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Good morning, I need help with this problem.

Find a basis and the dimension of the solution space $W $of the following homogeneous system

$\begin{cases} x+2y-2z+2s-t=0\\ x+2y-z+3s-2t=0\\ 2x+4y-7z+s+t=0 \end{cases} $

Martin Sleziak
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LilianaX ArangurenX
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  • What are your thoughts, what have you tried, where are you stuck? – Greg Martin Apr 18 '16 at 16:49
  • Hi @GregMartin I Try make a linear combination and solve the system by Gauss-Jordan, but really I do not understand very well what I have to find – LilianaX ArangurenX Apr 18 '16 at 16:58
  • Perhaps you can include that information in the question then, and ask specifically for help understanding the actual statement of the question. – Greg Martin Apr 18 '16 at 19:55
  • You can also have a look at some older posts asking about similar problem. For example: http://math.stackexchange.com/questions/603228/find-basis-of-solutions-of-this-linear-system, http://math.stackexchange.com/questions/702951/easiest-way-to-solve-system-of-linear-equations-involving-singular-matrix or http://math.stackexchange.com/questions/1382907/finding-the-kernel-of-a-linear-map (You can also try some of the questions listed among related questions in the sidebar.) – Martin Sleziak Apr 18 '16 at 23:33

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Note that $W$ is the null space of the matrix $$ A= \left[\begin{array}{rrrrr} 1 & 2 & -2 & 2 & -1 \\ 1 & 2 & -1 & 3 & -2 \\ 2 & 4 & -7 & 1 & 1 \end{array}\right] $$ Recall that $A$ and the reduced row-echelon form of $A$ have the same null space. We can compute $\DeclareMathOperator{rref}{rref}\rref A$ using row-reductions. In our case we have $$ \rref{A}= \left[\begin{array}{rrrrr} 1 & 2 & 0 & 4 & -3 \\ 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ Here, $\rref A$ has two pivot columns, so the rank-nullity theorem tells us that $W$ has dimension $$ \dim W=(\#\text{ columns of }A)-(\#\text{ pivot columns of }\rref A)=5-2=3 $$ Can you find three linearly independent vectors in the null space of $\rref A$?

Brian Fitzpatrick
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