The determinant of a 3×3 matrix can be computed using the formula below:
$$\det \left[ {\begin{array}{*{20}{c}} a&b&c \ d&e&f \ g&h&i \end{array}} \right] = (-1)^{a+b+c} \left| \begin{array}{cc} e&f \ h&i \end{array} \right| + a \left| \begin{array}{cc} d&f \ g&i \end{array} \right| + b \left| \begin{array}{cc} d&e \ g&h \end{array} \right|$$
In this case, we are given that the determinant of the matrix $\left[ {\begin{array}{*{20}{c}} 1&3&0 \ 2&6&4 \ { – 1}&0&2 \end{array}} \right]$ is -12. Substituting in the values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, and $h$, we get:
$$\det \left[ {\begin{array}{*{20}{c}} 1&3&0 \ 2&6&4 \ { – 1}&0&2 \end{array}} \right] = (-1)^{1+3+0} \left| \begin{array}{cc} 6&4 \ 0&2 \end{array} \right| + 1 \left| \begin{array}{cc} 2&4 \ { – 1}&2 \end{array} \right| + 0 \left| \begin{array}{cc} 2&6 \ { – 1}&0 \end{array} \right| = (-1)^{4} \left| \begin{array}{cc} 6&4 \ 0&2 \end{array} \right| = -12$$
Therefore, the determinant of the matrix $\left[ {\begin{array}{*{20}{c}} 2&6&0 \ 4&{12}&8 \ { – 2}&0&4 \end{array}} \right]$ is $-96$.
To see why, we can use the formula above to compute the determinant of the matrix $\left[ {\begin{array}{*{20}{c}} 2&6&0 \ 4&{12}&8 \ { – 2}&0&4 \end{array}} \right]$:
$$\det \left[ {\begin{array}{*{20}{c}} 2&6&0 \ 4&{12}&8 \ { – 2}&0&4 \end{array}} \right] = (-1)^{2+6+0} \left| \begin{array}{cc} {12}&8 \ 0&4 \end{array} \right| + 2 \left| \begin{array}{cc} 2&8 \ { – 2}&4 \end{array} \right| + 0 \left| \begin{array}{cc} 2&6 \ { – 2}&0 \end{array} \right| = (-1)^{8} \left| \begin{array}{cc} {12}&8 \ 0&4 \end{array} \right| = -96$$