Given the matrix:

Here's the solution to the problems:

(d) Show that the matrix $M$ is invertible and find $M^{-1}$

Given the matrix:

$$M = \begin{pmatrix} 3 & -2 & 5 \\ 7 & 4 & -8 \\ 5 & -3 & -4 \end{pmatrix}$$

Step 1: Calculate the determinant of $M$.
A matrix is invertible if and only if its determinant is non-zero.

$$\det(M) = 3 \begin{vmatrix} 4 & -8 \\ -3 & -4 \end{vmatrix} - (-2) \begin{vmatrix} 7 & -8 \\ 5 & -4 \end{vmatrix} + 5 \begin{vmatrix} 7 & 4 \\ 5 & -3 \end{vmatrix}$$ $$\det(M) = 3((4)(-4) - (-8)(-3)) + 2((7)(-4) - (-8)(5)) + 5((7)(-3) - (4)(5))$$ $$\det(M) = 3(-16 - 24) + 2(-28 - (-40)) + 5(-21 - 20)$$ $$\det(M) = 3(-40) + 2(-28 + 40) + 5(-41)$$ $$\det(M) = -120 + 2(12) - 205$$ $$\det(M) = -120 + 24 - 205$$ $$\det(M) = -96 - 205$$ $$\det(M) = -301$$

Since $\det(M) = -301 \neq 0$, the matrix $M$ is invertible.

Step 2: Calculate the cofactor matrix $C$.

$$C_{11} = \begin{vmatrix} 4 & -8 \\ -3 & -4 \end{vmatrix} = -16 - 24 = -40$$ $$C_{12} = -\begin{vmatrix} 7 & -8 \\ 5 & -4 \end{vmatrix} = -(-28 + 40) = -12$$ $$C_{13} = \begin{vmatrix} 7 & 4 \\ 5 & -3 \end{vmatrix} = -21 - 20 = -41$$ $$$$