Vtyjkyuk

Original matrix A is:

$$beginpmatrix
1 & 2 & 7& 1\
1 & 1 & 2& 0\
2&0&-6&-2 & \
1&1&2&0
endpmatrix$$

$w=(A*beginpmatrix
x1\
x2\
x3\
x4
endpmatrix=beginpmatrix
0\
0\
0\
0
endpmatrix)$
Find base set of vector for W.

Using Gaus method I get:

$$beginpmatrix
1 & 2 & 7&1\
0&-1&-5&-1 \
0 & 0 & 0&0\
0&0&0&0
endpmatrix$$ which has 2 leading elements. So the column of each elemnt suppose to be part of base, these are: $(1,1,2,1)$ , $(2,1,0,1)$ but seemingly it's wrong because the answer is written like this: $(3,-5,1,0),(1,-1,0,1)$

where is my mistake?

Take your reduced system

$$beginpmatrix1 &2 &7& 1\0 &-1 &-5 &-1\ 0 &0 &0 &0\ 0 &0 &0 &0endpmatrixbeginpmatrixx\y\z\wendpmatrix=beginpmatrixx+2y+7z+w\-y-5z-w\0\0endpmatrix=beginpmatrix0\0\0\0endpmatrix$$

You now just have to solve the equation for $v=(x,y,z,w)^top$. Working with the method of back-substitution, we first assign all the non-pivots(the variables not leading any row) a real constant, as they may vary in their value arbitrarily and we then can determine the pivot variables in dependence on them.

Thus, let $w=c,z=c'$(for constants $c,c'$). Then, by the second row of the vector equation, we may determine the value of $y$ in terms of the constants $c,c'$(values of $z,w$):

$$-y-5z-w=0Leftrightarrow-y-5c'-c=0Leftrightarrow y=-5c'-c$$

Continuing back-substitution, we now derive from the first row the value of $x$ in dependence of $c,c'$:

$$x+2y+7z+w=0Leftrightarrow x=-2y-7z-wLeftrightarrow x=10c'+2c-7c'-cLeftrightarrow x=3c'+c$$

Thus, your solutions are any vectors of the following form

$$(3c'+c,-5c'-c,c',c)=c'(3,-5,1,0)+c(1,-1,0,1)$$

for arbitrary $c,c'inmathbbR$.

The solution set of $Av=mathbf0$, i.e. the set $S(A,mathbf0)=vinmathbbR^4mid Av=mathbf0$, for which you want to determine a basis, is commonly called the kernel of the matrix $A$, written $mathrmker(A)$.

By the above, we've found that any solution to $Av=mathbf0$, i.e. any $vinmathrmker(A)$, can be written as

$$v=c'(3,-5,1,0)+c(1,-1,0,1)$$

Thus, in other terminology, we say that any $vinmathrmker(A)$ is a linear combination of $(3,-5,1,0)$ and $(1,-1,0,1)$. The set of all linear combination for a set of vectors $S$ is called the span(or linear hull) of the set $S$ and I denote this with $mathrmspan(S)$.

We may thus rephrase our result about the solutions of $mathrmker(A)$ as such, that

$$mathrmker(A)=mathrmspan((3,-5,1,0),(1,-1,0,1))$$

A basis, for any space $V$, is defined as linearly independent set of vectors spanning $V$, i.e. a linearly independent set $B$, s.t. $mathrmspan(B)=V$.

Now, as $(3,-5,1,0),(1,-1,0,1)$ are linearly independent(check this yourself), it follows that they also constitute a basis for this set(space).

Maybe, on a last note, try to generalize the process I've just outline. Toying with these notions and especially with the connections between subspaces, linear systems of equation and matrices in combination is a great exercise. E.g. show that for any $ntimes n$ real matrix $A$, $mathrmker(A)$ always forms a subspace. Is this true for any solution set $S(A,b)$ for a vector equation $Av=b$?

There is a standard method for finding a basis for the null space of a matrix. Compute the reduced row echelon form (RREF):
beginalign
A=beginpmatrix
1 & 2 & 7 & 1\
1 & 1 & 2 & 0\
2 & 0 &-6 &-2 \
1 & 1 & 2 & 0
endpmatrix
&to
beginpmatrix
1 & 2 & 7 & 1\
0 &-1 &-5 &-1\
0 &-4 &-20 &-4 \
0 &-1 &-5 & -1
endpmatrix
&&beginaligned R_2&gets R_2-R_1 \ R_3&gets R_3-2R_1 \ R_4&gets R_4-R_1endaligned
\[6px]
&to
beginpmatrix
1 & 2 & 7 & 1\
0 & 1 & 5 & 1\
0 &-4 &-20 &-4 \
0 &-1 &-5 & -1
endpmatrix
&& R_2gets -R_2
\[6px]
&to
beginpmatrix
1 & 2 & 7 & 1\
0 & 1 & 5 & 1\
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endpmatrix
&&beginaligned R_3&gets R_3+4R_2 \ R_4&gets R_4+R_2endaligned
\[6px]
&to
beginpmatrix
1 & 0 &-3 &-1\
0 & 1 & 5 & 1\
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
endpmatrix
&& R_1gets R_1-2R_2
endalign
The system $Ax=0$ is equivalent to the system $Ux=0$, where $U$ is the last matrix above (the RREF of $A$). This system can be rewritten as
begincases
x_1 = 3x_3+x_4 \[4px]
x_2 = -5x_3-x_4
endcases
and a basis for the null space can be obtained by choosing first $x_3=1$ and $x_4=0$, then $x_3=0$ and $x_4=1$, yielding the vectors
$$
beginpmatrix 3 \ -5 \ 1 \ 0 endpmatrix
qquadtextandqquad
beginpmatrix 1 \ -1 \ 0 \ 1 endpmatrix
$$