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I've had a little break in solving systems of equations and so I wanted to verify here that my own answer to this problem is 100% correct and done :) So I have the following problem:

Solve the system of equations:

$$3x_1+x_2-4x_3+5x_4=2$$ $$2x_1-3x_2-2x_3+3x_4=5$$

My attempt:

Because this is an underdetermined system we will have infinite amount of solutions. I will solve $x_1$ and $x_2$ explicitly in terms of $x_3$ and $x_4$. By multiplying the first equation by $3$ and adding it to the second we get:

$$11x_1-14x_3+18x_4=11$$

$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$

By substituting $x_2$ into the first equation we get:

$$3\left(1+\frac{14}{11}x_3-\frac{18}{11}x_4\right)+x_2-4x_3+5x_4=2$$

From which we get

$$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1$$

So the solution is:

$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$ $$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1,$$

where $x_3,x_4$ are free variables.

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  • $\begingroup$ i don't think you can arbitrarily pick the pivot variables as you have done here in picking $x_1$ and $x_2$. these variable are the by products of the row reduced system. $\endgroup$ – abel Jan 31 '15 at 19:05
  • $\begingroup$ @abel thank you, okay, what should I have done? :) $\endgroup$ – jjepsuomi Jan 31 '15 at 19:06
  • $\begingroup$ In this case, wouldn't any two work as pivots? Reduced row echelon form is only unique up to an ordering of the variables, so which variables are pivots isn't really an intrinsic property of the system. $\endgroup$ – G Tony Jacobs Jan 31 '15 at 19:09
  • $\begingroup$ you make an augmented matrix $[A|b]$ and row reduce. the variables corresponding to the pivot columns are the pivot variables and the rest are called free variables. you set all the free variables to zero to find a particular solution and then you set just one free variable to $1$ and solve them. $\endgroup$ – abel Jan 31 '15 at 19:10
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Note that

$$ \DeclareMathOperator{rref}{rref} \rref \begin{bmatrix} 3 & 1 & -4 & 5 & 2 \\ 2 & -3 & -2 & 3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -\frac{14}{11} & \frac{18}{11} & 1 \\ 0 & 1 & -\frac{2}{11} & \frac{1}{11} & -1 \end{bmatrix} $$ This implies \begin{align*} x_1 &= \frac{14}{11}\,x_3-\frac{18}{11}\,x_4+1 \\ x_2 &= \frac{2}{11}\,x_3-\frac{1}{11}\,x_4-1 \end{align*} which confirms your answer.

Do you know how to row-reduce a matrix into its reduced row-echelon form? This is a much more efficient method than substitution.

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  • $\begingroup$ Excellent! Thank you very much! Yes I do, thank you :) My first attempt was what I tried first. When you don't use it for a while you forget it you know ;D $\endgroup$ – jjepsuomi Jan 31 '15 at 19:48
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    $\begingroup$ For future reference: wolfram alpha can compute this. See wolframalpha.com/input/… This is a good way to check your answers! $\endgroup$ – Brian Fitzpatrick Jan 31 '15 at 19:58

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