Let A, B, C be three square matrices of order 3 x 3 such that A^2 + 2*I = 0, (2C  A^2) = 30 and satisfy the equation A^5  2A^3C + BA^2  2BC = 0. The value of (det(B))^2 is A) 64 B) 64 C) 512 D) 512
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
✨ Free AI Tools for You
Powered by Chatterbot AI
Answer
The correct answer is Option D: 512.
To solve for $(\det(B))^2$, we analyze the given information and equations:

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: $$ B = A^3 $$

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

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

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

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.
India's 1^{st} AI Doubt Solver for CBSE, JEE, and NEET
Ask a Question for Freeand then it's just ₹212 a month