I'm just beginning linear algebra at university and have a teacher who moves very fast and has pre-done slides so i can't actually see the problem worked out, he just talks it out. On top of this, he's also from China and heavily accented, making him hard to understand.
Anyway, i have an augmented matrix, and i want the values of $h$ that make it consistent:
$$\left[\begin{array}{cc|c} 1 & h & -5 \\ 2 & -8 & 6 \\ \end{array}\right]$$
and quite frankly, i'm not sure just how to start. I tried eliminating the 1 in the second row, but that made the second line $[0\;\;\; h+4\;\; -8]$ and i'm not even sure if that's the right direction or even allowed.
Thanks in advance.
$\endgroup$ 43 Answers
$\begingroup$A linear system is inconsistent is if it represents a contradiction, for instance the system
$$\left[\begin{array}{cc|c} 0 & 0 & -10 \\ 3 & -2 & 1 \\ \end{array}\right]$$
is inconsistent because the first line represents a linear equation $0x+0y=-10$, i.e. $0=-10$, which is a contradiction. Geometrically, when you solve a 2x2 linear system, you are finding the intersection between a pair of lines. If you reach a contradiction, like the system above, then your lines do not intersect, i.e. they must be parallel.
If you are being asked this question, you have probably already covered Gauss-Jordan ellimination. Inconsistencies in linear systems can be readily identified if the system is brought to reduced row echelon form (can you see why?), so I would start with that. The steps are simple:
$$\left[\begin{array}{cc|c} 1 & h & -5 \\ 2 & -8 & 6 \\ \end{array}\right]$$ Multiply the second row by $1/2$: $$\left[\begin{array}{cc|c} 1 & h & -5 \\ 1 & -4 & 3 \\ \end{array}\right]$$ Subtract the second row from the first: $$\left[\begin{array}{cc|c} 0 & h+4 & -8 \\ 1 & -4 & 3 \\ \end{array}\right]$$ Without even proceeding further, it is obvious that one way for the system to be inconsistent is if the first line is $0\, 0\, |\, -8$, since this would be equivalent to saying $0x+0y=-8$, that is $0=-8$, a contradiction. The first row would have this form only if $h=-4$, so $h=-4$ makes the system inconsistent.
Now it is pretty clear at this point that no other value of $h$ would make the system inconsistent, and after you are comfortable with Gauss-Jordan elimination this fact would be apparent to you as well, though you should really try to understand why first. So let's say $h\ne-4$. Then we can multiply the first row by $\frac 1 {h+4}$: $$\left[\begin{array}{cc|c} 0 & 1 & -\frac 8 {h+4} \\ 1 & -4 & 3 \\ \end{array}\right]$$ And now add 4 times the first row to the second: $$\left[\begin{array}{cc|c} 0 & 1 & -\frac 8 {h+4} \\ 1 & 0 & 3-\frac {32} {h+4} \\ \end{array}\right]$$ To really be precise, you can swap the two rows: $$\left[\begin{array}{cc|c} 1 & 0 & 3-\frac {32} {h+4} \\ 0 & 1 & -\frac 8 {h+4} \\ \end{array}\right]$$
Thus for any value of $h$ other than $-4$, we can solve the system - there is no way to make the system displayed above have a row which looks like $0\,0\,|\,c$, for any non-zero number $c$.
$\endgroup$ $\begingroup$A simpler solution is based on a theorem that any system $Ax = b$ is consistent iff rank of $[A \mid b]$ equal to rank of $A$.
To compute rank of $A$ perform elimination on $A$ to get: $ \pmatrix{1 & 0 \\ 0 & -8-2h} $ Hence $\text{rank}(A)=1$ if $h = -4$ and $\text{rank}(A) = 2$ otherwise.
To compute rank of $[A \mid b]$ perform elimination on $[A \mid b]$ to get: $$ \pmatrix{1 & 0 & \frac{3h-20}{h+4} \\ 0 & 1 & \frac{-8}{h+4}} $$ So for values other that $h=-4$ we have $\text{rank}([A \mid b]) = 2$.
Comparing two ranks, we have a consistent system other than $h=-4$.
$\endgroup$ $\begingroup$So it is consistent whenever there is at least one solution. That means that the two lines you have cannot be parallel to each other. Multiply the first row by $2$, and you get $[2, 2h, -10]$. The lines will be parallel for the equations $m_1x+n_1y=a$ and $m_2x+n_2y=b$ if $m_1=m_2$ and $n_1=n_2$. In this case, $m_1=2, m_2=2, n_1=2h, n_2=-8$. Since $m_1 = m_2, n_1 \neq n_2$, so $2h \neq -8, h\neq-4$.
Note that there are an infinite number of solutions (aka consistent) if $m_1=m_2, n_1=n_2,$ and $a=b$. Otherwise, you do not have to worry about the $a$ and $b$ values.
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