## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-6A+23B - 15C$
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, $det=\begin{vmatrix} A & B & C \\ -1 & 3 & 5 \\ 5 & 0 & -2 \\ \end{vmatrix}\\=A[(3)(-2)-(5)(0)]-B((-1)(-2)- (5) (5)]+C[(-1)(0)- (3) (5)] \\= -6A+23B - 15C$