# Determining the value of $h$ that makes a linear system consistent.

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.

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Is it clear to you what "consistent" means, or even what we mean by an "augmented matrix"? –  akkkk Aug 26 '12 at 22:13
I don't see the relevance of the ethnicity of your professor. I would wager he spends time to create those slides so that you read them. –  James S. Cook Aug 26 '12 at 23:02
I don't see the relevance of looking so much into it, I was just stating that I felt behind because he was from another country and heavily accented and I was having a hard time keeping up. The only difference was that I off-handedly mentioned where he was from. –  BMEdwards37 Sep 11 '12 at 16:17

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 subtract 4 times the first row from 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 $c$.

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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 $h=-4$.

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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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Wow, I was going in the complete wrong direction. I'm probably going to have another question here soon, but let me look at it first now after seeing this and i'll see if it helps, these variables are really throwing me off. Thank you though, and keep an eye out for another question here in about 20 minutes :p –  BMEdwards37 Aug 26 '12 at 22:28
@BMEdwards37 Sorry I messed up with the negatives, but i fixed my answer. You can think of an augmented matrix as a system of equations, where the each column represents a different dimension, and the last column represents a constant. I was wrong in using "x" and "y" as dimensions. Usually, people would use $X_1, X_2, X_3...X_n$ –  mathguy Aug 26 '12 at 22:30
@BMEdwards37 Wait, I want to let you know that you are not going in the wrong direction. You ended up with [0 h+4 -8], which is good! The only way to make sure a system is consistent is that ALL of the values cannot equal 0 (nevermind the last column). In your example, since the first column is 0, then the second column CANNOT be 0, so $h+4\neq0$. If all of the columns except the last equal 0, then there are no solutions. If all columns including the last equal 0, then there are infinite solutions =) –  mathguy Aug 26 '12 at 22:35
I understood your meaning, didn't catch the negatives either. I'm going to post another question because i'm not even sure what it's asking, i appreciate the help. –  BMEdwards37 Aug 26 '12 at 22:37
Ahh i see, i wasn't so far off. Thanks. –  BMEdwards37 Aug 26 '12 at 22:38