4
$\begingroup$

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}

  • 3
    Give $z$ a particular value, then solve the resulting $2 \times 2$ system.2011-12-18

2 Answers 2

1

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 $x_3=5/4$.

  • 0
    Thank you, I guess I forgot that having a free variable means that you can actually choose one freely :-)2011-12-18
0

Find a particular solution to $\mathbf{A}\boldsymbol{x} = \boldsymbol{y}$ where $\mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & -2 \\ \end{bmatrix} $ $\boldsymbol{y} = \begin{bmatrix} 4 \\ 3 \\ \end{bmatrix} $ if we try to find \begin{equation*} \begin{aligned} & \underset{\boldsymbol{x}}{\text{min}} & & \vert\vert\boldsymbol{x}\vert\vert_2 \\ & \text{s.t.} & & \mathbf{A}\boldsymbol{x} = \boldsymbol{y} \end{aligned} \end{equation*} this will give a unique solution (for underdetermined case): $\boldsymbol{x} = \mathbf{A}^\mathsf{T}(\mathbf{A}\mathbf{A}^\mathsf{T})^{-1}\boldsymbol{y}$