## Matrix 2-by-2 with fractions

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
GreenLantern
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Joined: Sat Mar 07, 2009 10:47 pm
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### Matrix 2-by-2 with fractions

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.

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

Does the above mean the following?

. . . . .$\left[\begin{array}{ccc}\frac{1}{1\, -\, 4x}&\frac{2}{1\, -\, 2y}&4\\\frac{3}{1\, -\, 4x}&\frac{-4}{1\, -\, 2y}&7\end{array}\right]$

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

. . . . .$\left[\begin{array}{ccc}1&2&4\\3&-4&7\end{array}\right]$

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

. . . . .$3\, =\, \frac{1}{1\, -\, 4x}$

. . . . .$\frac{1}{2}\, =\, \frac{1}{1\, -\, 2y}$

Solve the rational equations for the values of x and y.

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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### Re:

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.
. . . . .$\frac{3}{1}\, =\, \frac{1}{1\, -\, 4x}$

Then cross multiplication gets me:
. . . . .$1\, =\, 3\, -\, 12x$
. . . . .$-2\, =\, -12x$
. . . . .$1\, =\, 6x$
. . . . .$x\, =\, \frac{1}{6}$

Solving for Y (sparing the details):
. . . . .$y\, =\, \frac{-1}{2}$

So my final answer is:
. . . . .$(x,y)\, =\, (\frac{1}{6},\frac{-1}{2})$

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