Question

Let X and Y be two arbitary, 3 $\times$ 3 non – zero, skew – symmetric matrices and Z be an arbitary, 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 $X^T=-X,Y^T=-Y,Z^T=Z$
(a) Let $P=Y^3{Z}^4-Z^4{Y}^3$
Then, $P^T=(Y^3{Z}^4{)}^T-(Z^4{Y}^3)T$
$=\left(Z^T{)}^4\right(Y^T{)}^3-\left(Y^T{)}^3\right(Z^T{)}^4$
$=-Z^4{Y}^3+Y^3{Z}^4=P$
$\therefore$ P is symmetric matrix.
(b) Let $P=X^{44}+Y^{44}$
Then, $P^T=(X^T{)}^{44}+(Y^T{)}^{44}$
$=X^{44}+Y^{44}=P$
$\therefore$ P is symmetric matrix.
(c) Let $P=X^4{Z}^3-Z^3{X}^4\therefore P^T=(X^4{Z}^3{)}^T-(Z^3{X}^4{)}^T=\left(Z^T{)}^3\right(X^T{)}^4-\left(X^T{)}^4\right(Z^T{)}^3=Z^3{X}^4-X^4{Z}^3=-P$
$\therefore$ P is skew - symmetric matrix.
(d) Let $P=X^{23}+Y^{23}$
Then, $P^T=(X^T{)}^{23}+(Y^T{)}^{23}=-X^{23}-Y^{23}=-P$
$\therefore$ P is skew - symmetric matrix.