For getting the inverse of the given matrix $A$ by row elementary operations we may write the given matrix as

$$A=I A$$

(i) $\because\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

$$\begin{array}{lll} \Rightarrow \quad {\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2+R_1\right]} \\ \Rightarrow \quad\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] A & {\left[\because R_3 \rightarrow R_3-R_2\right]} \end{array}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] A} & {\left[\because R_1 \rightarrow R_1+R_2\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2-3 R_1\right]} \end{array}$$

$$\begin{aligned} & \Rightarrow \quad\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] A \quad\left[\because R_2 \rightarrow R_2-3 R_1\right] \\ & \Rightarrow \quad\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A \quad\left[\begin{array}{l} \because R_1 \rightarrow R_1+R_2 \\ \text { and } R_3 \rightarrow-1 \cdot R_3 \end{array}\right] \end{aligned}$$

$$\begin{array}{llc} \Rightarrow & {\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow R_1+10 R_3 \\ \text { and } R_2 \rightarrow R_2+17 R_3 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow-1 R_1 \\ \text { and } R_2 \rightarrow-1 R_2 \end{array}\right]} \end{array}$$

So, the inverse of $A$ is $\left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$.

$$\begin{aligned} &\text { (ii) } \therefore\\ &\begin{aligned} & {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \\ \Rightarrow\quad & {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] A} \quad \left[\begin{array}{l} \because R_2 \rightarrow R_2+R_3 \\ \text { and } R_1 \rightarrow R_1-2 R_3 \end{array}\right]\\ \Rightarrow\quad & {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 2 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right] A}\quad \left[\because R_2 \rightarrow R_2+R_1\right] \end{aligned} \end{aligned}$$

Since, second row of the matrix $A$ on LHS is containing all zeroes, so we can say that inverse of matrix $A$ does not exist.

$$\begin{aligned} &\begin{aligned} \text { (iii) } \therefore \quad & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \\ \Rightarrow\quad & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 3 & 1 & 1 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \quad \left[\because R_2 \rightarrow R_2-R_1\right] \end{aligned}\\ \end{aligned}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 2 & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_2 \rightarrow R_2-R_1 \\ \text { and } R_3 \rightarrow R_3+R_1 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 4 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_3 \rightarrow R_3+R_1 \\ \text { and } R_2 \rightarrow R_2-\frac{1}{2} R_1 \end{array}\right]} \end{array}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} & {\left[\because R_3 \rightarrow R_3-2 R_1\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] A} & {\left[\because R_3 \rightarrow R_3-R_2\right]} \end{array}$$

$$ \begin{array}{lll} \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 5 & -2 & 2 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow \frac{1}{2} R_1 \\ \text { and } R_3 \rightarrow 2 R_3 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] A} \end{array} \quad\left[\begin{array}{l} \because \rightarrow R_1+\frac{1}{2} R_3 \\ \text { and } R_2 \rightarrow R_2-\frac{5}{2} R_3 \end{array}\right] .$$

Hence, $\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ is the inverse of given matrix $A$.

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} & A=\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right] \\ \therefore\quad & A^{\prime}=\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} \text{Now,}\quad & \frac{A+A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right] \\ \text{and}\quad & \frac{A-A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right] \end{aligned}$$

$$\begin{aligned} &\therefore \quad \frac{A+A^{\prime}}{2}+\frac{A-A^{\prime}}{2}=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]\\ &\text { which is the required expression. } \end{aligned}$$