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I had a question from my homework Friday night which stumped me.

It reads, verbatim

Find all real values of $c$ such that the vectors $$ \begin{bmatrix}2\\c \\0 \end{bmatrix}, \begin{bmatrix}c\\c+3 \\0 \end{bmatrix} $$

are linearly independent.

I got stumped by this question; after trying to row reduce, I got

\begin{bmatrix} 2c &c^2 & |0 \\ 0& -\frac{1}{2}c^2+c+3& |0 \\ 0& 0& |0 \end{bmatrix}

I wasn't sure what to do at this point. I see the two quadratics; I'm just unsure of what to do and why.

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    $\begingroup$ Hint: two vectors $a,b \in \mathbb{R}^3$ are linearly independent if $a \times b \neq 0$ (cross product). $\endgroup$
    – arm1223
    Feb 21, 2021 at 21:02
  • $\begingroup$ The vectors are effectively $2$-vectors; just evaluate the two-by-two determinant. $\endgroup$
    – Lubin
    Feb 21, 2021 at 21:09

1 Answer 1

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Suppose the vectors are linearly dependent ie. $$ \begin{bmatrix}2\\c \\0 \end{bmatrix}=\lambda \begin{bmatrix}c\\c+3 \\0 \end{bmatrix} $$ with $\lambda \neq 0$. So $c^2 = 2(c+3)$ ie. $c=1 \pm \sqrt{7}$. And if $c=1 \pm \sqrt{7}$ the vectors are linearly dependent. So the vectors are linealy independent iff $c \in \Bbb R - \{1\pm \sqrt{7}\}$.

You can also use row-reduction: let $$ A=\begin{bmatrix}1 & c/2 \\ 0 & c+3-c^2/2 \end{bmatrix}, $$ and use the fact that the vectors are linearly independent iff $\ker A = \{0\}$.

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