1. **Problem Statement:** Solve the systems of linear equations represented by the augmented matrices given in parts (a) and (b).
2. **Matrix (a):**
$$\begin{pmatrix} 1 & -3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
This corresponds to the system:
$$\begin{cases} x_1 - 3x_2 = 0 \\ x_3 = 0 \\ x_4 = 1 \end{cases}$$
3. **Solving (a):**
- From the second and third equations, we have directly:
$$x_3 = 0$$
$$x_4 = 1$$
- From the first equation:
$$x_1 = 3x_2$$
- Here, $x_2$ is a free variable.
4. **General solution for (a):**
$$\left(x_1, x_2, x_3, x_4\right) = \left(3t, t, 0, 1\right), \quad t \in \mathbb{R}$$
5. **Matrix (b):**
$$\begin{pmatrix} 1 & 1 & -2 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}$$
This corresponds to the system:
$$\begin{cases} x_1 + x_2 - 2x_3 - x_5 = 1 \\ x_3 + x_4 = 0 \\ x_4 = 0 \\ x_5 = 0 \end{cases}$$
6. **Solving (b):**
- From the third and fourth equations:
$$x_4 = 0$$
$$x_5 = 0$$
- Substitute $x_4=0$ into the second equation:
$$x_3 + 0 = 0 \implies x_3 = 0$$
- Substitute $x_3=0$ and $x_5=0$ into the first equation:
$$x_1 + x_2 - 0 - 0 = 1 \implies x_1 + x_2 = 1$$
- Here, $x_2$ is a free variable.
7. **General solution for (b):**
$$\left(x_1, x_2, x_3, x_4, x_5\right) = \left(1 - s, s, 0, 0, 0\right), \quad s \in \mathbb{R}$$
**Summary:**
- For (a), the solution set is $$\{(3t, t, 0, 1) : t \in \mathbb{R}\}$$
- For (b), the solution set is $$\{(1 - s, s, 0, 0, 0) : s \in \mathbb{R}\}$$
**Maple code to solve these systems:**
```maple
# For (a)
A := Matrix([[1, -3, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]]):
vars_a := [x1, x2, x3, x4]:
linsolve(A, vars_a);
# For (b)
B := Matrix([[1, 1, -2, 0, -1, 1], [0, 0, 1, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0]]):
vars_b := [x1, x2, x3, x4, x5]:
linsolve(B, vars_b);
```
This code uses `linsolve` to find the solution sets for the augmented matrices.
Solve Augmented 565D74
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