If $A$ and $B$ are two matrices of order ‘3’ such that $2 A + 3 B B^{T} = I$ and $B^{- 1} = A^{T},$ then identify which of the following statement(s) is (are) correct?

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The value of $T r (A^{- 1} - 3 B^{3} + B A) = 9$

The value of $\det (A^{- 1} - 3 B^{3} + B A) = 27$

The value of $\det (A^{2} - 2 A^{3}) = 27$

The value of $T r (A^{- 1} - 3 B^{3} - B A + I) = 6$

answer is A, B, C, D.

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Detailed Solution

Since $2 A + 3 B B' = I$
$\Rightarrow 2 A = I - 3 B B'$ (symmetric matrix)
$\Rightarrow A$ is symmetric $\Rightarrow B^{- 1} = A' = A$
$\Rightarrow A & B$ are inverse of each other
Also, $A^{- 1} = B$ is symmetric thus, $2 A + 3 B B' = I$
$\Rightarrow 2 B^{- 1} + 3 B^{2} = I \Rightarrow 2 I + 3 B^{3} = B \Rightarrow 3 B^{3} = 2 I$
A) $t r (A^{- 1} - 3 B^{3} - B A + I) = t r (B - 3 B^{3}) = t r (2 I) = 6$
B) $t r (A^{- 1} - 3 B^{3} - B A) = t r (B - 3 B^{3} + I) = t r (3 I) = 9$
C) Since, $2 A + 3 B B' = I \Rightarrow 2 A + 3 {(A^{- 1})}^{2} = I$
$\Rightarrow 2 A^{3} + 3 I = A^{2} \Rightarrow A^{2} - 2 A^{3} = 3 I$
So, $\det (A^{2} - 2 A^{3}) = \det (3 I) = 27$