# Trying to apply matrices in Physics but having trouble

• Mar 22nd 2011, 02:34 PM
s3a
Trying to apply matrices in Physics but having trouble
Here is the original matrix:
matrix([20,10,0,0,549/100],[0,-10,30,30,41/50],[20,0,30,30,467/100],[1,-1,-1,0,0])

Here is the matrix in its echelon form:
matrix([1,1/2,0,0,549/2000],[0,1,2/3,0,183/1000],[0,0,1,9/11,303/11000],[0,0,0,0,1])

(The last column is for the constants on the right of the augmented line of the matrix).

Sorry for the dumb question but can someone please help me extrapolate the values of x1, x2, x3, and x4 given this information?

Any help would be greatly appreciated!
• Mar 22nd 2011, 04:28 PM
rtblue
If I have interpreted the question correctly, the solution is as follows:

$\displaystyle x_1+\frac{1}{2}x_2+0(x_3)+0(x_4) = \frac{549}{2000}$

$\displaystyle 0(x_1) + x_2+\frac{2}{3}x_3+0(x_4)= \frac{183}{1000}$

$\displaystyle 0(x_1)+0(x_2)+x_3+\frac{9}{11}(x_4) = \frac{303}{11000}$

Now, I run into a problem, but first let me explain to you what I've done. Each of the four columns (not including the constant column) corresponds to x1,x2,x3, and x4. The first column is x1, the second column is x2, the third column is x3, and the fourth column is x4. Look at the first row (That is the first equation). The first number in the first row is 1, and since this is in column one, we multiply this by x1. Now we add to that: the second number in the first row is 1/2 ,and since this is in the second column we multiply that by x2. We continue similarly all through the matrix.

The problem I encountered was in the last row. The fourth element should not be zero. In this case, only the first three elements of the last row should be zero.With the numbers you've given, we have 0+0+0+0=1 and this is obviously wrong. Have you made a typo?
• Mar 22nd 2011, 04:43 PM
s3a
Firstly, sorry if I sound rude for some reason, it's just I'm dead tired and burnt out so I'm a wreck.

On to the issue, I think I instead just made the matrix wrong. Your answer is wrong but it's very likely my fault.

I1 = x1
I2 = x2
I3 = x3
I4 = x4

Here are my equations. My friend said the fact that I3 = I4 (x3 = x4) means I HAVE TO treat it as one variable but I think that the matrix will "sort this out itself" if I just put them as two different variables even if they're the same. Again, I'm burnt out because of a lot of other work and I have this due tomorrow so I'm not in the best state to "argue" but I'm feeling like I'm right.

Here are the equations:
5.49 = 20I1 + 10I2
30I3 + 30I4 - 10I2=-0.82
30I3 + 30I4 + 20I1=4.67
I1 - I2 - I3 = 0

• Mar 22nd 2011, 05:19 PM
dwsmith
Quote:

Originally Posted by s3a
Firstly, sorry if I sound rude for some reason, it's just I'm dead tired and burnt out so I'm a wreck.

On to the issue, I think I instead just made the matrix wrong. Your answer is wrong but it's very likely my fault.

I1 = x1
I2 = x2
I3 = x3
I4 = x4

Here are my equations. My friend said the fact that I3 = I4 (x3 = x4) means I HAVE TO treat it as one variable but I think that the matrix will "sort this out itself" if I just put them as two different variables even if they're the same. Again, I'm burnt out because of a lot of other work and I have this due tomorrow so I'm not in the best state to "argue" but I'm feeling like I'm right.

Here are the equations:
5.49 = 20I1 + 10I2
30I3 + 30I4 - 10I2=-0.82
30I3 + 30I4 + 20I1=4.67
I1 - I2 - I3 = 0

Taking your equations and running rref yields:

$x_1=x_2=.183, \ x_3=-1\times 10^{-15}, \ x_4=.033666666666...$

However, from both your posts, I have no idea what is going on so maybe this will help.
• Mar 22nd 2011, 06:49 PM
Ackbeet
Quote:

...I have this due tomorrow...
Sounds like this problem is counting towards a final grade. Thread closed.
• Mar 23rd 2011, 04:03 AM
mr fantastic
Quote:

Originally Posted by Ackbeet
Sounds like this problem is counting towards a final grade. Thread closed.

Yes. Rule #6 here: http://www.mathhelpforum.com/math-he...hp?do=vsarules