Solve the following system using an augmented matrix. If this is dependent, write
using parameters. Write as a matrix and also as a sum of vectors.
x1 – 2x2 + x4 = 3
-x1 + x2 – x3 – 8x4 = 1

To solve the given system of equations using an augmented matrix, let's first rewrite the system clearly:

1. \( x_1 - 2x_2 + x_4 = 3 \)
2. \( -x_1 + x_2 - x_3 - 8x_4 = 1 \)

### Step 1: Represent the system as an augmented matrix.

We will express the coefficients of the variables and the constants in an augmented matrix form:

\[
\begin{pmatrix}
1 & -2 & 0 & 1 & | & 3 \\
-1 & 1 & -1 & -8 & | & 1
\end{pmatrix}
\]

### Step 2: Row reduction (Gaussian elimination).

We aim to simplify this matrix into row-echelon form using elementary row operations. The matrix is:

\[
\begin{pmatrix}
1 & -2 & 0 & 1 & | & 3 \\
-1 & 1 & -1 & -8 & | & 1
\end{pmatrix}
\]

- Add row 1 to row 2 to eliminate the first element in row 2:

\[
R_2 = R_2 + R_1
\]

\[
\begin{pmatrix}
1 & -2 & 0 & 1 & | & 3 \\
0 & -1 & -1 & -7 & | & 4
\end{pmatrix}
\]

- Now, multiply row 2 by \(-1\) to simplify:

\[
R_2 = -R_2
\]

\[
\begin{pmatrix}
1 & -2 & 0 & 1 & | & 3 \\
0 & 1 & 1 & 7 & | & -4
\end{pmatrix}
\]

### Step 3: Back substitution.

Now that the matrix is in row-echelon form, we can convert it back to equations:

1. From the first row: 
\[
x_1 - 2x_2 + x_4 = 3
\]

2. From the second row:
\[
x_2 + x_3 + 7x_4 = -4
\]

### Step 4: Solving the system.

We can express \( x_1 \) and \( x_2 \) in terms of free variables \( x_3 \) and \( x_4 \).

From the second equation:
\[
x_2 = -4 - x_3 - 7x_4
\]

Substitute this into the first equation:
\[
x_1 - 2(-4 - x_3 - 7x_4) + x_4 = 3
\]
\[
x_1 + 8 + 2x_3 + 14x_4 + x_4 = 3
\]
\[
x_1 = 3 - 8 - 2x_3 - 15x_4
\]
\[
x_1 = -5 - 2x_3 - 15x_4
\]

### Step 5: Write the solution using parameters.

Since \( x_3 \) and \( x_4 \) are free variables, we can parametrize the solution:

\[
x_1 = -5 - 2x_3 - 15x_4
\]
\[
x_2 = -4 - x_3 - 7x_4
\]
\[
x_3 = x_3 \quad \text{(free)}
\]
\[
x_4 = x_4 \quad \text{(free)}
\]

### Step 6: Write the solution as a matrix and as a sum of vectors.

The parametric solution can be written as:

\[
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{pmatrix}
=
\begin{pmatrix}
-5 \\
-4 \\
0 \\
0
\end{pmatrix}
+
x_3
\begin{pmatrix}
-2 \\
-1 \\
1 \\
0
\end{pmatrix}
+
x_4
\begin{pmatrix}
-15 \\
-7 \\
0 \\
1
\end{pmatrix}
\]

This represents the solution as a linear combination of vectors.

---

Would you like further details on any step? Here are some related questions:

1. How do elementary row operations affect the solution set of a system?
2. How can we tell if a system is dependent by looking at its augmented matrix?
3. What is the geometric interpretation of the solution as a sum of vectors?
4. Can this method be generalized to larger systems?
5. How does the number of free variables relate to the rank of a matrix?

**Tip:** When dealing with systems of equations, free variables correspond to the number of degrees of freedom in the solution, giving insight into whether the system has infinite solutions.

Solve the following system using an augmented matrix. If this is dependent, write using parameters. Write as a matrix and also as a sum of vectors.
x1 – 2x2 + x4 = 3
-x1 + x2 – x3 – 8x4 = 1

Learn how to solve the system of linear equations x1 – 2x2 + x4 = 3 and -x1 + x2 – x3 – 8x4 = 1 using an augmented matrix. This step-by-step solution covers Gaussian elimination, parametric solutions, and writing the final answer as a sum of vectors. Suitable for students in Grades 10-12 and early college.

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