Question

Let $A, B, C$ be three square matrices of order $3 \times 3$ such that $A^{2} + 2I = 0$, $\left| (2C - A^{2}) \right| = 30$ and satisfy the equation $A^{5} - 2A^{3}C + BA^{2} - 2BC = 0$.

The value of $(\det(B))^{2}$ is

A) 64

B) -64

C) 512

D) -512

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Answer

The correct answer is Option D: -512.

To solve for $(\det(B))^2$, we analyze the given information and equations:

  1. We have the matrix equation: $$ A^5 - 2A^3C + BA^2 - 2BC = 0 $$ Rearranging and factoring out common terms yields: $$ (A^3 + B)(A^2 - 2C) = 0 $$ Given that $\left| 2C - A^2 \right| = 30$, it infers that $\det(2C - A^2) \neq 0$, hence the matrix $2C - A^2$ is invertible. Therefore, we conclude: $$ A^3 + B = 0 $$ which implies: $$ B = -A^3 $$

  2. Knowing $A^2 + 2I = 0$, we substitute into the expression for $B$: $$ B = -A(-2I) = 2A $$

  3. To find $(\det(B))^2$, use properties of determinants: $$ \det(B) = \det(2A) = 2^3 \det(A) = 8 \det(A) $$ therefore: $$ (\det(B))^2 = (8 \det(A))^2 = 64 \det(A)^2 $$

  4. With $A^2 = -2I$, we compute $\det(A^2)$: $$ \det(A^2) = \det(-2I) = (-2)^3 = -8 $$ Since $A^2$ is a square of $A$, $\det(A)^2 = \det(-2I) = -8$.

  5. Placing the value of $\det(A)^2$ in the expression for $(\det(B))^2$: $$ (\det(B))^2 = 64 (-8) = -512 $$

Thus, the squared determinant of matrix $B$ is -512.


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