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 $c_1, c_2, c_3$.

matlabCopy code
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:

matlabCopy code
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.

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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.