To find the determinant of matrix $A = \begin{bmatrix} -5 & -4 & 8 \\ -3 & 0 & -6 \\ -7 & 5 & 9 \end{bmatrix}$, we will use the cofactor expansion method.

The determinant of a 3x3 matrix:

$\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$

Where the matrix elements are:

$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$

For matrix $A$, we have:

• $a = -5$
• $b = -4$
• $c = 8$
• $d = -3$
• $e = 0$
• $f = -6$
• $g = -7$
• $h = 5$
• $i = 9$

Now, let's compute it:

$\text{det}(A) = -5 \left(0 \times 9 - (-6) \times 5\right) - (-4) \left((-3) \times 9 - (-6) \times (-7)\right) + 8 \left((-3) \times 5 - 0 \times (-7)\right)$

Simplifying each term:

$\text{det}(A) = -5 \left(0 + 30\right) + 4 \left(-27 - 42\right) + 8 \left(-15\right)$

$\text{det}(A) = -5 \times 30 + 4 \times (-69) + 8 \times (-15)$

$\text{det}(A) = -150 + (-276) + (-120)$

$\text{det}(A) = -150 - 276 - 120 = -546$

Thus, the determinant of the matrix $A$ is $\boxed{-546}$.

Would you like more details or have any questions about this process?

Here are some related questions:

1. How does the cofactor expansion method work for higher-dimensional matrices?
2. What is the significance of a matrix having a determinant of zero?
3. How can determinants be used to solve systems of linear equations?
4. What properties of determinants can simplify their computation?
5. How does the determinant change if you swap two rows of a matrix?

Tip: Always check if a matrix is triangular (upper or lower), as its determinant is simply the product of the diagonal elements, simplifying the calculation.