Describe all solutions of Ax=0 [closed]

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I've gotten it into reduced echelon form, which was fairly simple. Adding a $b$ column that equaled zero then subtracting Line $2$ from Line $3$ times Line $1$. Thus, the $2$nd Line is all zeros and the only basic variable is $x_1$. There are infinitely many solutions given the fact that there are free variables.

I know $x_1$ can be written in terms of the free variables.$$x_1 = -4x_2+3x_3+2x_4$$

And I figured the coefficients might be the basis for these vectors but taking that equation and setting them equal to $x_2, x_3,$ and $x_4$ will only yield $3$ coefficients for each column, when I need $4$.

Blockquote$$x_1 = - 4x_2 + 3x_3 + 2x_4$$

$$x_1-3x_3-2x_4= -4x_2$$

$$-1/4x_1+3/4x_3+1/2x_4=x_2$$

Thus, it can't be the solution. What would be the correct way to go about this?

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1 Answer

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So you only have one equation, $$x_1+4x_2-3x_3-2x_4=0$$ and $$x_1=-4x_2+3x_3+2x_4$$

now take the column vector an substitute

$$\begin{pmatrix} x_1\\x_2\\x_3\\x_4\\\end{pmatrix}=\begin{pmatrix} -4x_2+3x_3+2x_4\\x_2\\x_3\\x_4\\\end{pmatrix}$$

Now separate the variables $$\begin{pmatrix} -4x_2+3x_3+2x_4\\x_2\\x_3\\x_4\\\end{pmatrix}= \begin{pmatrix} -4\\1\\0\\0\\\end{pmatrix}x_2+ \begin{pmatrix} 3\\0\\1\\0\\\end{pmatrix}x_3+\begin{pmatrix} 2\\0\\0\\1\\\end{pmatrix}x_4$$

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