Here is a sketch of the two equations as a

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Unformatted text preview: need to first get a 1 in the upper left corner and there isn’t going to be any easy way to do this that will avoid fractions so we’ll just divide the first row by 2. 1é 5 1ù 5 -1ù R1 ê 1 -ú é2 2 2 ê -10 -25 5ú 2 ê ú ë û ® -10 -25 5û ë Now, we can get a zero in the lower left corner. 5 1ù 5 1ù é é - ú R2 + 10 R1 ® R2 ê 1 -ú ê1 2 2 2 2 ê ú ê ú ® 5û 0û ë -10 -25 ë0 0 Now, as with the first part we are never going to be able to get a 1 in place of the red zero without changing the first zero in that row. However, this isn’t the nonsense that the first part got. Let’s convert back to equations. x+ 5 1 y=2 2 0=0 That last equation is a true equation and so there isn’t anything wrong with this. In this case we have infinitely many solutions. Recall that we still need to do a little work to get the solution. We solve one of the equations for one of the variables. Note however, that if we use the equation from the augmented matrix this is very easy to do. x=- 5 1 y2 2 We then write the solutio...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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