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$A$=\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}

I need to find the null space for this matrix. After performing row operations $R_3 + 3R_1$, then $R_{new row 3}$ + $R_2$, I got the reduced row echelon form. (the variable I chose were $x$, $y$, $z$, $w$). I identified the pivot variables as $x$ and $y$, after expressing both of these in terms of the free variables $z$ and $w$, I got $$\begin{bmatrix} x\\ y\\ z\\ w\\ \end{bmatrix} = z \begin{bmatrix} 23/2\\ 4\\ 1\\ 0\\ \end{bmatrix} + w\begin{bmatrix} -11/2\\ -2\\ 0\\ 1\\ \end{bmatrix}$$, and then chose both the column vectors alongside $z$ and $w$ as the basis for the nullspace. However the answer given is $$\begin{bmatrix} 23\\ 8\\ 2\\ 0\\ \end{bmatrix}\begin{bmatrix} 9\\ 4\\ 0\\ 2\\ \end{bmatrix}$$. What am I doing wrong?

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    $\begingroup$ Could you show your reduced row echelon form? Two row operations don't sound like enough to get to a reduced form for a matrix with that many nonzero entries. $\endgroup$ – Henning Makholm May 31 '15 at 10:26
  • $\begingroup$ The rank of the matrix is $3$, so the null-space has dimension $1$. The first vector you mentioned is member of the null-space. $\endgroup$ – Peter May 31 '15 at 10:27
  • $\begingroup$ If you multiply $A$ with the second vector, you see that it does not belong to the null-space. $\endgroup$ – Peter May 31 '15 at 10:28
  • $\begingroup$ The given answer is also incorrect. $\endgroup$ – Peter May 31 '15 at 10:30
  • $\begingroup$ Are all the signs of the last column of your matrix right? If the column was $[-1,-2,1,0]$ instead, the matrix would have rank 2, and $[9,4,0,2]$ would be in the null space. $\endgroup$ – Henning Makholm May 31 '15 at 10:30
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If matrix
$$ \begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix} $$ Row reduced form of matrix is $$ \begin{bmatrix}1 & 0 & -\frac{23}{2} & 0\\ 0 & 1 & -4 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &0 &0\end{bmatrix} $$ So basis is $$ \begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix} $$

If matrix is $$ \begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & -2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix} $$ Row reduced form of matrix is $$ \begin{bmatrix}1 & 0 & -\frac{23}{2} & -\frac{9}{2}\\ 0 & 1 & -4 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 &0 &0\end{bmatrix} $$ So basis is $$ \begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix} \begin{bmatrix}9 \\ 4 \\ 0 \\ 2\end{bmatrix} $$

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  • $\begingroup$ the last two basis vectors you wrote (for -2), is any scalar multiple of both of these also basis? $\endgroup$ – user140161 May 31 '15 at 10:57
  • $\begingroup$ Basis are independent vectors in any subspace and if you consider any scalar multiple of both these basis then that would be dependent on these vectors so they would not be called the basis. $\endgroup$ – vidhan May 31 '15 at 11:00
  • $\begingroup$ No I don't think you understand my question. I'm saying if we have $v_1$, $v_2$ as basis for a subspace, then is $kV_1$,$bV_2$ also a basis for that subspace? (b and k are scalars) $\endgroup$ – user140161 May 31 '15 at 11:04
  • $\begingroup$ A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every other vector in the vector space is linearly dependent on these vectors . If you multiply that vector with any scalar , say $kv_1=kv_1+0v_2$ so it is dependent on $v_1$ and $v_2$ therefore $kv_1$ cannot be a basis. Similarly $kv_2$. $\endgroup$ – vidhan May 31 '15 at 11:09
  • $\begingroup$ I think you're wrong. $\endgroup$ – user140161 May 31 '15 at 11:21

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