Finding $k$ such that the given vectors form a basis

Question

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:

$$ a_1 = \begin{bmatrix}k\\1-k\\k\end{bmatrix}, \quad a_2 = \begin{bmatrix}0\\3k-1\\2\end{bmatrix}, \quad a_3 = \begin{bmatrix}-k\\1\\0\end{bmatrix} $$

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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?

Answer 1

Consider $$\begin{vmatrix} k & 0 & -k \\ 1-k & 3k-1 & 1 \\ k & 2 & 0 \\ \end{vmatrix}$$

First, I will add the third column to the first column,

$$\begin{align}\begin{vmatrix} 0 & 0 & -k \\ 2-k & 3k-1 & 1 \\ k & 2 & 0 \\ \end{vmatrix}&=-k\begin{vmatrix} 2-k & 3k-1 \\k & 2\end{vmatrix}\\&=-k(4-2k-k(3k-1))\\ &=-k(4-2k-3k^2+k)\\&=k(3k^2+k-4)\\&=k(3k+4)(k-1)\end{align}$$

Hence the determinant is $0$ when $k \in \{ 0, -\frac{4}{3},1 \}$.

A santiy check is that if you consider the determinant, you will only obtain a cubic equation. We shouldn't have $4$ roots.

Answer 2

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.