$\begin{vmatrix} 1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{vmatrix}= x\begin{vmatrix}2 & 3\\ 8 & 9 \end{vmatrix}+ y \begin{vmatrix} 1 & 3\\7 & 9 \end{vmatrix}+z\begin{vmatrix}1 & 2\\7 & 8\end{vmatrix}$ Then $x+y + z $ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: -5
$\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}=0$
$\text{RHS minors:}$
$\begin{vmatrix}2&3\\8&9\end{vmatrix}=18-24=-6$
$\begin{vmatrix}1&3\\7&9\end{vmatrix}=9-21=-12$
$\begin{vmatrix}1&2\\7&8\end{vmatrix}=8-14=-6$
$0 = -6x -12y -6z$
$x+2y+z=0$
$\text{From cofactor expansion along second row: } x=-4,\ y=5,\ z=-6$
$x+y+z = -4 + 5 + -6 $
$x+y+z=-5$