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I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.

I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.

The first is $$ A= \begin{pmatrix} 1 & 1 & 0 & 2\\ 2 & 1 & 1 &-2\\ 2 & 2 & 2 & 1 \end{pmatrix},\quad \tilde{b}= \begin{pmatrix} 12\\0\\14 \end{pmatrix} $$ Proceeding with an augmented matrix, I have \begin{align*} \left( \begin{array}{rrrr|c} 1 & 1 & 0 & 2 & 12\\ 2 & 1 & 1 &-2 & 0\\ 2 & 2 & 2 & 1 & 14 \end{array} \right)&\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 &-1 & 1 &-6 & -24\\ 0 & 0 & 2 & -3 &-10 \end{array} \right)\\ &\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 & 1 &-1 & 6 & 24\\ 0 & 0 & 1 & -3/2 &-5 \end{array} \right) \end{align*} I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.

Solving this now, I have \begin{align*} x_4&=\alpha\in\mathbb{R}\\ x_3&=-5+\frac{3}{2}\alpha\\ x_2&=19-\frac{9}{2}\alpha\\ x_1&=-7+\frac{5}{2}\alpha \end{align*}

So from this, my solution will be $$ \tilde{x}= \begin{pmatrix} -7+\frac{5}{2}\alpha\\ 19-\frac{9}{2}\alpha\\ -5+\frac{3}{2}\alpha\\ \alpha \end{pmatrix} = \begin{pmatrix} -7\\19\\-5\\0 \end{pmatrix}+\beta \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}. $$

This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.

Edit: My professor's solution is $$ \tilde{x}= \begin{pmatrix} 3\\1\\1\\4 \end{pmatrix}+\alpha \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$

Can somebody please let me know what I've done incorrectly?

Thank you.

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  • $\begingroup$ I tried using WolframAlpha, but it didn't tell me what I did wrong, and it spat out something that looked different to both my solution and my professor's (although it did provide the same nullspace!). $\endgroup$ – Babe in the Woods Aug 10 '16 at 23:22
  • $\begingroup$ Oh, he went away.. $\endgroup$ – Babe in the Woods Aug 10 '16 at 23:22
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    $\begingroup$ Can you provide your professor's solution? $\endgroup$ – Aweygan Aug 10 '16 at 23:25
  • $\begingroup$ My apologies, my professor's solution has been added in the question. $\endgroup$ – Babe in the Woods Aug 10 '16 at 23:29
  • $\begingroup$ Your professor simply chose to set $x_4=2\alpha$ instead of $\alpha$ to avoid denominators in the general solution. $\endgroup$ – Bernard Aug 10 '16 at 23:33
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Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.

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  • $\begingroup$ You're welcome. $\endgroup$ – Aweygan Aug 10 '16 at 23:34
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This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found.

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