Answer
See below
Work Step by Step
Given $$ A=\left[ \begin{array}{cccc}{-2} & {0} & {1} & {6} \\ {-1} & {3} & {-1} & {-4} \\ {2} & {1} & {0} & {3} \\ {0} & {5} & {-4} & {-2}\end{array} \right]$$
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}}\\a_{41} &
a_{42} & a_{43}\end{array}\right]
\end{aligned}
So, we get
\begin{aligned} \operatorname{det}(A)&=\left|\begin{array}{cccc}{-2} & {0} & {1} & {6} \\ {-1} & {3} & {-1} & {-4} \\ {2} & {1} & {0} & {3} \\ {0} & {5} & {-4} & {-2}\end{array}\right|, \ \\
& \text{from fourth row, we obtain}\\
\operatorname{det}(A)&=-2\cdot\left|\begin{array}{ccc}{3} & {-1} & {-4} \\ {1} & {0} & {3} \\ {5} & {-4} & {-2}\end{array}\right|+1.\cdot\left|\begin{array}{ccc}{-1} & {3} & {-4} \\ {2} & {1} & {3} \\ {0} & {5} & {-2}\end{array}\right|-6\cdot\left|\begin{array}{ccc}{-1} & {3} & {-1} \\ {2} & {1} & {0} \\ {0} & {5} & {-4}\end{array}\right|\\
&=-2 \cdot(3.0 \cdot(-2) +(-1).3.5+(-4).1 \cdot (-4)-3. \cdot 3 \cdot (-4)-(-1) \cdot 1 \cdot (-2)-(-4) \cdot 0 \cdot 5)+1.(-1 \cdot 1 .(-2) +3.3.0+(-4).2.5-(-1).3.5-3.2.(-2)-(-1).1\cdot 0)-6.(-1.1.(-4)+3.0.0+(-1).2.5-(-1).0.5-3.2.(-4)-(-1).1.0)\\
& =-2.35+(-11)-6.18\\
&=-70-11-108\\
&=-189 \end{aligned}
