(i) Row-Reduction Steps to Find $A^{-1}$

(i) Row-Reduction Steps to Find $A^{-1}$

Form the augmented matrix $[A \mid I_3]$:

1. $R_1 \to \frac{1}{2} R_1$: Leading 1 in Row 1

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ -2 & 5 & 6 & 0 & 0 & 1 \end{array}\right]$$

2. $R_3 \to R_3 + 2R_1$: Clear Column 1 in Row 3

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ 0 & 11 & 3 & 1 & 0 & 1 \end{array}\right]$$

3. $R_2 \to -\frac{1}{2} R_2$: Leading 1 in Row 2

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 11 & 3 & 1 & 0 & 1 \end{array}\right]$$

4. $R_1 \to R_1 - 3R_2, \quad R_3 \to R_3 - 11R_2$: Clear Column 2

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 0 & -\frac{3}{2} & \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & 1 & \frac{11}{2} & 1 \end{array}\right]$$

5. $R_3 \to \frac{1}{3} R_3$: Leading 1 in Row 3

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 0 & -\frac{3}{2} & \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{1}{3} & \frac{11}{6} & \frac{1}{3} \end{array}\right]$$

6. $R_1 \to R_1 + \frac{3}{2} R_3$: Clear Column 3 $\implies [I_3 \mid A^{-1}]$

   $$\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & \frac{17}{4} & \frac{1}{2} \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{1}{3} & \frac{11}{6} & \frac{1}{3} \end{array}\right]$$