Question

Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 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. $X^4{Z}^3–Z^3{X}^4$
• C. $X^{23}+Y^{23}$
• D. $X^{44}+Y^{44}$

Answer 1

Answer: (B), (C)

$X^{23}+Y^{23}$

$X^T=–X,Y^T=–Yand{Z}^T=Z$
$P=Y^3{Z}^4–Z^4{Y}^3$
$PT=\left(Z^4{)}^T\right(Y^3{)}^T–\left(Y^3{)}^T\right(Z^4{)}^T$
$=\left(Z^T{)}^4\right(Y^T{)}^3–\left(Y^T{)}^3\right(Z^T{)}^4$
$=Z^4(–Y^3)–(–Y^3)Z^4$
$=Y^3{Z}^4–Z^4{Y}^3=P$(symmetric)
$Q=X^{44}+Y^{44}$
$Q^T=(X^T{)}^{44}+(Y^T{)}^{44}=X^{44}+Y^{44}=Q$(symmetric)
$R=X^4{Z}^3–Z^3{X}^4$
$R^T=\left(Z^3{)}^T\right(X^4{)}^T–\left(X^4{)}^T\right(Z^3{)}^T$
$=\left(Z^T{)}^3\right(X^T{)}^4–\left(X^T{)}^4\right(Z^T{)}^3$
$=Z^3(–X{)}^4–(–X{)}^4(Z{)}^3$
$=Z^3{X}^4–X^4{Z}^3=–R$(Skew symmetric)
$S=X^{23}+Y^{23}$
$S^T=(X^T{)}^{23}+(Y^T{)}^{23}=(–X{)}^{23}+(–Y{)}^{23}$
$=–X^{23}–Y^{23}=–S$ (Skew symmetric)