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 $C$. $$ C_{11} = \begin{vmatrix} 4 & -8 \\ -3 & -4 \end{vmatrix} = -16 - 24 = -40 $$ $$ C_{12} = -\begin{vmatrix} 7 & -8 \\ 5 & -4 \end{vmatrix} = -(-28 + 40) = -12 $$ $$ C_{13} = \begin{vmatrix} 7 & 4 \\ 5 & -3 \end{vmatrix} = -21 - 20 = -41 $$ $$ C_{21} = -\begin{vmatrix} -2 & 5 \\ -3 & -4 \end{vmatrix} = -(8 + 15) = -