However it is possible that the situation could be

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Unformatted text preview: isn’t possible to put it into one of these forms then we will know that we are in either the first or last possibility for the solution to the system. Before getting into some examples let’s first address how we knew what the solution was based on these forms of the augmented matrix. Let’s work with the two equation case. Since, é 1 0 hù ê0 1 k ú ë û is an augmented matrix we can always convert back to equations. Each row represents and equation and the first column is the coefficient of x in the equation while the second column is the coefficient of the y in the equation. The final column is the constant that will be on the right side of the equation. So, if we do that for this case we get, (1) x + ( 0 ) y = h ( 0 ) x + (1) y = k Þ x=h Þ y=k and this is exactly what we said the solution was in the previous section. © 2007 Paul Dawkins 335 College Algebra This idea of turning an augmented matrix back into equations will be important in the following examples. Speaking of which, let’s go ahead and work a couple o...
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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.

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