Determine for what parameter values 3 vectors are a base I am given the following vectors: 
$$
v_1 = (5-x,-1,-2)\\
v_2 = (-1,5-x,-2)\\
v_3 = (-2,-2,2-x)
$$
The task is to determine for what values for x the vectors constitutes a base in $R^3$.
My attempt: 
If they are not a base, I am able to form a linear combination of them equal to zero for a,b,c where at least one coefficient is not zero. So the following should be true.
$$
av_1 + bv_2 + cv_3 \neq 0\\
5a -xa - 3a -b + 5b - xb -2b -4c +2c -xc \neq 0\\
2a - xa +4b -xb -2c -xc \neq 0\\
a(2-x) + b(4-x) - c(2-x) \neq 0
$$
I am not sure how to get from here however. Explanations appreciated. Thanks in advance. 
 A: Take the determinant of the $3\times 3$ matrix formed by the three vectors that you have:
$$
\begin{pmatrix}
5-x & -1 & -2\\
-1 & 5-x & -2\\
-2 & -2 & 2-x\\
\end{pmatrix}
$$
For the vectors to form a basis, the determinant must be nonzero. Find the values of $x$ that make the determinant nonzero.
If you carry out the calculation, you will find that the determinant is
$$-x(x-6)^2,$$
 and it is equal to zero only when $x=0$ or $x=6$. So for any value of $x$ other than these two, you have a basis.
A: You are searching for a non-trivial (not all 0 scalars) linear combination of the given vectors that equal to the 0 vector. This is exactly what Gaussian elimination lets you find.
You can re-write your vectors as a single matrix, where each column of the matrix is one of your vectors. In your case this would be:
$$
        \begin{bmatrix}
        (5-x) & -1 & -2 \\
        -1 & (5-x) & -2 \\
        -2 & -2 & (2-x) \\
        \end{bmatrix}
$$
When you multiply this matrix by a vector, each component of the vector scales one the the matrix columns, and then the resulting three scaled vectors are added up to produce a final vector. In your case, this needs to be the 0 vector.
$$
        \begin{bmatrix}
        (5-x) & -1 & -2 \\
        -1 & (5-x) & -2 \\
        -2 & -2 & (2-x) \\
        \end{bmatrix}
        \begin{bmatrix}
        a \\
        b \\
        c \\
        \end{bmatrix}
=        \begin{bmatrix}
        0 \\
        0 \\
        0 \\
        \end{bmatrix}
$$
The above is equivalent to:
$$
        a\begin{bmatrix}
        (5-x)  \\
        -1 \\
        -2 \\
        \end{bmatrix}
    +
        b\begin{bmatrix}
        -1  \\
        (5-x) \\
        -2 \\
        \end{bmatrix}
    +
        c\begin{bmatrix}
        -2  \\
        -2 \\
        (2-x) \\
        \end{bmatrix}
    =
        \begin{bmatrix}
        0  \\
        0 \\
        0 \\
        \end{bmatrix}
$$
So what you're searching for is for the a,b,c vector. To do this you use simple Gaussian elimination on the matrix above augmented with the 0 vector. And then use elimination to find which values of x (if any) the system would have a solution.
