This preview shows page 1. Sign up to view the full content.
Unformatted text preview: o a zero that we had to
have as well was to also change the 1 in the lower right corner as well. This is okay. Sometimes
it will happen and trying to keep both ones will only cause problems.
Let’s finish the problem. 1
é 1 4 2 ù  R2 é 1 4 2 ù R1 + 4 R2 ® R1 é 1 0 2 ù
ê0 7 7 ú 7 ê0
ê0 1 1ú
1 1ú
®
ë
û®ë
û
ë
û
The solution to this system is then x = 2 and y = 1 .
[Return to Problems] (c) 3 x  6 y = 9
2 x  2 y = 12 Let’s first write down the augmented matrix for this system. é 3 6 9 ù
ê 2 2 12 ú
ë
û
Now, in this case there isn’t a 1 in the first column and so we can’t just interchange two rows as
the first step. However, notice that since all the entries in the first row have 3 as a factor we can
divide the first row by 3 which will get a 1 in that spot and we won’t put any fractions into the
problem.
© 2007 Paul Dawkins 329 http://tutorial.math.lamar.edu/terms.aspx College Algebra Here is the work for this system. 1
é 3 6 9 ù R1 é 1 2 3ù R2 + 2 R1 ® R2 é 1 2 3ù
ê 2...
View Full
Document
 Spring '12
 MrVinh

Click to edit the document details