Question
Without expanding the determinants, show that
$\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$
$\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$
Taking (-1) common from $R_1, R_2, R_3$, we get
$D=(-1)^3\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|$
Interchanging rows and columns, we get
$D=-1\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|$
$\begin{array}{ll}\therefore & D=-1(D) \\ \therefore & 2 D=0 \\ \therefore & D=0\end{array}$
$\therefore \quad\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$
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${ }^n P_r=720$ and ${ }^n C_{n-r}=120$
matrices.