Ax=b

$\left[ \begin{array}{cccc|c} 1&2&2&2&b1\\ 2&4&6&8&b2\\ 3&6&8&10&b3\\ \end{array} \right]$

Note: You may notice not all rows and columns are independent

Let's do elimination to start solving.

$\left[ \begin{array}{cccc|c} 1&2&2&2&b_1\\ 2&4&6&8&b_2\\ 3&6&8&10&b_3\\ \end{array} \right]$ $=>$ $\left[ \begin{array}{cccc|c} 1&2&2&2&b_1\\ 0&0&2&4&b_2-2b_1\\ 0&0&2&4&b_3-3b_1\\ \end{array} \right]$ $=>$ $\left[ \begin{array}{cccc|c} 1&2&2&2&b_1\\ 0&0&2&4&b_2-2b_1\\ 0&0&0&0&b_3-b_2-b_1\\ \end{array} \right]$

Well that makes sense. We can see intuitively we that the last row - the second - the first gives us 0.

Solvability condition on b

Solvable when b is in $C(A)$
Note: Column space of A $C(A)$ is all linear combinations of columns of A
If a combination of A gives zero row, then the same combinations of entries of b must give 0

Find Complete Solution

Find $x_{particular}$

Set all free variables to 0
Solve $Ax=b$ for pivot variables

Set all free variables to 0

Free variable are rows that do not have a pivot column. These free variables can be set to anything, so we set them to the most convenient thing 0 and do back substitution.

$\left[ \begin{array}{cccc} 1&2&2&2\\ 0&0&2&4\\ 0&0&0&0\\ \end{array} \right]$ $\left[ \begin{array}{cccc} x_1\\ 0\\ x_3\\ 0 \end{array} \right]$ $=$ $\left[ \begin{array}{ccc} 1\\ 3\\ 0\\ \end{array} \right]$

Solve $Ax-b$ for pivot variables.

$x_1 + 2x_3 = 1$ $2x_3 = 3$

This means

$x_3 = 3/2$ and $x_1 = -2$

Put those into out x and we have.

$x_p =$ $\left[ \begin{array}{cccc} -2\\ 0\\ 3/2\\ 0 \end{array} \right]$

Find all Solutions

Find anything in $x_{nullspace}$
$X=X_p + X_n$
Note: The nullspace are any values of x that solve for 0. So naturally the x particular solution + any combination of 0 is still the solution.
$x_{complete} = $ $\left[ \begin{array}{cccc} -2\\ 0\\ 3/2\\ 0 \end{array} \right]$ $+c_1$ $\left[ \begin{array}{cccc} -2\\ 1\\ 0\\ 0 \end{array} \right]$ $+c_1$ $\left[ \begin{array}{cccc} 2\\ 0\\ -2\\ 1 \end{array} \right]$