Evaluate the determinant of the matrix.$\begin{vmatrix}-9&0&1\\2&4&8\\-3&-4&-6\end{vmatrix}=$|−9 0 12 4 8−3 −4 −6|=​

The determinant of a 3x3 matrix can be calculated using the formula:

det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)

where: a, b, c are the entries of the first row, d, e, f are the entries of the second row, g, h, i are the entries of the third row.

For the given matrix, the entries are:

a = -9, b = 0, c = 1, d = 2, e = 4, f = 8, g = -3, h = -4, i = -6.

Substituting these values into the formula gives:

det(A) = -9(-24 - (-32)) - 0 + 1(8 - (-12)) = -9(-24 + 32) + 1(8 + 12) = -9*8 + 20 = -72 + 20 = -52

So, the determinant of the given matrix is -52.