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Matrices and Linear Equations

by: Melvina Keeling

Matrices and Linear Equations MATH 229

Marketplace > Colorado State University > Mathematics (M) > MATH 229 > Matrices and Linear Equations
Melvina Keeling
GPA 3.78

Daniel Bates

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Daniel Bates
Class Notes
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This 2 page Class Notes was uploaded by Melvina Keeling on Monday September 21, 2015. The Class Notes belongs to MATH 229 at Colorado State University taught by Daniel Bates in Fall. Since its upload, it has received 20 views. For similar materials see /class/210085/math-229-colorado-state-university in Mathematics (M) at Colorado State University.


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Date Created: 09/21/15
Supplementary notes for Math 229 Dan Bates Tuesday Sept 2 I ran short on time in class today and had to cut a couple of examples down to the bare minimum Here are the details on those two topics Nonnumerical matrices You can apply the techniques that we are discussing like Gaussian elimination to RREF form to nonnumerical matrices In fact I have heard that these are sometimes on the tests for 229 which I dont get to write Here is an example 2 7 x 1 2 1 2 7 x 2 1 272 2 R1HR2 272 1 0 1 1 3 i z W 1 1 3 i 2 At this point it is tempting to do R2 7 2 7zR1 7 R1 to make entry 21 zero In the case of a numerical matrix that would be ne In this case though check out what it would do to column 2 row 2it would make it even worse degree 2 instead of degree 1 Instead we follow a slightly different route that will make the entries of row 2 all divisible by the same polynomial Luckily the homework problem is a bit more clear 1 2795 2 1 272 2 RZ Rl RZ 172 271 0 RB Rl RS 172 271 0 1 1 372 6 0 9571172 Now I can divide the second and third rows by z 7 1 to make them nice numbers 1 1272 2 1 1272 2 ERZ RZ 1 71 0 ER37R3 1 71 0 0 95711795 6 0 1 i1 WARNING We multiplied by If x 1 this amounts to division by 0 7 bad news SO the solution that we get here will be valid for z 71 1 and we will need to do the z 1 case separately 1 272 2 1 732 72 Rzii RZ 1 732 72 R2R3 1 272 2 0 1 71 0 1 71 1 272 2 1 1 272 2 0 0 275 0 0 1 WARNING We did it again We now need to do the case of z 5 separately too At this point7 the matrix is in REF not RREF It is really simple from here to make RREF l7ll leave it to you7 but you should get the identity matrix 1s on the diagonal and 0s everywhere else Are we done Of course not nobody ever asks that if they are done We still need to handle the special cases of z 1 and z 5 Just plug those into the original matrix and work out the RREF I wont type this up since we have been over RREF for numerical matrices7 but you should get for z 1 and z 57 respectively 1 1 2 1 0 71 0 0 0 and 0 1 71 0 0 0 0 0 0 Generalized augment ed matrices This one is easier If you need to solve the same system repeatedly with different right hand sides7 allta7 constant vectors7 you can just toss all of the constant vectors into the augmented matrix For example7 to solve 2xy3 ziy1 and 2zy xiy simultaneously7 just form the following generalized augmented matrix and follow the steps for RREF as you would for a normal augmented matrix 2131R1lt gtR217112 17112 a 2131 R272R1lt gtR2 1 71 1 2 R2HR2 1 71 0 3173 0 1 1 1 39 Thus7 as I mentioned brie y in class7 the solution to the rst system of linear equations is 7 while the solution to the second system is 17 71 a nah t H to V a 71 R1R2lt gtR1 10 01 mwmm a


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