• 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.