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**Unformatted text preview: **1-3 0-1-2 1-4 1 1 9 4 I +3 II--- 1 0 0-1-12 1 0 1 0-4 1 0 0 0 1 9 4 0 0 0 I + III--- 1 0 0 0-3 5 0 1 0 0-4 1 0 0 0 1 9 4 0 0 0 0 The variables corresponding to non-pivot columns are the free variables. So, in this case, x 3 and x 5 are free. We can then solve the equations for the other variables and express them in terms of the free variables to obtain the general solution. x 1 = 5 + 3 x 5 x 2 = 1 + 4 x 5 x 3 is free x 4 = 4-9 x 5 x 5 is free We see that the solutions depend on two free variables. Since there are free variables, there are innitely many solutions to the system....

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