Let U = {[x, y, z]| x - 2y + z = 0, x + y - z = 0}. Which of the following is a basis for U?
Select one:
a. None of the other choices is correct
b. {[1, 2, 3]}
c. {[1, -1, 1]}
d. {[1, -1, 1], [0, 1, 1]}
e. {[1, 2, 3], [2, 4, 6]}

To determine a basis for \( U \), we first need to solve the system of linear equations defined by \( U \):

1. \( x - 2y + z = 0 \)
2. \( x + y - z = 0 \)

We can solve this system by eliminating variables. Let's subtract the second equation from the first:

\[
(x - 2y + z) - (x + y - z) = 0 - 0
\]

This simplifies to:

\[
-3y + 2z = 0 \quad \Rightarrow \quad 3y = 2z \quad \Rightarrow \quad y = \frac{2}{3}z
\]

Now, substitute \( y = \frac{2}{3}z \) into the second equation \( x + y - z = 0 \):

\[
x + \frac{2}{3}z - z = 0 \quad \Rightarrow \quad x + \frac{2}{3}z - \frac{3}{3}z = 0 \quad \Rightarrow \quad x - \frac{1}{3}z = 0 \quad \Rightarrow \quad x = \frac{1}{3}z
\]

Thus, the solution can be parameterized as:

\[
x = \frac{1}{3}z, \quad y = \frac{2}{3}z, \quad z = z
\]

This can be written in vector form as:

\[
[x, y, z] = \left[ \frac{1}{3}z, \frac{2}{3}z, z \right] = z \left[ \frac{1}{3}, \frac{2}{3}, 1 \right]
\]

The vector \(\left[ \frac{1}{3}, \frac{2}{3}, 1 \right]\) spans the solution space. To express this in a more familiar form, we can scale this vector by 3 to get:

\[
\left[ 1, 2, 3 \right]
\]

This means any vector in \( U \) is a scalar multiple of \([1, 2, 3]\), indicating that \( \{[1, 2, 3]\} \) is a basis for \( U \). Therefore, the correct answer is:

b. \(\{[1, 2, 3]\}\)

Would you like more details or have any questions? Here are some additional questions you might find interesting:

1. How do you determine the dimension of a vector space?
2. What is the null space of a matrix, and how do you find it?
3. Can you explain the concept of linear independence in vectors?
4. How do you find the rank of a matrix?
5. What is the difference between a vector space and a subspace?

**Tip:** To verify if a set of vectors forms a basis for a vector space, ensure the vectors are linearly independent and span the entire space.

Learn how to determine a basis for a vector space U by solving a system of linear equations. This problem covers linear algebra concepts, including linear independence and spanning sets. Suitable for advanced college-level students.

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