Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve for the full solution of $Ax=b $ where $$ A = \pmatrix{20 &40 &60\\ 30 &60 &90}, b = \pmatrix{10\\ 15}$$

The answers I am getting are $$ \pmatrix{-2\\1\\ 0}x_1 + \pmatrix{-3\\0\\ 1}x_2 + \pmatrix{1/2\\0\\ 0}$$

I'm not exactly sure if this is correct or not as the answers given to me by my professor is different and I don't know why. Can someone offer some insight?

Thanks

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Your answer is correct, as you can check doing these calculations:

$$ A \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} = A \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$

and

$$ A \begin{pmatrix} 1 / 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 10 \\ 15 \end{pmatrix} \ . $$

As for the answer given by your professor, you must take into account that those vectors $(-2 ,0 , 1), (-3,0, 1)$ and $(1/2, 0 ,0)$ are not unique. What is unique is the solution set

$$ (1/2, 0 ,0) + \ \mathrm{span}\left\{(-2 ,0 , 1), (-3,0, 1) \right\} \ . $$

But there are infinitely many ways to write it. For instance, another way to write the same solution set could be the following:

$$ (0, 0, 1/6) +\ \mathrm{span}\left\{(-1 ,0 , 1/2), (-1,0, 1/3) \right\} \ . $$

EDIT. I forgot: in order to fully check that your solution is right, you should also verify that the rank of your matrix $A$ is one (as it is); so the solution set has indeed dimension = number of unknowns - $\mathrm{rank}\ A = 3 -1 = 2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.