Answer
$10$
Work Step by Step
2 by 2 determinant
$D=\left|\begin{array}{ll}
a & b\\
c & d
\end{array}\right|=ad-bc$
3 by 3 determinant (cofactors of first row):
$\left|\begin{array}{lll}
{a_{11}}&{a_{12}}&{a_{13}}\\
{a_{21}}&{a_{22}}&{a_{23}}\\
{a_{31}}&{a_{32}}&{a_{33}}\end{array}\right|=a_{11}\left|\begin{array}{cc}
{a_{22}}&{a_{23}}\\
{a_{32}}&{a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{cc}
{a_{21}}&{a_{23}}\\
{a_{31}}&{a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{cc}
{a_{21}}&{a_{22}}\\{a_{31}}&{a_{32}}\end{array}\right|$
---
$D=3\left|\begin{array}{ll}
-1 & 5\\
2 & -2
\end{array}\right|-4\left|\begin{array}{ll}
1 & 5\\
1 & -2
\end{array}\right|+2\left|\begin{array}{ll}
1 & -1\\
1 & 2
\end{array}\right|$
$=3[(-1)(-2)-2(5)]-4[1(-2)-1(5)]$
$=3(-8)-4(-7)+2(3)$
$=-24+28+6$
$=10$