Question

Let A and B be $3\times 3$ matrices of real number, where A is symmetric, B is skew-symmetric and $(A+B)(A-B)=(A-B)(A+B).If(AB{)}^t=(-1{)}^{k}AB$, where $(AB{)}^t$ is the transpose of matrix AB, then the possible values of k are

• A. 1
• B. 3
• C. 5
• D. All of these

Answer 1

The correct option is D All of these

$(A+B)(A-B)=(A-B)(A+B)\Rightarrow A^2-AB+BA-B^2=A^2+AB-BA-B^2\Rightarrow AB=BA(AB{)}^t=(-1{)}^{k}AB\Rightarrow B^t{A}^t=(-1{)}^{k}AB\Rightarrow -B.A=(-1{)}^{k}AB\Rightarrow BA=(-1{)}^{k+1}AB\Rightarrow AB=(-1{)}^{k+1}AB(-1{)}^{k+1}=1$
$\therefore k+1$ is even, then k is odd.