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Unformatted text preview: ra 1
3 1ù - R2 é 1 3 1ù
ê0 -11 11ú 11 ê0 1 -1ú
Okay, we’re almost done. The final step is to turn the red three into a zero. Again, this almost
always requires the third row operation. Here is the operation for this final step. é 1 3 1ù R1 - 3R2 ® R1 é 1 0 4ù
ê0 1 -1ú
ê0 1 -1ú
We have the augmented matrix in the required form and so we’re done. The solution to this
system is x = 4 and y = -1 .
[Return to Problems] (b) -2 x + y = -3
x - 4 y = -2 In this part we won’t put in as much explanation for each step. We will mark the next number
that we need to change in red as we did in the previous part.
We’ll first write down the augmented matrix and then get started with the row operations. 1 -3ù R1 « R2 é 1 -4 -2 ù R2 + 2 R1 ® R2 é 1 -4 -2 ù
ê 1 -4 -2 ú ® ê -2
ê 0 -7 -7 ú
Before proceeding with the next step let’s notice that in the second matrix we had one’s in both
spots that we needed them. However, the only way to change the -2 int...
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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.
- Spring '12