$\begin{vmatrix} 1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{vmatrix}= x\begin{vmatrix}2 & 3\\ 8 & 9 \end{vmatrix}+ y \begin{vmatrix} 1 & 3\\7 & 9 \end{vmatrix}+z\begin{vmatrix}1 & 2\\7 & 8\end{vmatrix}$ Then $x+y + z $ is : |
15 5 -5 0 |
0 |
The correct answer is Option 4: 0 $\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}=0$ $\text{RHS minors:}$ $\begin{vmatrix}2&3\\8&9\end{vmatrix}=18-24=-6$ $\begin{vmatrix}1&3\\7&9\end{vmatrix}=9-21=-12$ $\begin{vmatrix}1&2\\7&8\end{vmatrix}=8-14=-6$ $0 = -6x -12y -6z$ $x+2y+z=0$ $\text{From cofactor expansion along first row: } x=1,\ y=-2,\ z=1$ $x+y+z=1-2+1=0$ $x+y+z=0$ |