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Let $W$ be the set of all $(x_1, x_2, x_3, x_4, x_5)$ in $\Bbb R^5$ which satisfy $2x_1-x_2+{4 \over 3}x_3 - x_4\qquad = 0$, $x_1\qquad+{2 \over 3}x_3\qquad- x_5 = 0$, $9x_1-3x_2+6x_3-3x_4-3x_5 = 0$. Find a finite set of vectors which spans $W$. I have found the solution set which is $({-2\over3}c + e, 2e-d, c, d, e)$. Then will the required vectors which spans the solution space be $({-2\over3}, 0, 1, 0, 0), (0, -1, 0, 1, 0), (1, 2, 0, 0, 1)$? Is this the right way of solving the problem or is there a better way?

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  • $\begingroup$ You haven't told us what is the way you used to obtain that solution. By the way, your first solution's vector doesn't satisfy the first equation. $\endgroup$ – Timbuc Mar 29 '15 at 14:13
  • $\begingroup$ I used the matrix method used to solve AX = 0, reducing it to row echelon form $\endgroup$ – In78 Mar 29 '15 at 14:22
  • $\begingroup$ SWo I thought, yet your first solution vector is not a solution of the first equation. Either you change the first coordinate to $\;\frac23\;$ or the third one to $\;-1\;$ (but not both!) $\endgroup$ – Timbuc Mar 29 '15 at 14:26
  • $\begingroup$ Oh! I wrote the question wrong.. it will be + 4/3 x_3 $\endgroup$ – In78 Mar 29 '15 at 14:33
  • $\begingroup$ Too bad, but it never matters: if you follow carefully my answer, you can see what to do with the change in sign. $\endgroup$ – Timbuc Mar 29 '15 at 14:37
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You can form a matrix and reduce it:

$$\begin{pmatrix} 1&0&\frac23&0&\!\!-1\\2&\!\!-1&\!\!-\frac43&\!\!-1&0\\ 9&\!\!-3&6&\!\!-3&\!\!-3\end{pmatrix}\longrightarrow\begin{pmatrix} 1&0&\frac23&0&\!\!-1\\ 0&\!\!-1&\!\!-\frac83&\!\!-1&2\\ 0&\!\!-3&0&\!\!-3&6\end{pmatrix}\longrightarrow\begin{pmatrix} 1&0&\frac23&0&\!\!-1\\ 0&\!\!-1&\!\!-\frac83&\!\!-1&2\\ 0&0&8&0&0\end{pmatrix}$$

And the solution space of the above is two dimensional, not three dimensional as what you got.

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