Question

If $\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|$ = $\frac{xyz}{k}(x-y)(y-z)(z-x)$, then what is $k$ ?

Answer 1

$\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|=\left|\begin{array}{ccc}x& \frac{x(x-1)}{2}& \frac{x(x-1)(x-2)}{6}\\ y& \frac{y(y-1)}{2}& \frac{y(y-1)(y-2)}{6}\\ z& \frac{z(z-1)}{2}& \frac{z(z-1)(z-2)}{6}\end{array}\right|=xyz\left|\begin{array}{ccc}1& \frac{(x-1)}{2}& \frac{(x-1)(x-2)}{6}\\ 1& \frac{(y-1)}{2}& \frac{(y-1)(y-2)}{6}\\ 1& \frac{(z-1)}{2}& \frac{(z-1)(z-2)}{6}\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$=xyz\left|\begin{array}{ccc}1& \frac{(x-1)}{2}& \frac{(x-1)(x-2)}{6}\\ 0& \frac{(y-x)}{2}& \frac{(y-1)(y-2)}{6}-\frac{(x-1)(x-2)}{6}\\ 0& \frac{(z-x)}{2}& \frac{(z-1)(z-2)}{6}-\frac{(x-1)(x-2)}{6}\end{array}\right|$
$=xyz(\frac{(y-x)}{2}(\frac{(z-1)(z-2)}{6}-\frac{(x-1)(x-2)}{6})-\frac{(z-x)}{2}(\frac{(y-1)(y-2)}{6}-\frac{(x-1)(x-2)}{6}))$
$=\frac{xyz}{12}(x-y)(y-z)(z-x)\therefore k=12$