# Thread: Basic Linear Algebra Problem

1. ## Basic Linear Algebra Problem

I'm just starting out with linear algebra, and in my book there is an exercise I don't really get. Until now all I have learned is to solve simple linear systems.

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

I've been messing around with it for a while but there just seems to be too many unknowns. What's the obvious thing I am missing here? Any tips on how to solve this type of problem?

2. Originally Posted by TwoPlusTwo
I'm just starting out with linear algebra, and in my book there is an exercise I don't really get. Until now all I have learned is to solve simple linear systems.

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

I've been messing around with it for a while but there just seems to be too many unknowns. What's the obvious thing I am missing here? Any tips on how to solve this type of problem?

Apply Gauss reduction algorithm to your matrix (Line 2 - 3 times Line 1) and get

$\displaystyle \begin{pmatrix}1&h&4\\0&6-3h&-4\end{pmatrix}$

From the above we get at once that the matrix represents a consistent linear system iff $\displaystyle 6-3h\neq 0$ , so...

Tonio

3. It might be a bit late for me but I think you want to say

$\displaystyle x+hy=4$ ...(1)
$\displaystyle 3x+6y=8$ ...(2)

$\displaystyle x=4-hy \implies 3(4-hy)+6y=8$
$\displaystyle 12-3hy+6y=8$
$\displaystyle 12-3y(h-2)=8$
$\displaystyle -3y(h-2)=-4$
$\displaystyle h-2=\frac{4}{3y}$
$\displaystyle h=\frac{4}{3y}+2$

4. Originally Posted by pickslides
It might be a bit late for me but I think you want to say

$\displaystyle x+hy=4$ ...(1)
$\displaystyle 3x+6y=8$ ...(2)

$\displaystyle x=4-hy \implies 3(4-hy)+6y=8$
$\displaystyle 12-3hy+6y=8$
$\displaystyle 12-3y(h-2)=8$
$\displaystyle -3y(h-2)=-4$
$\displaystyle h-2=\frac{4}{3y}$
$\displaystyle h=\frac{4}{3y}+2$
This gives you the same answer as tonio; $\displaystyle h \neq 2$ but $\displaystyle h$ can be anything else. Say you wanted $\displaystyle h$ to be some number $\displaystyle x$, then set $\displaystyle y={4}{3(x-2)}$. Clearly you can't choose x=2, but you can choose anything else.

Also, you may want to re-read rule 14 of the forum rules...

5. Thanks guys. Let me see if I can solve the next problem. I apply Gauss elimination and get the second line to say:

0 (2h+4) 0

Now, does this mean that h has to equal -2?

6. No. Since you got a 0 in the rightmost position, this augmented matrix represents a system which is always consistent.

All you need to know is the following:

After reducing your matrix to an echelon form, the last nontrivial row tells you if your system is consistent or inconsistent.

If it looks like 0 0 0 ... 0 a where a is not zero, then it is inconsistent. Otherwise it is consistent.

In particular, if the last entry is 0, then it's definately consistent.

In your last example, if h happens to be -2, then you get infinitely many solutions (because there is a free variable). If h is not -2, there is a unique solution.

7. Originally Posted by Swlabr

Also, you may want to re-read rule 14 of the forum rules...
Hi Swlabr you have misunderstood my post.

When I said "It might be a bit late for me" I actually meant that because it was 1AM and I was tired. Not late as in "someone else has answered first".

My apologies to TwoPlusTwo, tonio and yourself.

If you consider the time difference in mine and tonio's post you will notice there was nothing untoward meant in my post.