$$ (detA)(x_1)=detB_1\, or\, x_1=\frac{detB_1}{detA} $$
(Row 1 of $A$) times (column 1 of $C^T$) yields the first $det A$ on the right: $a_{11} C_{11} + a_{12}C_{12} + a_{13}C_{13} = det A$ This is exactly the cofactor rule!
also Row 2 of $A$ times column 2 of $C^T$
How to explain the zeros off the main diagonal? The rows of A are multiplying cofactors from different rows. Why is the answer zero?
You can think it create a new matrix with row 1 and row 2 same . As we all know ,
its determinant is zero.
If we know the corners $(x_1, y_1)$ and $(x_2, y_2)$ and $(x_3, y_3)$ of a triangle, what is the area?
The area of a triangle is half of a 3 by 3 determinant.
Why is $\frac{1}{2} |x_1y_2 - x_2 y_1|$ the area of this triangle? We can remove the factor $\frac{1}{2}$ for a parallelogram (twice as big, because the parallelogram contains two equal triangles).We now prove that the parallelogram area is the determinant $x_1y_2 - x_2 y_1.$
