Thread: finding the basis of a homogeneous system

1. finding the basis of a homogeneous system

Find a basis for and the dimensions of the solution space of the given homogeneous system.
$x_1-x_2+2x_3+3x_4+4x_5=0$
$-x_1+2x_2+3x_3+4x_4+5x_5=0$
$x_1-x_2+3x_3+5x_4+6x_5=0$
$3x_1-4x_2+1x_3+2x_4+3x_5=0$

So I make a matrix with these coefficients and rref to give
$
$\left[ {\begin{array}{ccccc} 1 & 0 & 0 & 0 & \frac{1}{3} \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{4}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{3} \\ \end{array} } \right]$$

(Here's the part I'm not sure if I did right)
Every solution is of the form $
$\left[ {\begin{array}{c} \frac{-1}{3}r \\ 0 \\ \frac{-4}{3}r \\ \frac{-1}{3}r \\ \end{array} } \right]$$
where r is any real number. The dimension is 1.

2. Originally Posted by Jskid
Find a basis for and the dimensions of the solution space of the given homogeneous system.
$x_1-x_2+2x_3+3x_4+4x_5=0$
$-x_1+2x_2+3x_3+4x_4+5x_5=0$
$x_1-x_2+3x_3+5x_4+6x_5=0$
$3x_1-4x_2+1x_3+2x_4+3x_5=0$

So I make a matrix with these coefficients and rref to give
$
$\left[ {\begin{array}{ccccc} 1 & 0 & 0 & 0 & \frac{1}{3} \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{4}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{3} \\ \end{array} } \right]$$

(Here's the part I'm not sure if I did right)
Every solution is of the form $
$\left[ {\begin{array}{c} \frac{-1}{3}r \\ 0 \\ \frac{-4}{3}r \\ \frac{-1}{3}r \\ \end{array} } \right]$$
where r is any real number. The dimension is 1.

It's correct, assuming your rref is correct.

Tonio