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Let A = \begin{pmatrix}-1&2&-13&-6\\ 2&1&1&-3\\ -3&0&-9&0\\ 7&2&11&-6\end{pmatrix}

Determine a basis for the null space of A."

So I reduced it and got the following matrix:

\begin{pmatrix}1&0&3&0\\ 0&1&-5&-3\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}

Writing it in terms of parameters I got:

x1 = -3x3

x2 = 5x3 + 3x4

So, I thought the vectors were:

\begin{pmatrix}-3\\ 5\\ 0\\ 0\end{pmatrix} and \begin{pmatrix}0\\ 3\\ 0\\ 0\end{pmatrix}

But this incorrect.

Any help?

sktsasus
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  • You are left with $x_2 =5x_3 +3x_4$ and $x_1 /(-3) =x_3$ so your solution should look like $(x_1 , 5x_1+3x_4, x_1/(-3), x_4 )$ Choosing $x_1=0$ and $x_4 >0$ then choosing $x_4=0$ and $x_1>0$ gives you your result – mm8511 Nov 30 '17 at 21:36
  • See here for a detailed explanation of how to read a basis for the null space from the rref. – amd Nov 30 '17 at 23:52

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You got $x_1=-3x_3, \; x_2=5x_3+3x_4$, so the general solution for $Ax=0$ is the vector $(-3x_3,5x_3+3x_4,x_3,x_4)'$, now rewrite it has $(-3x_3,5x_3,x_3,0)'+(0,3x_4,0,x_4)'$. Finally just choose 2 non-zero values for $x_3,x_4$ let's say $1$. We get $\{(-3,5,1,0)',(0,3,0,1)'\}$. If you want you can then normalize them just by dividing for the respective norm. You can check that it is a basis since the vectors are independent and both $\in Ker(A)$

chak
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