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Unformatted text preview: as a one in the lower right corner. What this tells us is that it isn’t
possible to put this augmented matrix form.
Now, go back to equations and see what we’ve got in this case. x y =6
0 = 13 ??? The first row just converts back into the first equation. The second row however converts back to
nonsense. We know this isn’t true so that means that there is no solution. Remember, if we reach
a point where we have an equation that just doesn’t make sense we have no solution.
Note that if we’d gotten é 1 1 6 ù
ê0 1 0ú
ë
û we would have been okay since the last row would return the equation y = 0 so don’t get
confused between this case and what we actually got for this system.
[Return to Problems]
© 2007 Paul Dawkins 336 http://tutorial.math.lamar.edu/terms.aspx College Algebra (b) 2 x + 5 y = 1
10 x  25 y = 5 In this case we know from the first section that there are infinitely many solutions to this system.
Let’s see what we get when we use the augmented matrix method for the solution.
Here is the augmented matrix for this system. 5 1ù
é2
ê 10 25 5ú
ë
û
In this case we’ll...
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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
 MrVinh

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