Solution:

(i) ${{\left( \text{AB} \right)}^{-1}}$ = ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

Solving Left Hand Side: ${{\left( \text{AB} \right)}^{-1}}$

$(AB)$ = $\left( \left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 4\times \left( -4 \right)+0\text{ }\!\!~\!\!\text{ }\times 1 & 4\times \left( -2 \right)+0\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) \\ -1\times \left( -4 \right)+2\text{ }\!\!~\!\!\text{ }\times 1 & \left( -1 \right)\times \left( -2 \right)+2\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -16+0 & -8+0 \\ 4+2 & 2-2 \\ \end{matrix} \right]$

$(AB)$ = $\left[ \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right]$

The multiplicative inverse of $(AB)$ is calculated as:

${{\left( \text{AB} \right)}^{-1}}$= $\frac{Adj~\left( \text{AB} \right)}{\left| \text{AB} \right|}$

$(AB)$ = $\left[ \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ }\left( \text{AB} \right)$ = $\left[ \begin{matrix} 0 & 8 \\ -6 & -16 \\ \end{matrix} \right]$

$\left| \text{AB} \right|$ = $\left| \begin{matrix} -16 & -8 \\ 6 & 0 \\ \end{matrix} \right|$

= $\left( -16 \right)\times 0-\left( -8 \right)\times 6$

= $0+48$

$\left|\text{AB} \right|$ = $48\ne 0$

Since $(AB)$ is non-singular therefore ${{\left( \text{AB} \right)}^{-1}}$ is possible

${{\left( \text{AB} \right)}^{-1}}$ = $\frac{\left[ \begin{matrix} 0 & 8 \\ -6 & -16 \\ \end{matrix} \right]}{48}$

= $\left[ \begin{matrix} \frac{0}{48} & \frac{8}{48} \\ \frac{-6}{48} & \frac{-16}{48} \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{-1}}$ = $\left[ \begin{matrix} 0 & \frac{1}{6} \\ -\frac{1}{8} & -\frac{1}{3} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(1)$

Solving Right Hand Side: ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

The multiplicative inverse of matrix B is calculated as:

${{\text{B}}^{-1}}$ = $\frac{Adj~B}{\left| \text{B} \right|}$

B = $\left[ \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -1 & 2 \\ -1 & -4 \\ \end{matrix} \right]$

$\left| \text{B} \right|$ = $\left| \begin{matrix} -4 & -2 \\ 1 & -1 \\ \end{matrix} \right|$

= $\left( -4 \right)\times \left( -1 \right)-\left( -2 \right)\times 1$

= $4+2$

$\left| \text{B} \right|$ = $6\ne 0$

Since B is non-singular therefore ${{\text{B}}^{-1}}$ is possible

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -1 & 2 \\ -1 & -4 \\ \end{matrix} \right]}{6}$

= $\left[ \begin{matrix} \frac{-1}{6} & \frac{2}{6} \\ \frac{-1}{6} & \frac{-4}{6} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} \frac{-1}{6} & \frac{1}{3} \\ \frac{-1}{6} & \frac{-2}{3} \\ \end{matrix} \right]$ 

Similarly, the multiplicative inverse of matrix A is calculated as:
${{\text{A}}^{-1}}$ = $\frac{Adj~\text{A}}{\left| \text{A} \right|}$

A = $\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ A}$ = $\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]$

$\left| \text{A} \right|$ = $\left| \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right|$

= $4\times 2-0\times \left( -1 \right)$

= $8+0$

$\left| \text{A} \right|$ = $8\ne 0$

Since A is non-singular therefore ${{\text{A}}^{-1}}$ is possible

${{\text{A}}^{-1}}$ = $\frac{\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]}{8}$

= $\left[ \begin{matrix} \frac{2}{8} & \frac{0}{8} \\ \frac{1}{8} & \frac{4}{8} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$ 

Now

${{\text{B}}^{-1}}{{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{-1}{6} & \frac{1}{3} \\ \frac{-1}{6} & \frac{-2}{3} \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$

= \[\left[ \begin{matrix} \left( \frac{-1}{6}\times \frac{1}{4} \right)+\left( \frac{1}{3}\times \frac{1}{8} \right) & \left( \frac{-1}{6}\times 0 \right)+\left( \frac{1}{3}\times \frac{1}{2} \right) \\ \left( \frac{-1}{6}\times \frac{1}{4} \right)+\left( \frac{-2}{3}\times \frac{1}{8} \right) & \left( \frac{-1}{6}\times 0 \right)+\left( \frac{-2}{3}\times \frac{1}{2} \right) \\ \end{matrix} \right]\]

= $\left[ \begin{matrix} -\frac{1}{24}+\frac{1}{24} & 0+\frac{1}{6} \\ -\frac{1}{24}-\frac{2}{24} & 0-\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ \frac{-1-2}{24} & -\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ \frac{-3}{24} & -\frac{2}{6} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & \frac{1}{6} \\ -\frac{1}{8} & -\frac{1}{3} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(2)$

From $(1)$ and $(2)$, it is clear that:

${{\left( \text{AB} \right)}^{-1}}$ = ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

(ii) ${{\left( \text{DA} \right)}^{-1}}$ = ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$

