This preview shows page 1. Sign up to view the full content.
Unformatted text preview: to minimize the work as much as
possible however.
So, since there is a one in the first column already it just isn’t in the correct row let’s use the first
row operation and interchange the two rows. é3 2 14 ù R1 « R2 é 1 3 1ù
ê 1 3 1ú ® ê3 2 14 ú
ë
û
ë
û
The next step is to get a zero below the 1 that we just got in the upper left hand corner. This
means that we need to change the red three into a zero. This will almost always require us to use
third row operation. If we add 3 times row 1 onto row 2 we can convert that 3 into a 0. Here is
that operation. 3 1ù
é 1 3 1ù R2  3R1 ® R2 é 1
ê3 2 14 ú
ê0 11 11ú
®
ë
û
ë
û
Next we need to get a 1 into the lower right corner of the first two columns. This means changing
the red 11 into a 1. This is usually accomplished with the second row operation. If we divide
the second row by 11 we will get the 1 in that spot that we need. © 2007 Paul Dawkins 328 http://tutorial.math.lamar.edu/terms.aspx College Algeb...
View Full
Document
 Spring '12
 MrVinh

Click to edit the document details