Answer
$det(A)=-30$
Work Step by Step
$ \begin{bmatrix}
2 & 4 & 6 \\
0& 3 & 1 \\
0 & 0 & -5
\end{bmatrix} $
$M_{11}= \begin{bmatrix}
3 &1 \\
0& -5\\
\end{bmatrix} = 3\times5-0(1)=-15$
$M_{12}= \begin{bmatrix}
0 &1 \\
0& -5\\
\end{bmatrix}=0\times(-5)-0\times1=0$
$M_{13}= \begin{bmatrix}
0 &3 \\
0& 0\\
\end{bmatrix}= 0\times0-0\times3=0$
To calculate the cofactors, use the cofactor definition: $C_{ij}=(-1)^{ij}\times M_{ij}$
$C_{11}=-15$
$C_{12}=0$
$C_{13}=0$
$det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$
$det(A)=2\times(-15)+4(0)+6(0)$
$det(A)=-30$