Answer
$det(A)=0.002$
Work Step by Step
$ \begin{bmatrix}
-0.4 & 0.4 & 0.3 \\
0.2 & 0.2 & 0.2 \\
0.3 & 0.2 & 0.2
\end{bmatrix} $
$M_{11}= \begin{bmatrix}
0.2 &0.2 \\
0.2& 0.2\\
\end{bmatrix} = 0.2(0.2)-0.2(0.2)=0$
$M_{12}= \begin{bmatrix}
0.2 &0.2 \\
0.3& 0.2\\
\end{bmatrix}=0.2\times0.2-(0.3)0.2=-0.02$
$M_{13}= \begin{bmatrix}
0.2 &0.2 \\
0.3& 0.2\\
\end{bmatrix}= 0.2\times0.2-(0.3)0.2=-0.02$
To calculate the cofactors, use the cofactor definition: $C_{ij}=(-1)^{ij}\times M_{ij}$
$C_{11}=0$
$C_{12}=0.02$
$C_{13}=-0.02$
$det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$
$det(A)=-0.4(0)+0.4(0.02)+0.3\times(-0.02)$
$det(A)=0.002$