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Find the basis for the null space of the given matrix and give nullity(A)?

$A=\left[\begin{array}{rrrrrr} 1 & 1 & 2 & 1 \\ 0 & 0 & 1 & -3 \end{array}\right] $

By the way, the answer is... the null space has a basis {${\begin{bmatrix} -7\\ 0\\ 3\\1\end{bmatrix}, \begin{bmatrix}-1\\ 1\\ 0\\0\end{bmatrix}}$} and nullity(A)=2

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    $\begingroup$ if you solve AX=0 , the solution space will be the required null-space of A $\endgroup$ – ketan Apr 11 '14 at 9:25
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Let $X = \begin{bmatrix} x \\ y \\ z \\ r \end{bmatrix}$. Then the null space of $A$ is given by those $X$ such that $AX=0$.
Satisfying $AX=0$ is equivalent to satisfying $x+y+2z+r=0 $ and $z-3r=0 $.

From the second equation you get $z=3r$. To get a basis for the null space of $A$ we have to distinguish two cases $r=0$ and $r=1$:

  1. Let $r=1$. Then $z=3$. Substituting with the first equation gives $x+y+7=0$, hence $x=-7-y$. Let $y=0$, so $x=-7$ and you get your first answer.
  2. Let $r=0$ and so on...
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