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All vectors in $\mathbb{R}^3$ with $3v_1 -v_3=0 , ~~~~2v_1 +3v_2 -4v_3 =0$. The solution of this problem is a $1$ dimesional vector space with basis $(3, 9, 10)$. However I do not know process of soultion. How do I solve this problem?

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  • $\begingroup$ Wait, what is the question...? $\endgroup$ – IAmNoOne Aug 28 '16 at 16:38
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Notice you are trying to find the kernel of $$\begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & -4 \\ 0 & 0 & 0 \end{bmatrix}.$$ Now, let's first find the dimension of the image, which is the span of the columns. It is easily seen they span the subspace of $\mathbb{R}^3$ with third coordinate equal to $0$, so it is two dimensional. By the rank nullity theorem the kernel of the matrix is one dimensional. Now you can do one of two things:

  1. Find (by some kind of clever trial and error) a nonzero solution to your equations. You now know all other solutions must be scalar multiples of the solution you found.
  2. Use gauss elimination to find the kernel of the matrix, however in this case, my above discussion will cost you extra time.
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Solve

$$ \left( \begin{array}{ccc|c} 3 &0 &-1 &0 \\ 2 &3 &-4 &0 \end{array} \right) $$

using $-(2/3)\rho_1+\rho_2$ to get this.

$$ \left( \begin{array}{ccc|c} 3 &0 &-1 &0 \\ 0 &3 &-10/3 &0 \end{array} \right) $$

Take $z$ as the parameter to get the solution set as this.

$$ \{ \left(\begin{array}{c} 1/3 \\ 10/9 \\ 1 \end{array} \right)z \;|\; z\in\Re \} $$

If you fancy it, you can multiply that vector through by $9$ to get entries of $3$, $10$, and $9$, which is something like what you are looking for. Now you just have to figure out who made the mistake, you or me.

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