Answer
$$ \operatorname{det}(A)=\left| \begin{array}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array} \right|=-144$$
Work Step by Step
Given $$ A=\left[ \begin{array}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array} \right]$$
since,if we have
\begin{aligned}A =\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]
\end{aligned}
So, we get
\begin{aligned}\operatorname{det}(A)& =\left|\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right|
&=a_{11}( a_{22} a_{33}-a_{23} a_{32})-a_{12}( a_{21} a_{33}-a_{23} a_{31})+a_{13}( a_{21} a_{32}-a_{22} a_{31})\\
& =a_{11}( a_{22} a_{33}-a_{23} a_{32})-a_{12}( a_{21} a_{33}-a_{23} a_{31})+a_{13}( a_{21} a_{32}-a_{22} a_{31})\\
&=a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32}-a_{11} a_{23} a_{32}-a_{12} a_{21} a_{33}-a_{13} a_{22} a_{31}\\
\end{aligned}
Therefore, we get
\begin{aligned} \operatorname{det}(A)&=\left|\begin{array}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array}\right|, \ \\
& \text{from fourth row, we obtain}\\
\operatorname{det}(A)&=-3 \cdot\left|\begin{array}{ccc}{4} & {1} & {8} \\ {0} & {-2} & {13} \\ {0} & {0} & {-6}\end{array}\right|\\
&=-3 \cdot(4 \cdot(-2) \cdot(-6)+1 \cdot 13 \cdot 0+8 \cdot 0 \cdot 0-4 \cdot 13 \cdot 0-1 \cdot 0 \cdot(-6)-8 \cdot(-2) \cdot 0)\\
& =3 \cdot(48+0+0-0+0+0)\\
&=-3 \cdot 48\\
&=-144 \end{aligned}
