0
$\begingroup$

I have here W:

$$ \left[ \begin{array}{rrrrr} 1 & 1 & 7 & 1 & 2 \\ 0 & 1 & 2 & 1 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ 1 & -1 & 0 & 0 & 2 \\ 2 & 1 & 6 & 0 & 1 \end{array} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right] $$

(Im sorry i dont know how to form matrices in here so i just wrote down these)

To find the basis, I first reduced W. And after some bloody reducing with 5x5, i found the reduced form of W: $$ \left[ \begin{array}{rrrrr|r} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] $$

so i got my solution set: \begin{array}{rcr} x_1 & = &- e \\ x_2 & = & e \\ x_3 & = & 0 \\ x_4 & = & -2e \\ x_5 & = & e \end{array}

and my conclusion: $$ \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix} \right] = \left[ \begin{array}{r} -e \\ e \\ 0 \\ -2e \\ e \end{array} \right] = e \left[ \begin{array}{r} -1 \\ 1 \\ 0 \\ -2 \\ 1 \end{array} \right] $$

The problem is that, i'm missing 1 basis. In the book the basis are = {[-2,-2,1,0,0],[-1,1,0,-2,1]}

As you can see, i got the second basis. but what about the first? am i missing something?

$\endgroup$
  • $\begingroup$ The system and conclusion seem right, check the reduction of the matrix $\endgroup$ – Leafar Feb 23 '15 at 15:35
  • $\begingroup$ my solution for reduction is correct. i double checked it online with matrix.reshish.com/gaussSolution.php $\endgroup$ – user66240 Feb 23 '15 at 15:41
  • 1
    $\begingroup$ I checked it too. Your solution and answer are correct which means that there is an error in your text. $\endgroup$ – Paul Sundheim Feb 23 '15 at 16:20
  • $\begingroup$ @user66240, why did you conclude that $x_5=e$? $\endgroup$ – Galc127 Feb 23 '15 at 16:22
  • $\begingroup$ @Galc127, its not part of the conclusion. i just changed the variables form the solution set just to make it clear. $\endgroup$ – user66240 Feb 23 '15 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.