1. **State the problem:** Solve the system of equations using Gauss-Jordan elimination: $$\begin{cases} -y + 2z - w = -1 \\ 2x + y - 2z - 2w = -2 \\ -x + 2y - 4z + w = 1 \\ 3x - 3w = -3 \end{cases}$$ 2. **Rewrite the system with all variables aligned:** $$\begin{cases} 0x - y + 2z - w = -1 \\ 2x + y - 2z - 2w = -2 \\ -x + 2y - 4z + w = 1 \\ 3x + 0y + 0z - 3w = -3 \end{cases}$$ 3. **Form the augmented matrix:** $$\left[\begin{array}{cccc|c} 0 & -1 & 2 & -1 & -1 \\ 2 & 1 & -2 & -2 & -2 \\ -1 & 2 & -4 & 1 & 1 \\ 3 & 0 & 0 & -3 & -3 \end{array}\right]$$ 4. **Swap row 1 and row 2** (to get a leading coefficient in first row): $$\left[\begin{array}{cccc|c} 2 & 1 & -2 & -2 & -2 \\ 0 & -1 & 2 & -1 & -1 \\ -1 & 2 & -4 & 1 & 1 \\ 3 & 0 & 0 & -3 & -3 \end{array}\right]$$ 5. **Make leading coefficient in row 1 equal to 1 by dividing row 1 by 2:** $$R1 \to \frac{1}{2} R1 = \left[1, \frac{1}{2}, -1, -1, -1\right]$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -1 & -1 & -1 \\ 0 & -1 & 2 & -1 & -1 \\ -1 & 2 & -4 & 1 & 1 \\ 3 & 0 & 0 & -3 & -3 \end{array}\right]$$ 6. **Eliminate x terms in rows 3 and 4:** - Row 3: $R3 + R1$ to eliminate $-1$ in $x$ position: $$R3 = (-1 + 1, 2 + \frac{1}{2}, -4 + (-1), 1 + (-1), 1 + (-1)) = (0, \frac{5}{2}, -5, 0, 0)$$ - Row 4: $R4 - 3 \times R1$ to eliminate 3 in $x$ position: $$R4 = (3 - 3 \times 1, 0 - 3 \times \frac{1}{2}, 0 - 3 \times (-1), -3 - 3 \times (-1), -3 - 3 \times (-1)) = (0, -\frac{3}{2}, 3, 0, 0)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -1 & -1 & -1 \\ 0 & -1 & 2 & -1 & -1 \\ 0 & \frac{5}{2} & -5 & 0 & 0 \\ 0 & -\frac{3}{2} & 3 & 0 & 0 \end{array}\right]$$ 7. **Make leading coefficient in row 2 equal to 1 by multiplying row 2 by -1:** $$R2 \to -1 \times R2 = (0, 1, -2, 1, 1)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -1 & -1 & -1 \\ 0 & 1 & -2 & 1 & 1 \\ 0 & \frac{5}{2} & -5 & 0 & 0 \\ 0 & -\frac{3}{2} & 3 & 0 & 0 \end{array}\right]$$ 8. **Eliminate y terms in rows 1, 3, and 4 using row 2:** - Row 1: $R1 - \frac{1}{2} R2$: $$R1 = (1, 0, -1 + \frac{1}{2} \times 2, -1 - \frac{1}{2} \times 1, -1 - \frac{1}{2} \times 1) = (1, 0, 0, -\frac{3}{2}, -\frac{3}{2})$$ - Row 3: $R3 - \frac{5}{2} R2$: $$R3 = (0, 0, -5 + \frac{5}{2} \times 2, 0 - \frac{5}{2} \times 1, 0 - \frac{5}{2} \times 1) = (0, 0, 0, -\frac{5}{2}, -\frac{5}{2})$$ - Row 4: $R4 + \frac{3}{2} R2$: $$R4 = (0, 0, 3 - \frac{3}{2} \times 2, 0 + \frac{3}{2} \times 1, 0 + \frac{3}{2} \times 1) = (0, 0, 0, \frac{3}{2}, \frac{3}{2})$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 0 & 0 & -\frac{3}{2} & -\frac{3}{2} \\ 0 & 1 & -2 & 1 & 1 \\ 0 & 0 & 0 & -\frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & 0 & \frac{3}{2} & \frac{3}{2} \end{array}\right]$$ 9. **Simplify rows 3 and 4 by dividing by their leading coefficients:** - Row 3: $R3 \to -\frac{2}{5} R3 = (0, 0, 0, 1, 1)$ - Row 4: $R4 \to \frac{2}{3} R4 = (0, 0, 0, 1, 1)$ Both rows 3 and 4 are identical, so row 4 can be removed or ignored. Matrix now: $$\left[\begin{array}{cccc|c} 1 & 0 & 0 & -\frac{3}{2} & -\frac{3}{2} \\ 0 & 1 & -2 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right]$$ 10. **Use row 3 to eliminate $w$ terms in rows 1 and 2:** - Row 1: $R1 + \frac{3}{2} R3$: $$R1 = (1, 0, 0, 0, -\frac{3}{2} + \frac{3}{2} \times 1) = (1, 0, 0, 0, 0)$$ - Row 2: $R2 - R3$: $$R2 = (0, 1, -2, 0, 1 - 1) = (0, 1, -2, 0, 0)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right]$$ 11. **Express variables from the matrix:** - From row 1: $x = 0$ - From row 3: $w = 1$ - From row 2: $y - 2z = 0 \Rightarrow y = 2z$ 12. **Since $z$ is free, let $z = t$ (parameter). Then:** $$x = 0, \quad y = 2t, \quad z = t, \quad w = 1$$ **Final solution:** $$\boxed{(x, y, z, w) = (0, 2t, t, 1) \text{ where } t \in \mathbb{R}}$$ This means infinitely many solutions depending on parameter $t$. **Slug:** gauss-jordan-system **Subject:** algebra **Desmos:** {"latex":"0x - y + 2z - w = -1","features":{"intercepts":true,"extrema":true}} **q_count:** 1