To find the determinant of the given $3 \times 3$ matrix:

$\text{Matrix} = \begin{bmatrix} -2 & 4 & -5 \\ 1 & 3 & -7 \\ -1 & 4 & -8 \end{bmatrix}$

We use the formula for the determinant of a $3 \times 3$ matrix:

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

Where the matrix elements are:

a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} -2 & 4 & -5 \\ 1 & 3 & -7 \\ -1 & 4 & -8 \end{bmatrix}$$ Substituting the values: $$\text{det}(A) = (-2)((3)(-8) - (-7)(4)) - 4((1)(-8) - (-7)(-1)) + (-5)((1)(4) - (3)(-1))$$ Now, let's compute the terms: 1. $$(3)(-8) - (-7)(4) = -24 + 28 = 4$$ 2. $$(1)(-8) - (-7)(-1) = -8 - 7 = -15$$ 3. $$(1)(4) - (3)(-1) = 4 + 3 = 7$$ So, the determinant is: $$\text{det}(A) = -2(4) - 4(-15) - 5(7)$$ $$= -8 + 60 - 35 = 17$$ Therefore, the determinant of the matrix is: $$\boxed{17}$$ Would you like more details on any specific part of this calculation? --- Here are five related questions you might find helpful: 1. How do you calculate the determinant of a $$2 \times 2$$ matrix? 2. What is the significance of the determinant of a matrix in linear algebra? 3. Can you find the determinant of a larger matrix, such as $$4 \times 4$$? 4. How can you use determinants to solve systems of linear equations? 5. What does it mean if the determinant of a matrix is zero? **Tip:** When calculating determinants of matrices, breaking down the computation step by step can help avoid errors.