Use Gaussian elimination to solve the following sets of equations: (a) \(2 x-y-z=3\) \(5 x+7 y+3 z=10\) \(\quad-3 x-y+4 z=-17\) (b) \(2 \alpha-\beta+\gamma=11\) \(\frac{\alpha}{2}+\beta+2 \gamma=6\) \(3 \beta-7 \gamma=-37\) $$ \text { (c) } \begin{aligned} &i_{1}+2 i_{2}-3 i_{3}=2 \\ &-3 i_{1}+2 i_{2}+i_{3}=10 \\ &2 i_{1}-3 i_{3}=-9 \end{aligned} $$

Write the set of equations as an augmented matrix: $$ \left[\begin{array}{ccc|c} 2 & -1 & -1 & 3 \\ 5 & 7 & 3 & 10 \\ -3 & -1 & 4 & -17 \end{array}\right] $$

Apply row operations to transform the augmented matrix into row echelon form: 1. Swap Row 1 with Row 3 2. Multiply Row 1 by \(\frac{1}{3}\) 3. Subtract \(\frac{5}{2}\) times Row 1 from Row 2. Add Row 1 to Row 3 to get the new matrix: $$ \left[\begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{4}{3} & \frac{9}{3} \\ 0 & \frac{17}{2} & \frac{39}{6} & \frac{7}{2} \\ 0 & -1 & \frac{1}{3} & 2 \end{array}\right] $$ 4. Multiply Row 3 by \(-\frac{17}{2}\) and add it to Row 2. Divide Row 3 by \(-1\) to obtain: $$ \left[\begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{4}{3} & 3 \\ 0 & 1 & \frac{3}{2} & \frac{25}{17} \\ 0 & 0 & \frac{8}{3} & \frac{29}{3} \end{array}\right] $$ 5. Divide Row 3 by \(\frac{8}{3}\). Now matrix is in row echelon form: $$ \left[\begin{array}{ccc|c} 1 & \frac{1}{3} & -\frac{4}{3} & 3 \\ 0 & 1 & \frac{3}{2} & \frac{25}{17} \\ 0 & 0 & 1 & \frac{29}{8} \end{array}\right] $$

Perform back-substitution to solve for \(x\), \(y\), and \(z\): 1. \(z = \frac{29}{8}\) 2. \(y = \frac{25}{17} - \frac{3}{2} \cdot \frac{29}{8} = \frac{5}{8}\) 3. \(x = 3 + \frac{4}{3} \cdot \frac{29}{8} - \frac{1}{3} \cdot \frac{5}{8} = \frac{79}{24}\) Solution: \((x, y, z) = \left( \frac{79}{24}, \frac{5}{8}, \frac{29}{8} \right)\). We will perform a similar process for the remaining sets of equations (b) and (c).