# Find base vectors and dim

Find base vectors and dim of a space described by the following system of equation: $$2x_1-x_2+x_3-x_4=0 \\ x_1+2x_2+x_3+2x_4=0 \\ 3x_1+x_2+2x_3+x_4=0$$ I did rref of the matrix and as a result i get:

$$\begin{pmatrix} 1 & 2 & 1 & 2 \\ 0 & -5 & -1 & -5 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Thus i think that the independent variables will be $x_1,x_2$ and base vectors are the solution of this sytem when $x_1=0,x_2=1$ and $x_1=1,x_2=0$, but in the answer to the question is that independent variables are $x_2,x_4$. Am i doing something wrong?

• I don't know what your definitino of rref is, but is that matrix even in rref form? – user160738 Feb 26 '15 at 10:40
• reduced row echelon form, simply reduced form of the system of equations matrix – kurkowski Feb 26 '15 at 10:41
• You can choose $x_3$ and $x_4$ free. Also you can choose $x_2$ and $x_4$ free. Probably this is meant with "independent variables". – Peter Feb 26 '15 at 10:44
• but if i will chose $x_1$ and $x_2$ thsi will give me a correct base vector, but different right? – kurkowski Feb 26 '15 at 10:45
• Not the standard way to make the choice because the resulting matrix is not a triangular matrix, but it should also work. – Peter Feb 26 '15 at 10:46

My definition of RREF is different from yours it seems, and I calculated RREF form of augmented matrix to be:

$$\left( \begin{array}{cccc|c} 1 & 0 & 3/5 & 0 & 0\\ 0 & 1 & 1/5 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array} \right)$$

From this you set $x_3=t, x_4=s$ where $t,s$ are (scalar) parameters, and you can describe your solution space with these two independent variables.

Then $x_1=-\frac{3}{5} t$, $x_2= -\frac{1}{5}t-s$. So

$$(x_1,x_2,x_3,x_4)=\left(-\frac{3}{5} t,\; -\frac{1}{5}t-s,\; t,\; s\right)=\left(-\frac{3}{5},\; -\frac{1}{5},\; 1,\; 0\right)t+\left(0,\; -1,\; 0,\; 1\right)s$$

and you easily see that dimension is $2$.

• my rref is simply not fully done i've stopped at some point and did not reduce the upper parts of the matrix – kurkowski Feb 26 '15 at 11:02