Introduction

• goal: give an algorthm to determine how many solutions (0, 1, $\infty$) does a lin. sys. have
  • and find out the value of solutions, when they exist
• such algorithm: row reduction to echelon form
  • formalize the way we solve lin. sys.also called: Gausiian elimination

Defining reduced echelon form

• row in matrix: nonzero if "not every entry in the row" is zero
  • $\exists A_{ij}\in A_{ij},A_{ij}\ne 0$
  • nonzero column: defined similarly
• leading entry in a row: first non-zero entry in the row
  • starting from left to right
• matrix of size $m\times n$ is in echelon form if:
  1. if a row is nonzero: every row above it is also nonzero
  2. the leading entry in a nonzero row: strictly right of the leading entry of any earlier row
     • 👨‍🎓 i.e. in somewhat triangular form?
  3. if a row is nonzero: every entry below its leading entry in the same column is zero
     • implied from the second principle
• by definition
  • every one-row matrix is in echelon form
  • one-column matrix in echelon form: where * is any number $$\left[\begin{array}{c}\ast \\ 0\\ 0\\ ⋮\\ 0\end{array}\right]$$

• a matrix in echelon form is reduced is
  1. each nonzero row has leading entry 1
  2. the leading 1 in each nonzero row is the only nonzero number in its column
• reduced echelon form: matrix in echelon form that is reduced
  • 👨‍🎓 $\approx$ solved?
• reduced echelon form looks like: $$\left[\begin{array}{cccccccccc}0& 0& 1& 0& \ast & \ast & 0& \ast & \ast & 0\\ 0& 0& 0& 1& \ast & \ast & 0& \ast & \ast & 0\\ 0& 0& 0& 0& 0& 0& 1& \ast & \ast & 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$
• ⭐ theorem: each matrix $A$ is row equivalent to exactly one matrix $RREF(A)$ in reduced echelon form
• $RREF(A)$: reduced echelon form of $A$

Solving a linear system from reduce echelon form

• pivot position in a matrix $A$: location containing a leading 1 in $RREF(A)$

  • aka: $pivot$

• pivot column in a matrix A: column containing a pivot position

• for lin. sys. in $x_1,x_2,\dots ,x_n$ with aug. matrix $A$

  • the var $x_i$ is ...
    • basic if $i$ is a pivot column of $A$
    • free if $i$ is not a pivot column of $A$

• case study

  $$RREF(A)=\left[\begin{array}{cccc}1& 0& -5& 1\\ 0& 1& 1& 4\\ 0& 0& 0& 0\end{array}\right]$$

  • which means system:

    $$\{\begin{array}{l}{x}_1-5x_3=1\\ x_2+x_3=4\\ 0=0\end{array}$$

    • $x_1,x_2$ are basic
    • and $x_3$ is free

  • to solve the system: choose any values for the free var., and solve for the basic var.

  • solution to above system: all in form

    • $(x_1,s_2,s_3)=(5a+1,4-a,a|a\in \mathbb{R})$

• theorem: consider a lin. sys. whose augmented matrix is A

  • the system has 0 solutions / is inconsistent
    • if the last column of $A$ contains a pivot
    • which the row will look like: $\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 1\end{array}\right]$
      • i.e. equivalent of saying $0=1$
  • the system has only 1 solution
    • if there are no free variables and the last column is not a pivot
  • otherwise: the system has infinitely many solutions

• once $RREF(A)$ is computed and free / basic var. are identified

  • we can write down all solutions to system as in the above example
  • let all free var. to be arbitrary, and solve it for basic var.

Computing reduced echelon form

• algorithm to compute $RREF(A)$ has two parts
  • row reduction to echelon form
  • reducing it to $RREF(A)$
• example
  • for general algorithm: input: $m\times n$ matrix $A$ $$A=\left[\begin{array}{cccc}0& 3& -6& 6\\ 3& -7& 8& -5\\ 3& -9& 12& -9\end{array}\right]$$
  • procedure

    1. begin with the leftmost nonzero column

       • this will be the pivot column, pivot being the top position
       • boxed: pivot position $$\left[\begin{array}{cccc}\boxed{0}& 3& -6& 6\\ 3& -7& 8& -5\\ 3& -9& 12& -9\end{array}\right]$$

    2. select a nonzero entry in the current pivot column

       • if needed: perform row operation to swap the row with the top one $$\left[\begin{array}{cccc}\boxed{0}& 3& -6& 6\\ 3& -7& 8& -5\\ 3& -9& 12& -9\end{array}\right]\to \left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& 3& -6& 6\\ 3& -9& 12& -9\end{array}\right]$$

    3. use row operations to create zeros below the boxed pivot position $$\left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& 3& -6& 6\\ 3& -9& 12& -9\end{array}\right]\to \left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& 3& -6& 6\\ 0& -2& 4& -4\end{array}\right]$$

    4. repeat steps 1-3 on the bottom right sub-matrix

       $$\left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& \boxed{3}& -6& 6\\ 0& -2& 4& -4\end{array}\right]\to \left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& \boxed{3}& -6& 6\\ 0& 0& 0& 0\end{array}\right]$$

    5. now: we have a echelon form: now reduce it (w/ replacements) $$\left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& \boxed{3}& -6& 6\\ 0& 0& 0& 0\end{array}\right]$$

    • now: rescale rightmost pivot, and cancel entries above rightmost pivot
      • repeat (heading left) $$\begin{gathered} \to \left[\begin{array}{cccc}\boxed{3}& -7& 8& -5\\ 0& \boxed{1}& -2& 2\\ 0& 0& 0& 0\end{array}\right]\to \left[\begin{array}{cccc}\boxed{3}& 0& -6& 9\\ 0& \boxed{1}& -2& 2\\ 0& 0& 0& 0\end{array}\right] \\ \to \left[\begin{array}{cccc}\boxed{1}& 0& -2& 3\\ 0& \boxed{1}& -2& 2\\ 0& 0& 0& 0\end{array}\right] \end{gathered}$$
    • final result $$\left[\begin{array}{cccc}\boxed{1}& 0& -2& 3\\ 0& \boxed{1}& -2& 2\\ 0& 0& 0& 0\end{array}\right]$$
• for a general matrix $A$: follow the algorithm below
  • input: $m\times n$ matrix $A$
  • procedure:
    1. begin with the leftmost nonzero column
       • i.e. the pivot column, position being the top position of column
    2. select a nonzero entry in current pivot column
       • swap rows as needed
    3. use row operations to create zeros in the entries below the pivot
    4. apply previous step for $(m-1)\times n$ sub-matrix that remains
       • repeat until the entire matrix is in echelon form
    5. start from the rightmost pivot position in matrix
       • rescale this row to have leading entry 1
       • use row operation to create zeros in entries in the same column as the pivot
       • repeat this for the pivot {above = left}
  • output: $RREF(A)$