I'm currently attempting to solve the following problem:
Find the value(s) of $k$ such that the vectors $\{\vec{a}_1, \vec{a}_2, \vec{a}_3\}$ do not span $\mathbb{R}^3$, where:
$$
a_1 = \begin{bmatrix}-1\\k\\7\end{bmatrix}, \quad
a_2 = \begin{bmatrix}4\\-2\\5\end{bmatrix}, \quad
a_3 = \begin{bmatrix}1\\-7\\2\end{bmatrix}
$$
To try and solve this, I first tried putting these vectors into an augmented matrix and attempted to reduce it:
$$
\begin{bmatrix}
-1 & 4 & 1 \\
k & -2 & -7 \\
7 & 5 & 2 \\
\end{bmatrix} \to
\begin{bmatrix}
-1 & 4 & 1 \\
k & -2 & -7 \\
0 & 33 & 9 \\
\end{bmatrix} \to
\begin{bmatrix}
-1 & 4 & 1 \\
0 & -2 + 4k & -7 + k \\
0 & 33 & 9 \\
\end{bmatrix}
$$
From here, I thought that perhaps what I needed to do was add up the elements in the second row, and set it equal to zero (solve for $-2 + 4k -7 + k = 0$) so that I could find the values for $k$ where the second row is equal to zero and so could not span $\mathbb{R}^3$. I then arrived at the solution $k = \frac{9}{5}$, but that did not turn out to be the answer.
I then wondered if I needed to account for the 'variables' — if the matrix can be expressed as $\vec{a}_1 x_1 + \vec{a}_2 x_2 + \vec{a}_3 x_3 = \vec{b}$, then perhaps I need to do $(-2 + 4k)x_2 + (-7 + k)x_3 = 0$. If so, then I thought that the solutions might be $k=\frac{2x_2+7x_3}{4x_2+x_3}$, but that also was not the solution (and in any case, I think I'm expected to provide actual numbers, not a formula).
I did look at this problem, which is attempting to solve a similar solution except in $\mathbb{R}^2$. In that posted question, it's very obvious that $h$ must be any value other then $3$, but I'm not seeing any obvious values for $k$ in my problem.
What am I doing wrong?
Best Answer
How about trying this: set up a matrix $A$ which has columns $a_i$, $1 \le i \le 3$, and find the value(s) of $k$ such that $\det A = 0$; for such $k$, the three columns will be linearly dependent and hence cannot span $\Bbb R^3$. So we take
$A = \begin{bmatrix} -1 & 4 & 1 \\ k & -2 & -7 \\ 7 & 5 & 2 \end{bmatrix} \tag{1}$
and then an easy calculation shows that
$\det A = -3k -221; \tag{2}$
setting
$-3k - 221 = 0 \tag{3}$
reveals precisely one value of $k$ such that $\det A = 0$, namely
$k = -\dfrac{221}{3} = -73\dfrac{2}{3}. \tag{4}$
Since my method is generally the same as David Holden's, but my arithmetic is different, I offer that I evaluated $\det A$ using the "six arrows" technique or Sarrus' rule for $3 \times 3$ matrices (see here), obtaining
$\det A = -4 -4(49) + 5k + 14 -35 -8k = -4(1 + 49) -3k - 21 = -3k -221. \tag{5}$
Hope this helps! Cheers,
and as always,
Fiat Lux!!!