Let {v1, v2, v3} = {[3 -3 -7 -2], [5 3 7 2], [4 3 7 2]}

If the set is linearly dependent, use disp or fprintf to express v1, v2, and v3 as a
non-trivial linear combination of the zero vector.

To express the vectors \( v_1, v_2, v_3 \) as a non-trivial linear combination of the zero vector, we need to check for linear dependence. The set \( \{v_1, v_2, v_3\} \) is **linearly dependent** if there exist scalars \( c_1, c_2, c_3 \) such that:

\[
c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3} = \mathbf{0}
\]

Where the zero vector \( \mathbf{0} \) is:

\[
\mathbf{0} = [0, 0, 0, 0]
\]

This equation translates to solving the following system of equations (one for each component):

\[
c_1 \times 3 + c_2 \times 5 + c_3 \times 4 = 0
\]
\[
c_1 \times (-3) + c_2 \times 3 + c_3 \times 3 = 0
\]
\[
c_1 \times (-7) + c_2 \times 7 + c_3 \times 7 = 0
\]
\[
c_1 \times (-2) + c_2 \times 2 + c_3 \times 2 = 0
\]

### MATLAB/Octave Code (using `disp` or `fprintf`)
You can use MATLAB to solve this system. Here’s how you can set it up using either `disp` or `fprintf`:

#### Step 1: Define the matrix of coefficients and solve for the scalars \( c_1, c_2, c_3 \).

```matlab
A = [3 5 4; -3 3 3; -7 7 7; -2 2 2];
b = [0; 0; 0; 0];

% Solve the homogeneous system
x = null(A, 'r');  % 'null' function will give the non-trivial solutions

% Display the result
disp('The non-trivial solution vector x is: ');
disp(x);
```

Alternatively, using `fprintf` for formatted output:

```matlab
A = [3 5 4; -3 3 3; -7 7 7; -2 2 2];
b = [0; 0; 0; 0];

% Solve the homogeneous system
x = null(A, 'r');

% Display the result using fprintf
fprintf('The non-trivial solution vector x is: \n');
fprintf('%f\n', x);
```

In the above code, `null(A, 'r')` finds a basis for the null space of \( A \), which gives us the linear dependence relations between \( v_1, v_2, \) and \( v_3 \). The non-trivial solution means there is a set of scalars \( c_1, c_2, c_3 \) that are not all zero and satisfy the linear combination that results in the zero vector.

### Explanation:
If \( x = [c_1, c_2, c_3]^T \) represents the solution vector, the system provides a combination of these vectors:

\[
c_1 v_1 + c_2 v_2 + c_3 v_3 = 0
\]

If the set of vectors is linearly dependent, the MATLAB solution will output a non-trivial solution for the scalars.

---

Would you like further details on any part of this? Here are a few related questions you might find interesting:

1. What is the geometric interpretation of linearly dependent vectors?
2. How does the `null` function in MATLAB work in finding solutions to a homogeneous system?
3. What are the necessary conditions for vectors to be linearly dependent?
4. How can I interpret the result of the null space in terms of vector spaces?
5. What are the implications of linear dependence in real-world applications?

**Tip:** In MATLAB, `null(A, 'r')` provides rational null space solutions, which is useful for detecting exact linear dependence in cases like this one.

Let {v1, v2, v3} = {[3 -3 -7 -2], [5 3 7 2], [4 3 7 2]}. If the set is linearly dependent, use disp or fprintf to express v1, v2, and v3 as a non-trivial linear combination of the zero vector.

Explore how to express the vectors {v1, v2, v3} = {[3 -3 -7 -2], [5 3 7 2], [4 3 7 2]} as a non-trivial linear combination of the zero vector. This problem involves key concepts in linear algebra, including linear dependence, homogeneous systems, and null space solutions, suitable for undergraduate-level study.

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