Question

Let $X$ and $Y$ be two arbitrary, $3\times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3\times 3$, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

• A. $Y^3{Z}^4-Z^4{Y}^3$
• B. $X^{44}+Y^{44}$
• C. $X^4{Z}^3-Z^3{X}^4$
• D. $X^{23}+Y^{23}$

Answer 1

Answer: (C), (D)

The correct options are
C $X^4{Z}^3-Z^3{X}^4$
D $X^{23}+Y^{23}$

Given: $X^T=-X,Y^T=-Y$ and $Z^T=Z$
Using Properties of transpose:
$(A+B{)}^T=A^T+B^T$ and $(AB{)}^T=B^T{A}^T$
Option A:
$(Y^3{Z}^4-Z^4{Y}^3{)}^T=(Y^3{Z}^4{)}^T-(Z^4{Y}^3{)}^T$
$\Rightarrow \left(Z^4{)}^T\right(Y^3{)}^T-\left(Y^3{)}^T\right(Z^4{)}^T=\left(Z^T{)}^4\right(Y^T{)}^3-\left(Y^T{)}^3\right(Z^T{)}^4=Y^3{Z}^4-Z^4{Y}^3$
Its a symmetric.
Option B:
$(X^{44}+Y^{44}{)}^T=(X^T{)}^{44}+(Y^T{)}^{44}=X^{44}+Y^{44}$
Its a symmetric.
Option C:
$(X^4{Z}^3-Z^3{X}^4{)}^T=(X^4{Z}^3{)}^T-(Z^3{X}^4{)}^T$
$\Rightarrow \left(Z^3{)}^T\right(X^4{)}^T-\left(X^4{)}^T\right(Z^3{)}^T=\left(Z^T{)}^3\right(X^T{)}^4-\left(X^T{)}^4\right(Z^T{)}^3=Z^3{X}^4-X^4{Z}^3=-(X^4{Z}^3-Z^3{X}^4)$
Its a skew symmetric.
Option D:
$(X^{23}+Y^{23}{)}^T=(X^T{)}^{23}+(Y^T{)}^{23}=-(X^{23}+Y^{23})$
Its a skew symmetric.
Hence, option C,D.