# Chapter 8 - Polar Coordinates; Vectors - Section 8.7 The Cross Product - 8.7 Assess Your Understanding - Page 651: 12

$2A+12B -6C$

#### 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, $det=\begin{vmatrix} A & B & C \\ 0 & 2 & 4 \\ 3 & 1 & 3 \\ \end{vmatrix}\\=A[(2)(3)-(4)(1)]-B((0)(3)- (4) (3)]+C[(0)(1)- (2) (3)] \\= 2A+12B -6C$

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