Solving Left Hand Side: ${{\left( \text{DA} \right)}^{-1}}$

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }$

= $\left[ \begin{matrix} 3\times 4+1\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) & 3\times 0+1\text{ }\!\!~\!\!\text{ }\times 2 \\ -2\times 4+2\text{ }\!\!~\!\!\text{ }\times \left( -1 \right) & -2\times 0+2\text{ }\!\!~\!\!\text{ }\times 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 12-1 & 0+2 \\ -8-2 & 0+4 \\ \end{matrix} \right]$

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right]$

The multiplicative inverse of $\left( \text{DA} \right)$ is calculated as:

${{\left( \text{DA} \right)}^{-1}}$= $\frac{Adj\text{ }\!\!~\!\!\text{ }\left( \text{DA} \right)}{\left| \text{DA} \right|}$

$\left( \text{DA} \right)$ = $\left[ \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ }\left( \text{DA} \right)$ = $\left[ \begin{matrix} 4 & -2 \\ 10 & 11 \\ \end{matrix} \right]$

$\left| \text{DA} \right|$ = $\left| \begin{matrix} 11 & 2 \\ -10 & 4 \\ \end{matrix} \right|$

= $11\times 4-2\times \left( -10 \right)$

= $44+20$

$\left| \text{DA} \right|$ = $64\ne 0$

Since $\left( \text{DA} \right)$ is non-singular therefore ${{\left( \text{DA} \right)}^{-1}}$ is possible

${{\left( \text{DA} \right)}^{-1}}$ = $\frac{\left[ \begin{matrix} 4 & -2 \\ 10 & 11 \\ \end{matrix} \right]}{64}$

= $\left[ \begin{matrix} \frac{4}{64} & \frac{-2}{64} \\ \frac{10}{64} & \frac{11}{64} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16} & \frac{-1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right]$

${{\left( \text{DA} \right)}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{16} & \frac{-1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right] \quad\quad\quad\quad\quad ...~(1)$

Solving Right Hand Side: ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$

The multiplicative inverse of matrix A is calculated as:

${{\text{A}}^{-1}}$ = $\frac{Adj~\text{A}}{\left| \text{A} \right|}$

A = $\left[ \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ A}$ = $\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]$

$\left| \text{A} \right|$ = $\left| \begin{matrix} 4 & 0 \\ -1 & 2 \\ \end{matrix} \right|$

= $4\times 2-0\times \left( -1 \right)$

= $8+0$

$\left| \text{A} \right|$ = $8\ne 0$

Since A is non-singular therefore ${{\text{A}}^{-1}}$ is possible

${{\text{A}}^{-1}}$ = $\frac{\left[ \begin{matrix} 2 & -0 \\ 1 & 4 \\ \end{matrix} \right]}{8}$

= $\left[ \begin{matrix} \frac{2}{8} & \frac{0}{8} \\ \frac{1}{8} & \frac{4}{8} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]$

Similarly, the multiplicative inverse of matrix D is calculated as:

${{\text{D}}^{-1}}$ = $\frac{Adj~\text{D}}{\left| \text{D} \right|}$

D = $\left[ \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ D}$ = $\left[ \begin{matrix} 2 & -1 \\ 2 & 3 \\ \end{matrix} \right]$

$\left| \text{D} \right|$ = $\left| \begin{matrix} 3 & 1 \\ -2 & 2 \\ \end{matrix} \right|$

= $3\times 2-1\times \left( -2 \right)$

= $6+2$

$\left| \text{D} \right|$ = \[8\ne 0\]

Since D is non-singular therefore ${{\text{D}}^{-1}}$ is possible

${{\text{D}}^{-1}}$ = \[\frac{\left[ \begin{matrix} 2 & -1 \\ 2 & 3 \\ \end{matrix} \right]}{8}\]

= $\left[ \begin{matrix} \frac{2}{8} & -\frac{1}{8} \\ \frac{2}{8} & \frac{3}{8} \\ \end{matrix} \right]$

${{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & -\frac{1}{8} \\ \frac{1}{4} & \frac{3}{8} \\ \end{matrix} \right]$

Now

${{\text{A}}^{-1}}{{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{4} & 0 \\ \frac{1}{8} & \frac{1}{2} \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{4} & \frac{-1}{8} \\ \frac{1}{4} & \frac{3}{8} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( \frac{1}{4}\times \frac{1}{4} \right)+\left( 0\times \frac{1}{4} \right) & \left( \frac{1}{4}\times \frac{-1}{8} \right)+\left( 0\times \frac{3}{8} \right) \\ \left( \frac{1}{8}\times \frac{1}{4} \right)+\left( \frac{1}{2}\times \frac{1}{4} \right) & \left( \frac{1}{8}\times \frac{-1}{8} \right)+\left( \frac{1}{2}\times \frac{3}{8} \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16}+0 & -\frac{1}{32}+0 \\ \frac{1}{32}+\frac{1}{8} & -\frac{1}{64}+\frac{3}{16} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{16} & -\frac{1}{32} \\ \frac{1+4}{32} & \frac{-1+12}{64} \\ \end{matrix} \right]$

${{\text{A}}^{-1}}{{\text{D}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{16} & -\frac{1}{32} \\ \frac{5}{32} & \frac{11}{64} \\ \end{matrix} \right]\quad\quad\quad\quad\quad ...~(2)$ 

From $(1)$ and $(2)$, it is clear that:

${{\left( \text{DA} \right)}^{-1}}$ = ${{\text{A}}^{-1}}{{\text{D}}^{-1}}$