Question

Let $X$ and $Y$ be two arbitrary, $3\times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3\times 3$, nonzero, 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)

$X^T=-X{Y}^T=-Y{Z}^T=Z$
$\left(A\right){\left(Y^3{Z}^4-Z^4{Y}^3\right)}^T={\left(Y^3{Z}^4\right)}^T-{\left(Z^4{Y}^3\right)}^T$
$={\left(Z^4\right)}^T{\left(Y^3\right)}^T-{\left(Y^3\right)}^T{\left(Z^4\right)}^T$
$={\left(Z^T\right)}^4{\left(Y^T\right)}^3-{\left(Y^T\right)}^3{\left(Z^T\right)}^4$
$=Z^4{\left(-Y\right)}^3-{\left(-Y\right)}^3{\left(Z\right)}^4$
$=-Z^4{Y}^3+Y^3{Z}^4$
$=Y^3{Z}^4-Z^4{Y}^3$
Hence it is symmetric matrix.
$\left(B\right){\left(X^{44}+Y^{44}\right)}^T={\left(X^T\right)}^{44}+{\left(Y^T\right)}^{44}$
$=X^{44}+Y^{44}$
Hence it is symmetric matrix.
$\left(C\right){\left(X^4{Z}^3-Z^3{X}^4\right)}^T={\left(X^4{Z}^3\right)}^T-{\left(Z^3{X}^4\right)}^T$
$={\left(Z^T\right)}^3{\left(X^T\right)}^4-{\left(X^T\right)}^4{\left(Z^T\right)}^3$
$=Z^3{X}^4-X^4{Z}^3$
$=-\left(X^4{Z}^3-Z^3{X}^4\right)$
Hence it is skew symmetric matrix.
$\left(D\right){\left(X^{23}+Y^{23}\right)}^T={\left(X^T\right)}^{23}+{\left(Y^T\right)}^{23}$
$=-\left(X^{23}+Y^{23}\right)$
Hence it is skew symmetric matrix.