Matrix 2-by-2 with fractions  TOPIC_SOLVED

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Matrix 2-by-2 with fractions

Postby GreenLantern on Sun May 03, 2009 7:29 pm

I took a quiz on Friday, and this one matrix problem really gave me a hard time. I'm going to try and recreate it...

{(1/1-4x)+(2/1-2y)=4
{(3/1-4x)-(4/1-2y)=7

The important things I remember from this question is that the denominator of both the X's was the same and the denominator of both the Y's was the same.

I honestly have no idea where to start this, even though I do really well with most 2-by-2's. I'm thinking I'm missing some simplification skills from an earlier class; if that turns out to be the case, moving this may be a good idea.
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Postby stapel_eliz on Sun May 03, 2009 8:52 pm

GreenLantern wrote:{(1/1-4x)+(2/1-2y)=4
{(3/1-4x)-(4/1-2y)=7

Does the above mean the following?

. . . . .

If so, then it might help to let 1/(1 - 4x) = X and 1/(1 - 2y) = Y, so you get:

. . . . .

Solving leads to X = 3 and Y = 1/2, so:

. . . . .

. . . . .

Solve the rational equations for the values of x and y. :wink:
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Re:  TOPIC_SOLVED

Postby GreenLantern on Sun May 03, 2009 10:48 pm

Okay. I think the fractions in the matrix are what freaked me out about this problem. I didn't realize substitution could be applied here to simplify the matrix so quickly. After solving, I got the same results as you: (x,y)=(3,1/2). Converting it back to the original fractions with those answers looks much easier than before, and significantly less scary!

Since I'm dealing with fractions again, I made both sides into fractions.
. . . . .

Then cross multiplication gets me:
. . . . .
. . . . .
. . . . .
. . . . .

Solving for Y (sparing the details):
. . . . .

So my final answer is:
. . . . .

After that I have some bad news. I got that quiz question wrong! :lol: But now I understand it for the big test! Thank you Liz! :mrgreen:
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