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Unformatted text preview: re we see that x3 and x4 are free. Which means we will assign them each a different free variable. In this case let us call them x3=t1 and x4=t2. Right away we can see that the number of free variables will match the number of parameters we have. If we continue to solve for the basic variables we get: x1=‐2t1 and x2=‐2t1‐t2. This gives us the solution set as follows: 2t1 t 2 2
0 2 1 2t1 t2 S t1
0 t1 1 t 2 1 0 We see that the free variables will be written with a 1 in its corresponding parameter vector. You will also know we do not...
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