Question

$\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|$$=\frac{xyz}{C}(x-y)(y-z)(z-x)$

Find the value of C?

Answer 1

$\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|=\left|\begin{array}{ccc}x& \frac{x(x-1)}{2}& \frac{x(x-1)(x-2)}{6}\\ y& \frac{y(y-1)}{2}& \frac{y(y-1)(y-2)}{6}\\ z& \frac{z(z-1)}{2}& \frac{z(z-1)(z-2)}{6}\end{array}\right|$

$=\frac{xyz}{12}\left|\begin{array}{ccc}1& x-1& (x-1)(x-2)\\ 1& y-1& (y-1)(y-2)\\ 1& z-1& (z-1)(z-2)\end{array}\right|$

applying $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$ gives

$=\frac{xyz}{12}\left|\begin{array}{ccc}0& x-y& (x-y)(x+y-2)\\ 0& y-z& (y-z)(y+z-2)\\ 1& z-1& (z-1)(z-2)\end{array}\right|$

expanding along $C_1$ gives

$=\frac{xyz}{12}(x-y)(y-z)(z-x)$

$\therefore C=12$