Answer
$-11A+2B+5C$
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, $det=\begin{vmatrix}
A & B & C \\
2 & 1 & 4 \\
1 & 3 & 1 \\
\end{vmatrix}\\=A[(1)(1)-(4)(3)]-B((2)(1)- (4) (1)]+C[(2)(3)- (1) (1)] \\=-11A+2B+5C$
