Evaluate the determinant of the matrix.$\begin{vmatrix}1&0&5\\6&2&0\\3&8&9\end{vmatrix}=$|1 0 56 2 03 8 9|=​

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, we have:

a = 1, b = 0, c = 5, d = 6, e = 2, f = 0, g = 3, h = 8, i = 9.

Substituting these values into the formula, we get:

det(A) = 1*(29 - 08) - 0*(69 - 03) + 5*(68 - 23) = 118 - 0 + 542 = 18 + 210 = 228

So, the determinant of the given matrix is 228.