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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}

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Give $z$ a particular value, then solve the resulting $2 \times 2$ system. –  David Mitra Dec 18 '11 at 15:26

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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 $x_3=5/4$.

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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. –  user21385 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 :-) –  user21385 Dec 18 '11 at 16:03

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