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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Cuntryboner on January 06, 2010, 10:16:42 am

Title: Holiday Help
Post by: Cuntryboner on January 06, 2010, 10:16:42 am: Hi all, I am having difficulty with Systems of simultaneous equations, using matrices.
The question goes:
for what values of m, does
mx+2y=8 and 4x-(2-m)y=2m
have
1. No solutions
2. Infinitely no solutions
I have acquired the correct answer for part 1, but I always get stuck for finding the values of m to make the two equations the same line.
make things quicker
x = 4/(m+2) and y = 2(m+4)/(m+2)
Any help will be greatly appreciated
CB
Title: Re: Holiday Help
Post by: Stroodle on January 06, 2010, 10:43:06 am: Hi there.

I skipped the chapter on matrices, but there are infinitely many solutions when both equations are the same. If you equate the coefficients (i think that's what it's called) after arranging both equations into the same form (e.g. arrange them both into the form of ax+by=c) you can see that both equations are the same where m=4.

Hope that helps, and that I'm not just stating the obvious...

*edit - assuming you meant "infinitely many solutions", not "infinitely no solutions" :)
Title: Re: Holiday Help
Post by: brightsky on January 06, 2010, 11:17:13 am: What book are you using? I don't think I've done that before..

I am lost at how to incoorperate matrices in it, but this site may help: http://www.onlinemathlearning.com/simultaneous-equations-matrices.html
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Without using matrices, this is what I got (prob wrong though)
a.

$mx + 2y = 8 \implies y = \frac {-mx + 8}{2}$ so gradient is equal to $-\frac{m}{2}$

$4x - (2 - m)y = 2m \implies y = \frac {4x -2m}{2-m}$ so gradient is equal to $\frac {4}{2 - m}$

Simultaneous equations have no solutions when the two are parallel, i.e. have the same gradient.

So the simultaneous equation has no solutions when:

$-\frac{m}{2} = \frac{4}{2-m} \implies m = -2$ or $m = 4$
Title: Re: Holiday Help
Post by: Stroodle on January 06, 2010, 01:07:02 pm: I don't think there is a way to use matrices other than to find the values of "m" that give you zero when you subtract the value directly above or below in each row.
Title: Re: Holiday Help
Post by: brightsky on January 06, 2010, 01:19:34 pm: Again, not sure if this is right, but following logic...

b.

The simultaneous equations would have infinitely many solutions when the two are equal.

$y = f(x) = \frac{-mx+8}{2} = \frac{4x-2m}{2-m}$

And you work out m from that...
Title: Re: Holiday Help
Post by: cipherpol on January 06, 2010, 01:30:24 pm: Using matrix methods; det = $m^2-2m+8$

For infinite or no solutions, det = 0

$m^2-2m+8=(m-4)(m+2)=0$

So for infinite or no solutions, m = 4, or m = -2

Just sub m in; if new equations are the same, then there are infinite solutions for that value of m.
Title: Re: Holiday Help
Post by: brightsky on January 10, 2010, 11:26:13 am: Don't know if this is legit, but for part b, a quicker way would be:

$\begin{bmatrix} m \\ 2\\ 8 \end{bmatrix} = \begin{bmatrix} 4\\ -(2-m)\\ 2m \end{bmatrix}$

$\implies m = 4, 2 = -(2-m), 8 = 2m$

$\implies m = 4, m = 4, m = 4$