To find the determinant \(\displaystyle \left|\begin{array}{ccc} {_2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|\), seeing that "0" on the right of the first row, I would "expand" on either the first row or the third column.
Expanding on the first row,
\(\displaystyle \left|\begin{array}{ccc} {-2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|= -2\left|\begin{array}{cc} 2 & 4 \\ -1 & 5\end{array}\right|- \left|\begin{array}{cc} a & 4 \\ -2 & 5\end{array}\right|= -2(10+ 4)- (5a+ 8)= -28- 5a- 8= -36- 5a= -51\) so -5a= 36- 51= -15. a= 3.
Expanding on the third column,
\(\displaystyle \left|\begin{array}{ccc} {-2} & 1 & 0 \\ a & 2 & 4 \\ {-2} & -1 & 5 \end{array}\right|= -4 \left|\begin{array}{cc}a & 2 \\ -2 & -1\end{array}\right|+ 5\left|\begin{array}{cc}-2 & 1 \\ a & 2 \end{array}\right|= -4(-a+ 4)+ 5(-4- a)= 4a- 16- 20- 5a= -36- 5a= -51\) so, again, 5a= 51- 36= 15 and a= 3.
(Does anyone know why I am getting the "-" in "-2" moved like that?)