Question

Let A and B be two $3 \times 3$ real matrices such that $(A^2 - B^2)$ is invertible matrix. If $A^5=B^5$ and $A^3B^2=A^2B^3$, then the value of the determinant of the matrix $A^3+B^3$ is equal to :

• A. 0
• B. 1
• C. 2
• D. 4

Hint:

When dealing with matrix equations, look for ways to factor expressions to isolate the term you are interested in. Remember that if a product of two matrices is the zero matrix ($MN=0$) and one of the matrices ($N$) is invertible, then the other matrix ($M$) must be the zero matrix. Similarly, if $det(MN)=0$ and $det(N) \neq 0$, then $det(M)$ must be 0.

Answer 1

The Correct Option is A

From $A^5 = B^5$, we have \[ A^5 - B^5 = 0 \] Add and subtract $A^3B^2$: \[ A^5 - A^3B^2 + A^3B^2 - B^5 = 0 \] Using the given condition $A^3B^2 = A^2B^3$: \[ A^3(A^2 - B^2) + B^3(A^2 - B^2) = 0 \] \[ (A^3 + B^3)(A^2 - B^2) = 0 \] Since $(A^2 - B^2)$ is invertible, \[ \det(A^2 - B^2) \neq 0 \] Taking determinants: \[ \det(A^3 + B^3)\det(A^2 - B^2) = 0 \] Hence, \[ \det(A^3 + B^3) = 0 \]