+12 let x and y be distinct real numbers such that
\( \begin{vmatrix} 1 & 4 & 9 \\ 3 & x & y \\ 3 & y & x \end{vmatrix} = 0.\) find x+y
Here is the Aops way of doing it, same way as it did it:
Expanding the determinant, we obtain:
\begin{align*}
\begin{vmatrix} 1 & 4 & 9 \\ 3 & x & y \\ 3 & y & x \end{vmatrix} &= \begin{vmatrix} x & y \\ y & x \end{vmatrix} - 4 \begin{vmatrix} 3 & y \\ 3 & x \end{vmatrix} + 9 \begin{vmatrix} 3 & x \\ 3 & y \end{vmatrix} \\
&= (x^2 - y^2) - 4(3x - 3y) + 9(3y - 3x) \\
&= x^2 - y^2 - 39x + 39y \\
&= (x - y)(x + y) - 39(x - y) \\
&= (x - y)(x + y - 39).
\end{align*}
Since this is 0, either x - y = 0 or x + y - 39 = 0. But x and y are distinct, so \(x + y = \boxed{39}.\)
