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(D) Given the equation $A^T + 2A = I$. Taking the transpose of both sides,we get $(A^T)^T + 2A^T = I^T$,which simplifies to $A + 2A^T = I$. Now we hav

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$27$

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(D) Given the equation $A^T + 2A = I$. Taking the transpose of both sides,we get $(A^T)^T + 2A^T = I^T$,which simplifies to $A + 2A^T = I$. Now we have a system of two equations: $1) A^T + 2A = I$ $2) 2A^T + A = I$ Multiply equation $(1)$ by $2$: $2A^T + 4A = 2I$. Subtract equation $(2)$ from this: $(2A^T + 4A) - (2A^T + A) = 2I - I$,which gives $3A = I$,so $A = \frac{1}{3}I$. Then $A^{-1} = (\frac{1}{3}I)^{-1} = 3I$. Since $A$ is a $3 	imes 3$ matrix,$A^{-1} = 	ext{diag}(3, 3, 3)$. Therefore,$\det(A^{-1}) = 3 	imes 3 	imes 3 = 27$.

Let $A$ be a $3 \times 3$ matrix such that $A^T + 2A = I$. Then $\det(A^{-1})$ is equal to:

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