1. **Solve the system by Gauss-Jordan elimination:** Given system: $$\begin{cases} x - y - z + 2p = 2 \\ 2x + y - 2z + p = 0 \\ -x + 2y - z - 4p = -1 \end{cases}$$ Form the augmented matrix: $$\left[ \begin{array}{cccc|c} 1 & -1 & -1 & 2 & 2 \\ 2 & 1 & -2 & 1 & 0 \\ -1 & 2 & -1 & -4 & -1 \end{array} \right]$$ 2. Make the first pivot 1 (already 1). Next, eliminate below and above: - R2 = R2 - 2*R1 \rightarrow (2-2*1,1-2*(-1),-2-2*(-1),1-2*2,0-2*2) = (0,3,0,-3,-4) - R3 = R3 + R1 \rightarrow (-1+1,2-1,-1-1,-4+2,-1+2) = (0,1,-2,-2,1) Matrix now: $$\left[ \begin{array}{cccc|c} 1 & -1 & -1 & 2 & 2 \\ 0 & 3 & 0 & -3 & -4 \\ 0 & 1 & -2 & -2 & 1 \end{array} \right]$$ 3. Make pivot in R2 column 2 a 1 by dividing R2 by 3: $$R2 = \frac{1}{3}R2 = (0,1,0,-1,-\frac{4}{3})$$ 4. Eliminate other entries in column 2: - R1 = R1 + R2 \rightarrow (1,-1+1,-1+0,2-1,2-4/3) = (1,0,-1,1,2/3) - R3 = R3 - R2 \rightarrow (0,1-1,-2-0,-2+1,1+4/3) = (0,0,-2,-1,7/3) Matrix now: $$\left[ \begin{array}{cccc|c} 1 & 0 & -1 & 1 & \frac{2}{3} \\ 0 & 1 & 0 & -1 & -\frac{4}{3} \\ 0 & 0 & -2 & -1 & \frac{7}{3} \end{array} \right]$$ 5. Make pivot in R3 column 3 a 1 by dividing R3 by -2: $$R3 = -\frac{1}{2} R3 = (0,0,1,\frac{1}{2}, -\frac{7}{6})$$ 6. Eliminate other entries in column 3: - R1 = R1 + R3 \rightarrow (1,0,0,1+\frac{1}{2}, \frac{2}{3}-\frac{7}{6}) = (1,0,0,\frac{3}{2}, -\frac{5}{6}) Final matrix: $$\left[ \begin{array}{cccc|c} 1 & 0 & 0 & \frac{3}{2} & -\frac{5}{6} \\ 0 & 1 & 0 & -1 & -\frac{4}{3} \\ 0 & 0 & 1 & \frac{1}{2} & -\frac{7}{6} \end{array} \right]$$ 7. Express variables: - From R1: $x + \frac{3}{2}p = -\frac{5}{6} \Rightarrow x = -\frac{5}{6} - \frac{3}{2}p$ - From R2: $y - p = -\frac{4}{3} \Rightarrow y = -\frac{4}{3} + p$ - From R3: $z + \frac{1}{2}p = -\frac{7}{6} \Rightarrow z = -\frac{7}{6} - \frac{1}{2}p$ Since $p$ is free, parametrize solution in terms of $p$: $$\boxed{ \left(x,y,z,p\right) = \left(-\frac{5}{6} - \frac{3}{2}p,-\frac{4}{3} + p,-\frac{7}{6} - \frac{1}{2}p,p\right) }$$ --- 8. **For part (b), determine number of solutions for:** $$\begin{cases} x + y = 1 \\ 3x + 3y = -5 \end{cases}$$ The second equation is $3$ times the first left side but right side differs ($3*1 = 3 \neq -5$), so equations are inconsistent. So, **there are no solutions.** --- 9. **Solve the homogeneous system:** $$\begin{cases} 3x - 2y - z + 3w = 0 \\ x - y + 2z - 2w = 0 \end{cases}$$ Form augmented matrix: $$\left[ \begin{array}{cccc|c} 3 & -2 & -1 & 3 & 0 \\ 1 & -1 & 2 & -2 & 0 \end{array} \right]$$ 10. Use row operations: - R1 = R1 - 3*R2: $$(3 - 3*1, -2 - 3*(-1), -1 - 3*2, 3 - 3*(-2), 0 - 3*0) = (0,1,-7,9,0)$$ New matrix: $$\left[ \begin{array}{cccc|c} 1 & -1 & 2 & -2 & 0 \\ 0 & 1 & -7 & 9 & 0 \end{array} \right]$$ 11. Eliminate term in R1 column 2: - R1 = R1 + R2: $$(1,-1+1,2-7,-2+9,0+0) = (1,0,-5,7,0)$$ Matrix now: $$\left[ \begin{array}{cccc|c} 1 & 0 & -5 & 7 & 0 \\ 0 & 1 & -7 & 9 & 0 \end{array} \right]$$ 12. Express $x,y$ in terms of free variables $z,w$: From R1: $$x - 5z + 7w = 0 \Rightarrow x = 5z - 7w$$ From R2: $$y - 7z + 9w = 0 \Rightarrow y = 7z - 9w$$ 13. Parametric solution: $$\boxed{ \left(x,y,z,w\right) = \left(5z - 7w, 7z - 9w, z, w\right) }$$ where $z,w$ are free parameters. --- Final answers: - (a) parametric solution with free $p$ - (b) no solutions - (c) parametric solution with free $z,w$