Answer
\begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{5} & {4} & {3} \\ {-2} & {9} & {12} \\ {1} & {-1} & {0}\end{array}\right| =87 \end{aligned}
Work Step by Step
Given $$ A=\left[ \begin{array}{ccc}{5} & {4} & {3} \\ {-2} & {9} & {12} \\ {1} & {-1} & {0}\end{array} \right]$$
Since, if we have$$A=\left[ \begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array} \right]$$
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}{ccc}{5} & {4} & {3} \\ {-2} & {9} & {12} \\ {1} & {-1} & {0}\end{array}\right| \\ &=5 \cdot 9 \cdot 0+4 \cdot 12 \cdot 1+3 \cdot(-2) \cdot(-1)-5 \cdot 12 \cdot(-1)-4 \cdot(-2) \cdot 0-3 \cdot 9 \cdot 1 \\ &=0+48+6+60+0-27=87 \end{aligned}
