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11-29-2008, 10:10 PM
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Solving matrices on a TI-83 plus
Need to remember how I did it in HS for a Physics final this week and I cant for the life of me figure out how to use the matrix feature to get a set of answers for 3 variables.
Anyone remember how to?
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11-29-2008, 11:06 PM
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Post up the matrices.
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11-29-2008, 11:16 PM
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I don't remember it being able to solve matrices, only being able to do simple operations with them.
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11-29-2008, 11:17 PM
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Hit 2nd and the X to the negative one button. Then go to edit and set the amount of rows and columns you have. Then just fill in the chart.
You will have to enter A and B separately. To add them or perform whatever function you need, go back into MATRIX, highlight for example A, tap enter, hit plus, go back into MATRIX, then highlight which ever other letter you need and tap enter.
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2012' Scion tC 6-speed M/T. Last edited by russiankid; 11-29-2008 at 11:23 PM. |
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11-30-2008, 04:33 PM
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Ill find a example later, I was just looking for a general directions. The teacher wants us to do it free-hand something like this.
2x+3y-4z=9 1x+2y+7z=2 x-4y+z= -4 Solve for x, y, z. IDK, if that problem will actually work out. But its the type of problem he wants me to spend 4 minutes on doing it by hand. I would rather get half credit and not show any work then waste the time and get it wrong by hand lol.
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Last edited by Drftpretty; 11-30-2008 at 04:37 PM. |
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11-30-2008, 04:52 PM
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Enter the matrices like russiankid said. If [A][x]=[B], then [A]^-1 * [B] = [x]. A is your coefficients (3x3), B is the right hand side values(3x1). Find A^-1, then multiply it times B, there is your solution values for X, Y and Z.
I assume you know how to set it up in matrix format. If not, ask me or PM me, I'm sitting around sick and have nothing better to do... Does he want you to solve it by hand using matrices or longhand? It's real easy to teach you to find the inverse of a 3x3 by hand if you want full credit. Edit: in case you were wondering, there is a solution for your made-up example. Edit 2: missed that your 4 was negative. 1.76, 1.35, -.35
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11-30-2008, 05:16 PM
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Here, you basically set up a 3x4 matrix and then find the reduced row echelon form rref():
http://www.msubillings.edu/asc/PDF-8..._Equations.pdf Solving by hand is easy though...  1) 2x+3y-4z=9 2) 1x+2y+7z=2 3) x-4y+z= -4 Set 2 and 3 equal to each other: 2y + 7z -2 = -x -4y + z + 4 = -x 2y + 7z -2 = -4y + z + 4 6y = 6 -6z y = 1 - z Substitute into 3: x - 4 + 4z + z = -4 x = -5z Substitute for z into 1: -10z +3 - 3z - 4z = 9 -17z = 6 z = -6/17 Substitute for z back into the simplified equations: x = -5z x = 30/17 y = 1 - z y = 23/17
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11-30-2008, 06:44 PM
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^I have no idea what you did there lol. Did I mention I hate math?
EDIT: Is it really that easy? My prof is a total D-bag! He has us setting the matrices to zeros using some Gauss-Jordan method that is hard as hell, I thought I remember substitution somehow. Thanks!
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Last edited by Drftpretty; 11-30-2008 at 06:49 PM. |
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