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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.

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    Get used to moving fast. It's just the way things are. At the level of mathematics that you are in, they expect you to keep up or get out. I've had many instructors with foreign accents. The trick is to study in such a way that you already know what they are talking about. It makes a huge difference. Since you don't know what 'consistent' means, my guess is you don't know your way around a text book. Your instructor isn't your problem. You might also want to ask your instructor if you can get copies of his notes after the lecture.2016-09-21

3 Answers 3

7

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 $c$.

2

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 $h=-4$.

1

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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    Ahh i see, i was$n$'t so far off. Tha$n$ks.2012-08-26