Answer
The proof is below.
Work Step by Step
We take the cross product to find:
$=\vec{A}\times\vec{B}\\=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\times(B_x\hat{i}+B_y\hat{j}+B_z\hat{k}) \\ = (A_yB_z-A_zB_y)\hat{i}+(A_zB_x-A_xB_z)\hat{j}+(A_xB_y-A_yB_x)\hat{k}$
When we take the determinant, we find:
$(A_yB_z-A_zB_y)\hat{i}+(A_zB_x-A_xB_z)\hat{j}+(A_xB_y-A_yB_x)\hat{k}$
Thus, the determinant and the cross product of the two expressions are the same.
