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$$ A = \begin{pmatrix} 1 & 2 & 1 & 2 \\ 2 & 3 & 3 & 5 \\ 1 & 2 & 1 & 3 \\ 3 & 1 & 8 & 1 \end{pmatrix} \to B = \begin{pmatrix} 1 & 0 & 3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ Where $A$ reduces to $B$ using Gaussian Elimination. I would like to find the null space of $A$.

What I see here is that $x_3 = 0$, which means that the first and second row also equals zero (since x = -3x and x2 = x3. While we have our free variable (the zero row) which I can denote as S. Which means that we have a matrice s(0,0,0,1) left. However the book says that x4 is zero and that x3 is our free variable.

What am I doing wrong here? I guess I could switch row 3 and 4 and it would work but why does it give me a different answer that way? Is there any rule that makes requires me to put rows that EQUAL zero in the last and not the rows with 0 inputs that im missing?

amWhy
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Blue
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3 Answers3

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Looking at the third row: [0 0 0 1], you get that $x_4 = 0$. (Not $x_3$, as you claimed!)

The "rule" is that the columns of the matrix correspond to the $x$ you need. Indeed, switching two rows should not change anything (and it doesn't.) Can you take it from there?

Wizact
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Assuming you did the elimination correctly, you are left with the equations $x_1 + 3x_3 = 0, x_2-x_3=0,x_4=0$. Thus, you can solve for everything in terms of $x_3$, getting $$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} -x_3 \\ x_3 \\ x_3 \\ 0 \end{pmatrix} = x_3 \begin{pmatrix} -1 \\ 1 \\ 1 \\ 0 \end{pmatrix} $$ so the null space looks like a line through the origin in the direction of the vector $(-1,1,1,0)^T$.

gt6989b
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You’re misinterpreting the third row of $B$. It doesn’t tell you that $x_3=0$. For that the row would look like $(0,0,1,0)$. Compare this to the last row of $B$, which does say that $x_4=0$. What the third row actually tells you is that $0x_3=0$, i.e., that $x_3$ can have any value whatsoever: it’s a free variable.

amd
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