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

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Solution 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.

Solution 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.

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Updated on August 01, 2022

Comments

  • Truth-seek
    Truth-seek over 1 year

    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?

    • Paul
      Paul about 6 years
      Your reduction operations include dividing by $3k-1$, which is not valid when $k=1/3$.
    • Truth-seek
      Truth-seek about 6 years
      Yeah. I so that just now. Nevertheless, thn u very much
  • Truth-seek
    Truth-seek about 6 years
    Thn u. Thn u so much. But why does the determinant thing work?
  • Truth-seek
    Truth-seek about 6 years
    Thnx. Btw, why does the determinant thing work?
  • Siong Thye Goh
    Siong Thye Goh about 6 years
    A square matrix is nonsingular if and only if the determinant is non-zero.
  • S. Venkataraman
    S. Venkataraman about 6 years
    From the definition of rank, a $n\times n$ square matrix has rank $n$ if and only if the determinant is not zero. The rank is $n$ if and only if the $n$ columns are linearly independent.
  • S. Venkataraman
    S. Venkataraman about 6 years
    Sorry, that is not very clear. There are two ways of defining the rank of a matrix. One is the row(column) rank which is the dimension of the subspace spanned by the rows(columns). The other is determinantal rank which is the size of the largest non-zero minor. You want the column rank to be 3. The determinantal rank is 3 if there is a $3\times 3$ non-zero minor. The matrix is $3\times 3$, so this is just saying that the determinant is non-zero.
  • Truth-seek
    Truth-seek about 6 years
    Thank you for your help.