Answer
$det(A)=-0.022$
Work Step by Step
$ \begin{bmatrix}
0.1 & 0.2 & 0.3 \\
-0.3 & 0.2 & 0.2 \\
0.5& 0.4 & 0.4
\end{bmatrix} $
$M_{11}= \begin{bmatrix}
0.2 &0.2 \\
0.4& 0.4\\
\end{bmatrix} = 0.2\times0.4-0.4(0.2)=0$
$M_{12}= \begin{bmatrix}
-0.3 &0.2 \\
0.5& 0.4\\
\end{bmatrix}=-0.3\times0.4-0.5\times0.2=-0.22$
$M_{13}= \begin{bmatrix}
-0.3 &0.2 \\
0.5& 0.4\\
\end{bmatrix}= -0.3\times0.4-0.2\times0.5=-0.22$
To calculate the cofactors, use the cofactor definition: $C_{ij}=(-1)^{ij}\times M_{ij}$
$C_{11}=0$
$C_{12}=0.22$
$C_{13}=-0.22$
$det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$
$det(A)=0.1\times0+0.2\times0.22+0.3(-0.22)$
$det(A)=-0.022$
