Answer
$-169$
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):
$\left|\begin{array}{lll}
{a_{11}}&{a_{12}}&{a_{13}}\\
{a_{21}}&{a_{22}}&{a_{23}}\\
{a_{31}}&{a_{32}}&{a_{33}}\end{array}\right|=a_{11}\left|\begin{array}{cc}
{a_{22}}&{a_{23}}\\
{a_{32}}&{a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{cc}
{a_{21}}&{a_{23}}\\
{a_{31}}&{a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{cc}
{a_{21}}&{a_{22}}\\{a_{31}}&{a_{32}}\end{array}\right|$
---
$D=1\left|\begin{array}{rr}
1 & -5\\
2 & 3
\end{array}\right|-3\left|\begin{array}{rr}
6 & -5\\
8 & 3
\end{array}\right|+(-2)\left|\begin{array}{rr}
6 & 1\\
8 & 2
\end{array}\right|$
$=1[1(3)-2(-5)]-3[6(3)-8(-5)]$
$=1(13)-3(58)-2(4)$
$=13-174-8$
$=-169$
