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The column space of A is a plane. I need to find the equation of the plane.

Matrix $A$
$\begin{bmatrix} 2&4&6&4 \\2&5&7&6 \\2&3&5&2 \end{bmatrix}$

My solution:

my logic is that vector b which is in the Col(A)

$\begin{bmatrix} b_1 \\b_2 \\b_3 \end{bmatrix}$

Then Ax = b has a solution:

then,

$\begin{bmatrix} 2&4&6&4 = b_1 \\2&5&7&6 = b_2 \\2&3&5&2 = b_3 \end{bmatrix}$

when I reduced the matrix I produced(not sure if I did this part correctly)

$\begin{bmatrix} 1&2&3&2 = 1/2b_1 \\0&1&1&2 = b_2-b_1 \\0&0&0&0 = 2b_1-2b_2-b_3 \end{bmatrix}$

Not quite sure where to go from here or if my workings are correct to this point. Looking for some guidance

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What you've determined is that $Ax=b$ has a solution if and only if $2b_1 - 2b_2 - b_3=0$. However, $Ax=b$ has a solution if and only if $b$ is in the column space of $A$.

So, an equation for the column space of $A$ is $$ 2x-2 y - z=0 $$ ...or at least, this would be a way to find the equation of the plane if you had row-reduced correctly.

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  • $\begingroup$ okay so I didn't need to go through all those workings ? $\endgroup$ – jh123 Sep 19 '16 at 20:58
  • $\begingroup$ You did have to go through that work. I'm using your work. Apparently, however, you reduced the matrix incorrectly. $\endgroup$ – Omnomnomnom Sep 19 '16 at 20:59
  • $\begingroup$ okay I was not sure if my worings were correct when I row reduced $\endgroup$ – jh123 Sep 19 '16 at 21:01

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