Show that in any 3 x 3 determinant if one row is 0, the value of the determinant is 0.
Well, there might be an elegant way to do it, but since there are only three rows to check you can just do three problems. I'll do two of them.
Find the determinant of
$\displaystyle \left( \begin{array}{ccc} 0 & 0 & 0 \\ d & e & f \\ g & h & i \end{array} \right )$
How do you expand to find the determinant? By minors, or course. I'll do the easy row and expand along the top row:
0*(e*i - f*h) - 0*(d*i - f*g) + 0*(d*h - e*g) = 0.
So the determinant is 0.
Find the determinant of
$\displaystyle \left( \begin{array}{ccc} a & b & c \\ 0 & 0 & 0 \\ g & h & i \end{array} \right )$
Again I'll expand along the row with the zeros, but remember, since this is the second row down, we have an extra minus sign on each term:
-0*(b*i - c*h) + 0*(a*i - c*g) - 0*(a*h - b*g) = 0.
I'll let you do the last row.
-Dan