I'm currently attempting to solve the following problem:
Find the value(s) of $k$ such that the vectors $\{\vec{a}_1, \vec{a}_2, \vec{a}_3\}$ form a basis in $\mathbb{R}^3$, where:
$$
a_1 = \begin{bmatrix}k\\1-k\\k\end{bmatrix}, \quad
a_2 = \begin{bmatrix}0\\3k-1\\2\end{bmatrix}, \quad
a_3 = \begin{bmatrix}-k\\1\\0\end{bmatrix}
$$
To try and solve this, I first tried putting these vectors into an augmented matrix and attempted to reduce it:
$$
\begin{bmatrix}
k & 0 & -k \\
1-k & 3k-1 & 1 \\
k & 2 & 0 \\
\end{bmatrix} \to
\begin{bmatrix}
k & 0 & -k \\
0 & 3k-1 & 2-k \\
0 & 2 & k \\
\end{bmatrix} \to
\begin{bmatrix}
k & 0 &-k \\
0 & 3k-1 & 2-k \\
0 & 0 & \frac{(k-1)(3k+4)}{3k-1} \\
\end{bmatrix}
$$
Since the vectors are linearly independent, this matrix must have full rank, giving us:
$$k\not=0, 1, \frac{-4}{3} , \frac{1}{3}$$
However, out of these $k=\frac{1}{3}$ when substituted into the matrix, the rank is still 3. So, what did I do wrong? More importantly, how do I do this process of finding $k$ without making such errors? Did I miss any values of $k$? If any, how to find them?
Best Answer
You have divided by $3k-1$. You have to eliminate the case $3k-1=0$ before you do that. Check if you have done `division by zero' previously also. The determinant turns out to be $3k^3 + k^2 - 4k=k(k-1)(3k+4)$, so $\frac{1}{3}$ is a value for which the vectors form a basis. The other values are correct. Also, we can see that the determinant is a polynomial of degree 3 without computing it, so we can guess there can be at most 3 distinct values for which the vectors do not form a basis.