The column space of A is a plane. I need to find the equation of the plane. Matrix $A$
$\begin 2&4&6&4 \\2&5&7&6 \\2&3&5&2 \end$ My solution: my logic is that vector b which is in the Col(A) $\begin b_1 \\b_2 \\b_3 \end$ Then Ax = b has a solution: then, $\begin 2&4&6&4 = b_1 \\2&5&7&6 = b_2 \\2&3&5&2 = b_3 \end$ when I reduced the matrix I produced(not sure if I did this part correctly) $\begin 1&2&3&2 = 1/2b_1 \\0&1&1&2 = b_2-b_1 \\0&0&0&0 = 2b_1-2b_2-b_3 \end$ Not quite sure where to go from here or if my workings are correct to this point. Looking for some guidance
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.
answered Sep 19, 2016 at 20:57 Ben Grossmann Ben Grossmann 228k 12 12 gold badges 172 172 silver badges 338 338 bronze badges $\begingroup$ okay so I didn't need to go through all those workings ? $\endgroup$ Commented Sep 19, 2016 at 20:58$\begingroup$ You did have to go through that work. I'm using your work. Apparently, however, you reduced the matrix incorrectly. $\endgroup$