Question

Find all values of h and k such that the system
hx + 6y = 2
x + (h + 1)y = 2k
has (a) no solutions (b) a unique solution (c) innitely many solutions.

Answer 1

well this is a system of linear equations.
(a) it has no solutions if we have one equation is y=mx+b and the other is y=mx+(not b)
that is same slope and different y-intercept
(c) has infinitely many solutions if both equations can be written as y=mx+b
that is same everything (same slope and same y-intercept)
(b) has unique solution if the slopes are different

Answer 2

Thank you for the help. I'm aware of all those conditions though. Anything you could add that would help me out a bit more?

Answer 3

As a matrix, you would set it up as an augmented matix. If it is incositent then it has no solution. If you set it up with a pivot in each column of the left side, then it is unique solutions. If you have a pivotless column and therefore free variable it is infinite solutions.

Answer 4

I set it up as an augmented matrix, but I got stuck on reducing it to RREF. I want to get it down to
1 0 [bunch of stuff]
0 1 [bunch of stuff]

But I'm having trouble doing that. The h is confusing me when I want to get stuff back to 0.

Answer 5

The h will stay. You will be lookong for values of h that do 1 or 0 as needed.

Answer 6

Does that entail multiplying the rows by -h so I can subtract them?

Answer 7

You get it as close to rref as you can, then look for values to put in for h,k to finish the job, make it unworkable, etc.

Answer 8

Here is a similar problem, well, with just k to deal with in 3 equations, but shows the concept I am talking about:
http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.111688.html

Answer 9

Okay thanks. I'll try that and see what happens. <3

Answer 10

This is an exam with some answers/notes on a more similar question:
http://www.math.unl.edu/~bharbourne1/M314F05/M314FinalExamF2002Sols.pdf

Answer 11

I was looking for the theorem involved. Found this:

The Gauss-Jordan elimination procedure allows us to prove the following important theorem.

Theorem. A system of linear equations either has no solutions or has exactly one solution or has infinitely many solutions. A system of linear equations has infinitely many solutions if and only if its reduced row echelon form has free unknowns and the last column of the reduced row echelon form has no leading 1's. It has exactly one solution if and only if the reduced row echelon form has no free unknowns and the last column of the reduced row echelon form has no leading 1. It has no solutions if and only if the last column of the reduced row echelon form has a leading 1

On this site:
http://www.math.vanderbilt.edu/~msapir/msapir/jan10.shtml

That is what your answers are based on.

Answer 12

So for:

\(\left[ \begin{array}{cc|c}
1 & 0 & b_1\\
0 & 1 & b_2\\
\end{array}\right] \)

There is one and only one solution... so what h and k would do something like that?

\(\left[ \begin{array}{cc|c}
1 & 0 & b_1\\
0 & 0 & b_2\\
\end{array}\right] \) or \(\left[ \begin{array}{cc|c}
0 & 0 & b_1\\
0 & 1 & b_2\\
\end{array}\right] \)

It is inconsistent because you have a 0x or 0y = a number... which can't happen.

And for the infitinte, it is when you have a 0, 0 , 0 row so free variable.

Answer 13

I am heading out now, but I hope that gives you enough to work with.

Answer 14

My problem was just row reduction. But I think I got it now. Thanks again!