Question

The determinant $\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|=$

• A. $\frac{1}{3}xyz(x+y)(y+z)(z+x)$
• B. $\frac{1}{4}xyz(x+y-z)(y+z-x)$
• C. $\frac{1}{12}xyz(x-y)(y-z)(z-x)$
• D. $None$

Answer 1

Answer: (A)

$\frac{1}{3}xyz(x+y)(y+z)(z+x)$

$\left|\begin{array}{ccc}{}^x{C}_1& {}^x{C}_2& {}^x{C}_3\\ {}^y{C}_1& {}^y{C}_2& {}^y{C}_3\\ {}^z{C}_1& {}^z{C}_2& {}^z{C}_3\end{array}\right|$

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

$=xyz\left|\begin{array}{ccc}1& \frac{x-1}{2}& \frac{(x-1)(x-2)}{6}\\ 1& \frac{y-1}{2}& \frac{(y-1)(y-2)}{6}\\ 1& \frac{z-1}{2}& \frac{(z-1)(z-2)}{6}\end{array}\right|$

$=xyz\left|\begin{array}{ccc}1& \frac{x-1}{2}& \frac{(x-1)(x-2)}{6}\\ 0& \frac{y-1}{2}\frac{x-1}{2}& \frac{(y-1)(y-2)}{6}-\frac{(x-1)(x-2)}{6}\\ 0& \frac{z-1}{2}\frac{x-1}{2}& \frac{(z-1)(z-2)}{6}-\frac{(x-1)(x-2)}{6}\end{array}\right|$

$=xyz[\left(\frac{y-x}{2}\right)[\frac{(z-1)(z-2)}{6}-\frac{(x-1)(x-2)}{6}]-\left(\frac{z-x}{2}\right)[\frac{(y-1)(y-2)}{6}-\frac{(x-1)(x-2)}{6}]$

On solving

$=\frac{1}{3}xyz(x+y)(y+z)(z+x)$