Matrices and Determinants 1 Question 11
11. 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) $X^{23}+Y^{23}$
(2015 Adv.)
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Answer:
Correct Answer: 11. (c,d)
Solution:
- Given, $X^{T}=-X, Y^{T}=-Y, Z^{T}=Z$
(a) Let $\quad P=Y^{3} Z^{4}-Z^{4} Y^{3}$
Then, $\quad P^{T}=\left(Y^{3} Z^{4}\right)^{T}-\left(Z^{4} Y^{3}\right)^{T}$
$ =\left(Z^{T}\right)^{4}\left(Y^{T}\right)^{3}-\left(Y^{T}\right)^{3}\left(Z^{T}\right)^{4} $
$ =-Z^{4} Y^{3}+Y^{3} Z^{4}=P $
$\therefore \quad P$ is symmetric matrix.
(b) Let $P =X^{44}+Y^{44}$
Then, $P^{T} =\left(X^{T}\right)^{44}+\left(Y^{T}\right)^{44} $
$=X^{44}+Y^{44}=P$
$\therefore \quad P$ is symmetric matrix.
(c) Let $\quad P=X^{4} Z^{3}-Z^{3} X^{4}$
Then, $\quad P^{T}=\left(X^{4} Z^{3}\right)^{T}-\left(Z^{3} X^{4}\right)^{T}$
$ =\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}=-P$
$\therefore P$ is skew-symmetric matrix.
(d) Let
$P=X^{23}+Y^{23}$
Then, $\quad P^{T}=\left(X^{T}\right)^{23}+\left(Y^{T}\right)^{23}$
$=-X^{23}-Y^{23}=-P$
$\therefore P$ is skew-symmetric matrix.
