Question

Let $A$ and $B$ be $3\times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skew-symmetric, and $(A+B)(A-B)=(A-B)(A+B)$ if ${\left(AB\right)}^T={(-1)}^{n}AB$, then

• A. $nϵZ$
• B. $nϵN$
• C. $n$ is an even natural number
• D. $n$ is an odd natural number

Answer 1

Answer: (D)

$n$ is an odd natural number

Given $A$ is symmetric, $B$ is skew-symmetric.

$A^T=A,B^T=-B$
Also, ${\left(AB\right)}^T={(-1)}^{n}AB$

$\Rightarrow B^T{A}^T={(-1)}^{n}AB$

$\Rightarrow -BA={(-1)}^{n}AB$

Here, if n is even natural, $AB=-BA$

and n is odd natural, $AB=BA$
Now, $(A+B)(A-B)$

$=A^2-AB+BA-B^2$

If n is even natural number then $=A^2-2AB-B^2$

If n is an odd natural number then $=A^2-B^2$
SImilarly $(A-B)(A+B)$

$=A^2+AB-BA-B^2$

If n is even natural number then $=A^2+2AB-B^2$

If n is an odd natural number then $=A^2-B^2$
Hence, when n is odd natural number then $(A+B)(A-B)=(A-B)(A+B)$