Answer
$-407$
Work Step by Step
The given determinant is
$\begin{vmatrix}
\begin{vmatrix}
5&0 \\
4 & -3
\end{vmatrix}& \begin{vmatrix}
-1&0 \\
0 &-1
\end{vmatrix} \\
\begin{vmatrix}
7& -5\\
4&6
\end{vmatrix} & \begin{vmatrix}
4 & 1\\
-3 &5
\end{vmatrix}
\end{vmatrix}$
First we solve the inner determinant.
$\Rightarrow \begin{vmatrix}
5 & 0 \\
4&-3
\end{vmatrix}=(5)(-3)-(4)(0)=-15-0=-15$
$\Rightarrow \begin{vmatrix}
-1 & 0 \\
0&-1
\end{vmatrix}=(-1)(-1)-(0)(0)=1-0=1$
$\Rightarrow \begin{vmatrix}
7 & -5 \\
4&6
\end{vmatrix}=(7)(6)-(4)(-5)=42+20=62$
$\Rightarrow \begin{vmatrix}
4 & 1 \\
-3&5
\end{vmatrix}=(4)(5)-(-3)(1)=20+3=23$
Substitute all values into the given determinant and solve.
$\Rightarrow \begin{vmatrix}
-15 & 1 \\
62&23
\end{vmatrix}=(-15)(23)-(62)(1)=-345-62=-407$
