Find the determinant of the 3x3 matrix.

What  is  the determinant of  C?
$C=\begin{bmatrix} 1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{bmatrix}$

$\begin{array}{|rcll|} \hline C_{determinant} &=& \begin{vmatrix}1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{vmatrix} \\ &=& 1\cdot 0 \cdot 2 + 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 \\ && - 5\cdot 0 \cdot (-2) -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& -10 -18 -12 +3 \\ &=& -37 \\ \hline \end{array}$

Extend it like this:$\begin{bmatrix} 1 && 2 &&-2 &&1&&2\\ 3 && 0 && -1&&3&&0\\ 5 && 3 && 2&&5&&3 \end{bmatrix}$

Multiply the numbers if they can be linked top-to-bottom diagonally with 3 numbers: 

Example: 1, 0, and 2 is in the same diagonal and they are 3 numbers.

Then add up the products obtained from multiplying the number of top-left to bottom-right. Call that 'answer 1'

(1)(0)(2) + (2)(-1)(5) + (-2)(3)(3) = 0 - 10 - 18 = -28 -----(1)

Then add up the products obtained from multiplying the number of top-right to bottom-left. Call that 'answer 2'

(-2)(0)(5) + (1)(-1)(3) + (2)(3)(2) = -10 - 3 + 12 = -1 -----(2)

Add up answer 1 and answer 2: -28 - 1 = -29.

Therefore det(C) = -29. Best method ever , I don't have a proof of it though but I got through the determinant questions correctly when I use this method.