College

**Solution Spaces of Homogeneous Linear Algebraic Systems**

Solve the linear systems given in Problems 60-62 and determine their solution spaces.

[tex] \text{60.} \begin{cases}
x_1 - x_2 + 4x_4 + 2x_5 - x_6 = 0 \\
2x_1 - 2x_2 + x_3 + 2x_4 + 4x_5 - x_6 = 0
\end{cases} [/tex]

Answer :

Let's solve the given system of linear equations to determine the solution space. The system is given as:

[tex]\[
\begin{array}{r}
x_1 - x_2 + 4x_4 + 2x_5 - x_6 = 0 \\
2x_1 - 2x_2 + x_3 + 2x_4 + 4x_5 - x_6 = 0
\end{array}
\][/tex]

To solve this system, we'll proceed by expressing some variables in terms of the others.

1. First Equation:
[tex]\[
x_1 - x_2 + 4x_4 + 2x_5 - x_6 = 0
\][/tex]
Rearrange it to express [tex]\(x_1\)[/tex] in terms of the other variables:
[tex]\[
x_1 = x_2 - 4x_4 - 2x_5 + x_6
\][/tex]

2. Second Equation:
[tex]\[
2x_1 - 2x_2 + x_3 + 2x_4 + 4x_5 - x_6 = 0
\][/tex]

Substitute [tex]\(x_1\)[/tex] from the first equation into the second equation:
[tex]\[
2(x_2 - 4x_4 - 2x_5 + x_6) - 2x_2 + x_3 + 2x_4 + 4x_5 - x_6 = 0
\][/tex]

3. Simplification:
[tex]\[
2x_2 - 8x_4 - 4x_5 + 2x_6 - 2x_2 + x_3 + 2x_4 + 4x_5 - x_6 = 0
\][/tex]
Combine like terms:
[tex]\[
-6x_4 + x_6 + x_3 = 0
\][/tex]

Rearrange to express [tex]\(x_3\)[/tex] in terms of [tex]\(x_4\)[/tex] and [tex]\(x_6\)[/tex]:
[tex]\[
x_3 = 6x_4 - x_6
\][/tex]

Thus, we have obtained:
[tex]\[
x_1 = x_2 - 4x_4 - 2x_5 + x_6
\][/tex]
[tex]\[
x_3 = 6x_4 - x_6
\][/tex]

So, the solution space can be expressed in terms of free variables [tex]\(x_2, x_4, x_5, x_6\)[/tex]. In summary:
- [tex]\(x_1 = x_2 - 4x_4 - 2x_5 + x_6\)[/tex]
- [tex]\(x_3 = 6x_4 - x_6\)[/tex]
- [tex]\(x_2, x_4, x_5, x_6\)[/tex] are free variables.

Therefore, the solution set of the homogeneous system is given by the following parameterization:
[tex]\[
\{(x_1, x_2, x_3, x_4, x_5, x_6) \mid x_1 = x_2 - 4x_4 - 2x_5 + x_6, x_3 = 6x_4 - x_6, x_2, x_4, x_5, x_6 \in \mathbb{R}\}
\][/tex]

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