Given the set of vectors in matrix form and taking into consideration this is for the homogeneous system:
$$
\left(\begin{matrix}
1 & -3 & 5 \\
-3 & 9 & -7 \\
2 & -6 & h \\
\end{matrix}\right)
$$
where $v_3$ is the column vector:
$$
\left(\begin{matrix}
5 \\
-7 \\
h \\
\end{matrix}\right)
$$
First off to determine if the vector is in the span all you have to do is check for what values make the system consistent correct? And to find what value h makes the system linearly dependent is to find the value of h where we have a free variable, or a row of all zeros? Here is my work and reasoning
$$
\left(\begin{matrix}
1 & -3 & 5 \\
-3 & 9 & -7 \\
2 & -6 & h \\
\end{matrix}\right)
$$
then perform the row operation $3r_1+r_2$ giving:
$$
\left(\begin{matrix}
1 & -3 & 5 \\
0 & 0 & 8 \\
2 & -6 & h \\
\end{matrix}\right)
$$
then perform the row operation: $r2 \leftrightarrow r_3$
$$
\left(\begin{matrix}
1 & -3 & 5 \\
2 & -6 & h \\
0 & 0 & 8 \\
\end{matrix}\right)
$$
then perform the row operation: $-2r_1+r_2$ giving:
$$
\left(\begin{matrix}
1 & -3 & 5 \\
0 & 0 &-10 + h\\
0 & 0 & 8 \\
\end{matrix}\right)
$$
This is where my work stopped: The book says that there are no values of $h$ that make the system consistent and that the system is linearly dependent for all $h$. Now for determining what value of $h$ is the vector in the span I have technically mastered that on my online hw. I got a 100% on all of those problems and then proceeded to do a mastery test where a 100% was needed in one go. My work and reasoning which has worked in the past for determining what value of h puts the vector in the span is:
(after all $-10 + h$ is just a constant)
$-10+h x_3 = 0$
dividing over $-10 + h$ gives:
$x3 = 0/-10+h$
where any value of $h$ gives a row of all zeros since the numerator is $0$ and $0$ divided by anything is zero.
For determining what value makes the system linearly dependent I was under the assumption it would be just be $10$ because $0=-10+h$ a value of $10$ would give the row all zeros but I could see where my above reasoning would just as well be appropriate for justifying that any value of h makes the system linearly dependent. Did I do something wrong in my calculations is my reasoning incorrect? where did I go wrong?
Please help as this is frustrating to think that I got by on luck all this time.
Best Answer
It is easiest to note that the second column is just $-3$ times the first, so the vectors are always linearly dependent.
For $v_3$ to be in the span of the other two is now just equivalent to being in the span of the first, so you are looking for $\alpha$ such that $\alpha v_1 = v_3$. Let's try to find such an $\alpha$:
$$ \alpha v_1 = v_3 \Rightarrow \alpha (1, - 3, 2) = (5, -7, h)\\ \Rightarrow \alpha = 5,\; \text{and} \; \alpha = \frac 73 $$ but this is a contradiction, so $v_3$ is always linearly independent of $v_1$ and $v_2$.
To see where you went wrong, just observe that in your final matrix, the second row is just a multiple of the third, so you can row reduce further, and you see that you can obtain a row of zeros for all values of $h$. That is subtract $\frac{-10 + h}{8}$ times the third row from the second if $h$ is not $0$, and if $h=0$ then as you said, $v_3$ is not in the span of the other two.