Linear functions and Matrices

Marisca · 2011-10-06 20:59:28

Hi, I'm having trouble with these questions and was wondering if anyone could help:

Consider the system of equations;
x+2y-z=2
2x+5y-(a+2)z=3
-x+(a-5)y+z=1

a) Solve the equations in terms of 'a', for suitable values of 'a'.
b) State the values of 'a' for which there is a unique solution.

Is this the type of question where I would need to set the three equations as matrices?

bobbym · 2011-10-06 21:48:17

Hi Marisca;

There is some confusion about that. You can always format the problem in terms of matrices. This is just formal though. It does not assist in the solution of the problem.

For that you must either invert a matrix ( a purely computational task ) or use iterative means to get the actual answer.

For schoolwork and small systems hand methods are used.

Marisca · 2011-10-06 21:57:06

Okay that cleared some stuff up, so would I start off by rearranging the three equations?

Sorry, I'm really terrible at these questions.

bobbym · 2011-10-06 22:00:19

Hi;

Do you want to try it by hand?

Marisca · 2011-10-06 22:07:45

I'll try; do I arrange them for y?

bobbym · 2011-10-06 22:11:16

Hi;

First, do not be ashamed about finding this one hard. This question is the act of a sadist. If I were his/her student I would deliver a sharp blow to their probably ugly proboscis, causing them extreme discomfort.

Try to eliminate one variable, use the first equation and solve for z.

Marisca · 2011-10-06 22:18:19

So, would it be wrong if I removed x and solved for z in the first equation and got

z=2y+2 ?

bobbym · 2011-10-06 22:19:55

Hi;

Yes, that would be very wrong. You just can not make variables disappear no matter how much we hate them. Just get z by itself and show me what you get.

Marisca · 2011-10-06 22:23:12

Okay I got
z=x+2y-2

bobbym · 2011-10-06 22:26:13

That is correct!

This is a little bit tough. In the next two equations substitue for z in both of them.

Need help?

Marisca · 2011-10-06 22:34:57

I'm not sure but is it;

2x+5y-ax-2ay+2a=3

ay-3y=3

Edit: oh ok, I see that I wasn't supposed to expand it.

Last edited by Marisca (2011-10-06 22:37:08)

bobbym · 2011-10-06 22:38:23

Hi Marisca;

The second one is correct, the first one is not. You forgot to multiply by the 2 in (a+2).

Let me show you a trick. You should get this:

Notice how I did not simplify or combine. Many of the greatest moves in math were made by people who resisted the urge to simplify every time they can.

Marisca · 2011-10-06 22:43:55

ok that makes sense.
Not really sure what I need to do now though...

bobbym · 2011-10-06 22:47:27

Hi;

Remember, we do not simplify or combine unless there is a good reason to. Look at the second equation. If we combine the x and -x they will cancel and we will be left with an equation with one variable y! These we can solve easily. Do that and show me what you get.

Marisca · 2011-10-06 22:51:55

I got y=3/a-3

bobbym · 2011-10-06 22:53:59

Hi;

I think you mean y = 3 / ( a - 3 ). I see a lot of students taking the shorthand, poor notation of writing it like y=3/a-3. Resist that urge and you are ahead of them.

Marisca · 2011-10-06 22:58:01

oops, yes I should have done that.

From this point would I need to substitute this equation into the first equation, and then solve for a?

bobbym · 2011-10-06 23:04:10

Okay, let's see what we have:

We started with these:

Worked our way to these:

Now we have this:

Let's take the first equation up there and try to get another variable by itself. If you substitute for y in it what do you get?

Marisca · 2011-10-06 23:09:03

I got
x-z+ 6/(a-3) =2

bobbym · 2011-10-06 23:13:53

Hi;

Sorry, I meant the first equation in the second group. This one right here.

Marisca · 2011-10-06 23:18:06

Ok is it

-ax- 15/(a-3) +2a-2=3

bobbym · 2011-10-06 23:24:08

Hi Marisca;

Very good! That is correct! Now get x by itself.

Marisca · 2011-10-06 23:26:09

Thanks, I got;

x=2a-11/(a-3)

Last edited by Marisca (2011-10-06 23:28:53)

bobbym · 2011-10-06 23:29:39

Hi;

Mathematically,
x=2a-11/(a-3) means:

Which is not what you intended. When you write it like this:

x=(2a-11)/(a-3)

That means :

Which is what you want and is correct!

Marisca · 2011-10-06 23:34:28

ah, sorry. I'll have to keep that in mind.

Ok so would I now use this to substitute back into the equation and solve for a?

Math Is Fun Forum

#1 2011-10-06 20:59:28

Linear functions and Matrices

#2 2011-10-06 21:48:17

Re: Linear functions and Matrices

#3 2011-10-06 21:57:06

Re: Linear functions and Matrices

#4 2011-10-06 22:00:19

Re: Linear functions and Matrices

#5 2011-10-06 22:07:45

Re: Linear functions and Matrices

#6 2011-10-06 22:11:16

Re: Linear functions and Matrices

#7 2011-10-06 22:18:19

Re: Linear functions and Matrices

#8 2011-10-06 22:19:55

Re: Linear functions and Matrices

#9 2011-10-06 22:23:12

Re: Linear functions and Matrices

#10 2011-10-06 22:26:13

Re: Linear functions and Matrices

#11 2011-10-06 22:34:57

Re: Linear functions and Matrices

#12 2011-10-06 22:38:23

Re: Linear functions and Matrices

#13 2011-10-06 22:43:55

Re: Linear functions and Matrices

#14 2011-10-06 22:47:27

Re: Linear functions and Matrices

#15 2011-10-06 22:51:55

Re: Linear functions and Matrices

#16 2011-10-06 22:53:59

Re: Linear functions and Matrices

#17 2011-10-06 22:58:01

Re: Linear functions and Matrices

#18 2011-10-06 23:04:10

Re: Linear functions and Matrices

#19 2011-10-06 23:09:03

Re: Linear functions and Matrices

#20 2011-10-06 23:13:53

Re: Linear functions and Matrices

#21 2011-10-06 23:18:06

Re: Linear functions and Matrices

#22 2011-10-06 23:24:08

Re: Linear functions and Matrices

#23 2011-10-06 23:26:09

Re: Linear functions and Matrices

#24 2011-10-06 23:29:39

Re: Linear functions and Matrices

#25 2011-10-06 23:34:28

Re: Linear functions and Matrices

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