To determine if the vector $\mathbf{v} = \begin{bmatrix} 2 \\ -1 \\ 3 \\ 0 \end{bmatrix}$ is a linear combination of the vectors $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 3 \\ 2 \\ 0 \end{bmatrix}$, $\mathbf{v}_2 = \begin{bmatrix} 5 \\ 1 \\ 2 \\ 4 \end{bmatrix}$, and $\mathbf{v}_3 = \begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \end{bmatrix}$, we need to solve the following system of linear equations:

$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{v}$

This translates to the matrix equation:

$c_1 \begin{bmatrix} 1 \\ 3 \\ 2 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 5 \\ 1 \\ 2 \\ 4 \end{bmatrix} + c_3 \begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 3 \\ 0 \end{bmatrix}$

Or, more explicitly:

$\begin{bmatrix} 1 & 5 & -1 \\ 3 & 1 & 0 \\ 2 & 2 & 2 \\ 0 & 4 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 3 \\ 0 \end{bmatrix}$

Now, we solve this system using either Gaussian elimination or matrix inversion.

I'll proceed with solving it for you.The solution to the system of equations is approximately:

$c_1 \approx -0.0067, \quad c_2 \approx 0.2937, \quad c_3 \approx 0.5232$

However, the system has a non-zero residual of about 7.54, which indicates that the vector $\mathbf{v} = \begin{bmatrix} 2 \\ -1 \\ 3 \\ 0 \end{bmatrix}$ is not an exact linear combination of $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$.

Would you like a deeper explanation of these results or further details?

Here are some related questions you might explore:

1. What does it mean for a vector to be a linear combination of others?
2. How does Gaussian elimination work for solving linear systems?
3. What is the significance of the residual in a least squares solution?
4. Can we always solve any system of linear equations?
5. How does matrix rank affect the solvability of systems?

Tip: Non-zero residuals in least squares indicate the system is overdetermined or inconsistent.