Answer
$-6A+23B - 15C$
Work Step by Step
The determinant for a $3 \times 3$ matrix can be found as:
$det = \begin{vmatrix}
a & b & c \\
d &e & f \\
g &h & i \\ \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$
As per the problem,
$det=\begin{vmatrix}
A & B & C \\
-1 & 3 & 5 \\
5 & 0 & -2 \\
\end{vmatrix}\\=A[(3)(-2)-(5)(0)]-B((-1)(-2)- (5) (5)]+C[(-1)(0)- (3) (5)] \\= -6A+23B - 15C$
