Find the determinant of matrix A:

$$A=[[2, 6, -3],[4, 11, 3], [-2, 5, 0]] $$

To find the determinant of a 3x3 matrix, the formula below is used:

$$\begin{gathered} A = \left.open\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right] \\ \left.openA\right| = a·\left.open\begin{array}{cc}e& f\\ h& i\end{array}\right|-b·\left.open\begin{array}{cc}d& f\\ g& i\end{array}\right|+c·\left.open\begin{array}{cc}d& e\\ g& h\end{array}\right| \end{gathered}$$

Essentially what we are doing is taking a term in the first row, and multiplying it by the determinant of the 2x2 matrix that is formed when you eliminate all of the terms in the same row and in the same column. The determinant of a 2x2 matrix is solved as follows:

$A= \left.open\begin{array}{cc}a& b\\ c& d\end{array}\right], \left.openA\right|=ad-bc$

We can write the determinant of our matrix A in the problem as

$$\begin{gathered} \left.openA\right|=2·\left.open\begin{array}{cc}11& 3\\ 5& 0\end{array}\right|-6·\left.open\begin{array}{cc}4& 3\\ -2& 0\end{array}\right|-3\left.open\begin{array}{cc}4& 11\\ -2& 5\end{array}\right| \\ \left.openA\right|=2·(0-15)-6·(0+6)-3(20+22) \\ \left.openA\right|=2·(-15)-6·\left(6\right)-3\left(42\right) \\ \left.openA\right|=-30-36-126 \\ \left|A\right|\mathbf{=}\mathbf{-}\mathbf{192} \end{gathered}$$