• We’ll reuse our example matrix to explore $$Ax=b$$:
$\begin{bmatrix} \begin{array}{rrrr} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \\ \end{array} \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$
• Let’s use an augmented matrix to deal with the right hand side and do elimination:
$\begin{bmatrix} \begin{array}{rrrrr} 1 & 2 & 2 & 2 & b_1 \\ 2 & 4 & 6 & 8 & b_2 \\ 3 & 6 & 8 & 10 & b_3 \\ \end{array} \end{bmatrix}$
• End up with, the condition for a solution is $$0 = -b_1 - b_2 + b_3$$ (e.g. $$b = (1, 5, 6)$$):
$\begin{bmatrix} \begin{array}{rrrrr} 1 & 2 & 2 & 2 & b_1 \\ 0 & 0 & 2 & 4 & -2b_1 + b_2 \\ 0 & 0 & 0 & 0 & -b_1 - b_2 + b_3 \\ \end{array} \end{bmatrix}$
• Solvability: a condition on $$b$$. $$Ax=b$$ is solvable when $$b$$ is in the columnspace of $$A$$, $$C(A)$$. Alternatively, if a combination of the rows of $$A$$ give the zero row, then the same combination of the components of $$b$$ have to give a zero.

• To find complete solution to $$Ax=b$$:

• Find a particular solution. One way to find a particular solution: set all free variables to zero and then solve for pivot variables.
$x_p = \begin{bmatrix} \begin{array}{r} -2 \\ 0 \\ \frac{3}{2} \\ 0 \\ \end{array} \end{bmatrix}$
• Add in the nullspace $$x_n$$ (aka special solutions).
• The complete solution to $$Ax=b$$ is $$x_{complete} = x_p + x_n$$

$x_{complete} = \begin{bmatrix} \begin{array}{r} -2 \\ 0 \\ \frac{3}{2} \\ 0 \\ \end{array} \end{bmatrix} + c_1 * \begin{bmatrix} \begin{array}{r} -2 \\ 1 \\ 0 \\ 0 \\ \end{array} \end{bmatrix} + c_2 * \begin{bmatrix} \begin{array}{r} 2 \\ 0 \\ -2 \\ 1 \\ \end{array} \end{bmatrix}$
• Note $$A * (x_p + x_n) = A * x_p + A * x_n = A * x_p + 0 = A * x_p = b$$.

• $$x_c$$ in this case is a subspace (the nullspace) shifted by the $$x_p$$. Instead of going through zero, the subspace goes through $$x_p$$.

• Relations between rank $$r$$ in a $$m \times n$$ matrix:

• $r \leq m$
• $r \leq n$
• Full column rank $$r = n$$: no free variables, nullspace is only the zero vector, one or zero solutions to $$Ax=b$$. Matrix will look tall and thin. Each zero row will be an additional condition on $$b$$.

• Full row rank $$r = m$$: One or many solutions for every $$b$$ (depending if there are free variables. There will be $$n-r = n-m$$ free variables). Matrix will be short and fat. $$m > n$$ results in free variables.

• Full row and column rank $$r = m = n$$. Square matrix and invertible, one solution for every b. Row reduced echelon form is the identity matrix.

• If $$r < m$$ and $$r < n$$, then you will have 0 solutions (if the conditions of the zero row are not satisifed) or infinite solutions if the conditions of the zero row are satisfied.