Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If one asked to solve the set of equation below with the associated homogenous system, I'd know how to do it.

$$S \leftrightarrow \begin{cases} 3x + 5y + z = 8\\\ x + 2y - 2z = 3 \end{cases}$$

$$S' \leftrightarrow \begin{cases} 3x + 4y + z = 0\\\ x + 2y - 2z = 0 \end{cases}$$

You'd find the solution of the homogeneous system $S'$ to be: \begin{equation} (x, y, z) = \{ k\cdot (-12, 7, 1) | k \in \mathbb{R} \} \end{equation}

With the particular solution of $S$... \begin{equation} (x, y, z) = (1, 1, 0) \end{equation}

You can count them up and you'd find: \begin{equation} (x, y, z) = \{(1 - 12k, 1+ 7k, k)|k \in \mathbb{R}\} \end{equation}

And your original system of equations $S$ is solved.

Now I've got one question: how do you find such a particular solution to a non-homogeneous system of equations. How do you find $(1, 1, 0)$ in this case?

Another example:
How do I find one particular solution to this non homogeneous system? \begin{cases} x_1 + x_2 +x_3 =4\\ 2x_1 + 5x_2 - 2x_3 = 3 \end{cases}

share|improve this question
2  
Give $z$ a particular value, then solve the resulting $2 \times 2$ system. –  David Mitra Dec 18 '11 at 15:26
add comment

1 Answer

up vote 1 down vote accepted

Just set $z=0$, say. With a bit of luck, you'll be able to solve the resulting system: $$ \eqalign{ 3x+5y&=8\cr x+2y&=3 } $$

The solution of the above system is $y=1 , x=1 $; so, a solution to the original equation is $(1, 1 , 0)$.

For your second question, do a similar thing. Set $x_2=0$. Then you can conclude $x_1=11/4$ and $x_3=5/4$.

share|improve this answer
    
Some times, you won't be lucky, in which case you could select a different value, perhaps for a different variable. Of course, you could just solve the system using the usual augmented matrix techniques, then pick one solution. –  David Mitra Dec 18 '11 at 15:51
    
Yes my particular solution is wrong: I accidentally wrote 4 instead of 5 with the $y$-value of the first equation. –  user3.1415 Dec 18 '11 at 15:55
    
@Ief2 I edited my answer to correspond to the correct system of equations. –  David Mitra Dec 18 '11 at 15:58
    
Thank you, I guess I forgot that having a free variable means that you can actually choose one freely :-) –  user3.1415 Dec 18 '11 at 16:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.