The reason for this will be apparent soon enough 1 1 3

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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...
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