 1} (I - B) = (I + B) {((I - B) (I - B))}^{- 1} (I - B) = (I + B) {((I - B) (I + B))}^{- 1} (I - B)$

$(∵ I + B, I - B a r e c o m m u t a t i v e)$

$∴ A A^{T} = (I + B) {(I + B)}^{- 1} {(I - B)}^{- 1} (I - B) = I$

$∴ A A^{T} = I$

${| A A |}^{T} = | I | = 1 \Rightarrow {| A |}^{2} = 1 a s | A | > 0 \Rightarrow | A | = 1$

det (2A) – det(adj(A)) = 8 det(A) – $\det {(A)}^{2}$ = $8 (1) - {(1)}^{2} = 7$

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