Solution:

(i) A = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

The multiplicative inverse of matrix A is calculated as:

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

$\text{A}$ = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

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

$\left| \text{A} \right|$ = $\left| \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right|$

$\left| \text{A} \right|$ = $\left( -1 \right)\times 0-2\times 3$

$\left| \text{A} \right|$ = $0-6$

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

Since A is a non-singular matrix therefore solution is possible

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

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

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

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

(ii) B = $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

The multiplicative inverse of matrix B is calculated as:

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

$\text{B}$ = $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

$\left| B \right|$ = $\left| \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right|$

$\left| B \right|$ = $1\times \left( -5 \right)-2\times \left( -3 \right)$

$\left| B \right|$ = $-5+6$

$\left| B \right|$ = $1\ne 0$

Since B is a non-singular matrix therefore solution is possible

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]}{1}$

= $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} -5 & -2 \\ 3 & 1 \\ \end{matrix} \right]$

(iii) C = $\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]$

The multiplicative inverse of matrix C is calculated as:

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

$\text{C}$ = $\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]$

$Adj~\text{C}$ = $\left[ \begin{matrix} -9 & -6 \\ -3 & -2 \\ \end{matrix} \right]$

$\left| C \right|$ = $\left| \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right|$

$\left| C \right|$ = $\left( -2 \right)\times \left( -9 \right)-6\times 3$

$\left| C \right|$ = $18-18$

$\left| C \right|$ = $0$

Since C is a singular matrix therefore solution is not possible.

Therefore, ${{\text{C}}^{-1}}$ does not exist

(iv) D =$\left[ \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right]$

The multiplicative inverse of matrix D is calculated as:

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

$\text{D}$ =$\left[ \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right]$

$Adj\text{ }\!\!~\!\!\text{ D}$ = $\left[ \begin{matrix} 2 & \frac{-3}{4} \\ -1 & \frac{1}{2} \\ \end{matrix} \right]$

$\left| D \right|$ = $\left| \begin{matrix} \frac{1}{2} & \frac{3}{4} \\ 1 & 2 \\ \end{matrix} \right|$

$\left| D \right|$ = $\frac{1}{2}\times 2-\frac{3}{4}\times 1$

$\left| D \right|$ = $1-\frac{3}{4}$

$\left| D \right|$ = $\frac{4-3}{4}$

$\left| D \right|$ = $\frac{1}{4}\ne 0$

Since D is a non-singular matrix therefore solution is possible.

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

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

= $\left[ \begin{matrix} 2~\times 4 & \frac{-3}{4}~\times 4 \\ -1~\times 4 & \frac{1}{2}~\times 4 \\ \end{matrix} \right]$

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