https://file.aiquickdraw.com/mathbot/file/1734274090341d7x4ikz6.jpg

Let's solve the given problems step-by-step.

### Given Matrix \( A \)
The transformation \( T: \mathbb{R}^5 \rightarrow \mathbb{R}^4 \) is defined by the matrix \( A \):
\[ A = \begin{bmatrix}
1 & -3 & 1 & 2 & 2 \\
-1 & 3 & 0 & -2 & -2 \\
1 & -3 & 3 & 3 & 4 \\
0 & 0 & 5 & -4 & -8 
\end{bmatrix} \]

The Reduced Row Echelon Form (RREF) of \( A \) is given as:
\[ \text{RREF}(A) = \begin{bmatrix}
1 & -3 & 0 & 0 & -2 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 
\end{bmatrix} \]

### Part (a): Solution Set of the Given System

The system of linear equations is:
\[
\begin{cases}
x_1 - 3x_2 + x_3 + 2x_4 + 2x_5 = 0 \\
-x_1 + 3x_2 + 0x_3 - 2x_4 - 2x_5 = 0 \\
x_1 - 3x_2 + 3x_3 + 3x_4 + 4x_5 = 0 \\
0x_1 + 0x_2 + 5x_3 - 4x_4 - 8x_5 = 0 
\end{cases}
\]

We use the RREF form to find the solutions:
\[
\begin{cases}
x_1 - 3x_2 - 2x_5 = 0 \\
x_3 = 0 \\
x_4 + 2x_5 = 0 
\end{cases}
\]

From the second and third equations, we have:
\[ x_3 = 0 \]
\[ x_4 = -2x_5 \]

Substitute \( x_3 = 0 \) and \( x_4 = -2x_5 \) into the first equation:
\[ x_1 - 3x_2 - 2x_5 = 0 \]
\[ x_1 = 3x_2 + 2x_5 \]

Thus, the solution set can be parameterized by \( x_2 \) and \( x_5 \):
\[ x_1 = 3x_2 + 2x_5 \]
\[ x_3 = 0 \]
\[ x_4 = -2x_5 \]

Therefore, the general solution is:
\[
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 
\end{pmatrix}
=
\begin{pmatrix}
3x_2 + 2x_5 \\
x_2 \\
0 \\
-2x_5 \\
x_5 
\end{pmatrix}
=
x_2
\begin{pmatrix}
3 \\
1 \\
0 \\
0 \\
0 
\end{pmatrix}
+
x_5
\begin{pmatrix}
2 \\
0 \\
0 \\
-2 \\
1 
\end{pmatrix}
\]

### Part (b): Spanning Set and Basis

The set \( S \) is:
\[ S = \left\{ \begin{pmatrix}
1 \\
-1 \\
1 \\
0 
\end{pmatrix}, 
\begin{pmatrix}
-3 \\
3 \\
-3 \\
0 
\end{pmatrix}, 
\begin{pmatrix}
1 \\
0 \\
3 \\
5 
\end{pmatrix}, 
\begin{pmatrix}
2 \\
-2 \\
3 \\
-4 
\end{pmatrix} \right\} \]

To determine if \( S \) spans \( \mathbb{R}^4 \), we need to check the rank of the matrix formed by these vectors as columns:
\[
B = \begin{bmatrix}
1 & -3 & 1 & 2 \\
-1 & 3 & 0 & -2 \\
1 & -3 & 3 & 3 \\
0 & 0 & 5 & -4 
\end{bmatrix}
\]

We see from the RREF of \( A \):
\[
\text{RREF}(A) = \begin{bmatrix}
1 & -3 & 0 & 0 & -2 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 
\end{bmatrix}
\]

Since the rank of \( A \) (and hence \( B \)) is 3 (there are 3 non-zero rows), the set \( S \) does not span \( \mathbb{R}^4 \) because \( \mathbb{R}^4 \) requires 4 linearly independent vectors.

Would you like further details or have any questions?

Here are some related questions for practice:
1. How do you determine the rank of a matrix?
2. What is the significance of the RREF of a matrix in solving linear systems?
3. How can you determine if a set of vectors spans a given space?
4. What does it mean for vectors to be linearly independent?
5. How do you find a basis for the null space of a matrix?

**Tip:** To check if a set of vectors spans a space, always consider the number of vectors and their linear independence.

Find the solution set of the given linear system and explain why the set of vectors does or does not span R^4.

This problem involves solving a linear system using the RREF of a matrix and determining whether a given set of vectors spans R^4. It explores concepts such as matrix rank, linear independence, and general solutions for systems of equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation