Reed Eaton

2022-06-25

I am given the following matrix A and I need to find a nullspace of this matrix.
$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$
I have found a row reduced form of this matrix, which is:
${A}^{\prime }=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$
And then I used the formula ${A}^{\prime }x=0$, which gave me:
${A}^{\prime }x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$
Hence I obtained the following system of linear equations:
$\left\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$
So I just said that ${x}_{3}=\alpha$, ${x}_{4}=\beta$ and the nullspace is:

Is my thinking correct?

Zayden Wiley

Expert

Since ${x}_{1}=2{x}_{4}-3{x}_{3}$ and ${x}_{2}=2{x}_{3}-3{x}_{4}⇒$
if $\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)\in$ nullspace(A):
$\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)=\left(2{x}_{4}-3{x}_{3},2{x}_{3}-3{x}_{4},{x}_{3},{x}_{4}\right)={x}_{3}\left(-3,2,1,0\right)+{x}_{4}\left(2,-3,0,1\right)$
So Nullspace $\left(A\right)=⟨\left(-3,2,1,0\right),\left(2,-3,0,1\right)⟩$.

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