If $A + B = \text{BA}$ and $A^2 - B^2 = I$, then the value of the determinant of matrix $A^T + B$ (where $A$ and $B$ are square matrices of order $3 \times 3$)
Step-by-Step Solution
Key Concept: For commuting matrices satisfying $A^2 = B^2$, the determinant condition det$(A+B) = 0$ holds when $A \neq B$.
Given $A^2 - B^2 = (A-B)(A+B) = 0$ with $AB = BA$, if $A \neq B$ then $A - B$ is not a null matrix. For the product to be zero, either $A + B$ is a null matrix (meaning $A = -B$) or $\det(A-B) = \det(A+B) = 0$. In both scenarios, $\det(A+B) = 0$.
Correct Answer: 1