Gaussian Elimination

Gaussian Elimination or Row Reduction is a method for solving a System of Linear Equations

  • it corresponds to elimination of variables in the system
  • if a matrix $A$ that we reduce is non-singular and invertible, then we always have a solution
  • a by-product of Gaussian Elimination is LU Factorization


Row Elimination

  • First, do the forward elimination to reduce the matrix to triangular
  • Then we do the back-substitution to find the solution
  • When we multiply a row on some constant $c$ and subtract from any other row, we get an equivalent system


Elimination by Example

We have the following system with 3 equations and 3 unknowns:

$\left\{\begin{array}{l} x + 2y + z = 2\\ 3x + 8y + z = 12 \\ 4y + z = 2 \end{array}\right.$

Matrix form:

  • $\underbrace{\begin{bmatrix}

1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}}_{A} \cdot \underbrace{\begin{bmatrix} x \\ y \\ z \end{bmatrix}}_{\mathbf x} = \underbrace{\begin{bmatrix} 2 \\ 12 \\ 2 \end{bmatrix}}_{\mathbf b}$


Forward elimination

So, first let's see how elimination works

  • i.e. concentrate only on the matrix $A$
  • $A = \begin{bmatrix}

1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}$


Let's proceed

  • at each step
    • multiply the $i$th equation by the right number and subtract it from the equations below
    • s.t. there's 0 for the first column in all other rows
    • so we want to knock out the $x_i$ part of the equation, or "eliminate" $x_i$
  • $\begin{bmatrix}

\boxed 1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}$. $\boxed 1$ is the 1st pivot

  • step 2.1: clean cell $a_{21}$:
    • $\begin{bmatrix}

\boxed 1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix} \sim \begin{bmatrix} \boxed 1 & 2 & 1\\ 0 & 2 & -2\\ 0 & 4 & 1 \end{bmatrix}$

  • multiplying row 1 by 3 will knock out $x$ from the 2nd equation
  • step 3.1: clean cell $a_{21}$
    • already 0, continuing
  • $\begin{bmatrix}

\boxed 1 & 2 & 1\\ 0 & \boxed 2 & -2\\ 0 & 4 & 1 \end{bmatrix}$, $\boxed 2$ is the 2nd pivot

  • step 3.2
    • $\begin{bmatrix}

\boxed 1 & 2 & 1\\ 0 & \boxed 2 & -2\\ 0 & 4 & 1 \end{bmatrix} \sim \begin{bmatrix} \boxed 1 & 2 & 1\\ 0 & \boxed 2 & -2\\ 0 & 0 & \boxed 5 \end{bmatrix}$

  • $\begin{bmatrix}

\boxed 1 & 2 & 1\\ 0 & \boxed 2 & -2\\ 0 & 0 & \boxed 5 \end{bmatrix}$, $\boxed 5$ - it's 3rd pivot


The matrix is now in the upper-triangular form $U$


When Does if Fails?

Not always we are able to do the forward elimination

0 at a Pivot position:

  • suppose the first pivot is 0:
  • $\begin{bmatrix}

\boxed 0 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}$

  • But here it doesn't mean that we can't solve the system: we can switch the rows and continue:
  • $\begin{bmatrix}

\boxed 3 & 8 & 1\\ 0 & 2 & 1\\ 0 & 4 & 1 \end{bmatrix}$

  • so, if there's a 0 at the pivot position, try to exchange rows
  • if there's no rows with non-zero values at the pivot position - the elimination fails
    • there's no solution to the system


Augmented Matrix

But we shouldn't forget about $\mathbf b$!

  • $A$ augmented is $\begin{bmatrix}

\mathop{a_1}\limits_|^| \mathop{a_2}\limits_|^| ... \mathop{a_n}\limits_|^| \ \Bigg| \ \mathop{\mathbf b}\limits_|^| \end{bmatrix}$: matrix $A$ with column $\mathbf b$ stacked to the right

  • $A$ augmented: $\left[\begin{array}{ccc|c}

1 & 2 & 1 & 2\\ 3 & 8 & 1 & 12\\ 0 & 4 & 1 & 2 \end{array}\right]$

  • the right part $\mathbf b$ also changes as we go through elimination
  • $\left[\begin{array}{ccc|c}

1 & 2 & 1 & 2\\ 3 & 8 & 1 & 12\\ 0 & 4 & 1 & 2 \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 0 & 2 & -6 & 6\\ 0 & 4 & 1 & 2 \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 0 & 2 & -6 & 6\\ 0 & 0 & 5 & -10 \end{array}\right]$

  • let $\mathbf c$ be the result of applying elimination to $\mathbf b$


Back Substitution

After we forward-eliminated variables of augmented $A$, we can do back substitution:

  • Our matrix is

$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 0 & 2 & -6 & 6\\ 0 & 0 & 5 & -10 \end{array}\right]$

  • It means that our system is


$\left\{\begin{array}{rl} x + 2y + z & = 2\\ 2y - 2z & = 6 \\ 5z & = -10 \end{array}\right.$

Now we just go backwards and solve: first for $z$, then for $y$, and finally for $x$

  • we get $z = -2, y = -1, x =2$
  • it's easy to solve because the matrix is triangular


Elimination: Matrix Form

We can write these elimination steps in matrix form

  • $\begin{bmatrix}

1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}$

  • the first step of the elimination was to take first and last rows unchanged, and subtract 3 times 1st row from the second.
  • Can we write it with Matrix Multiplication?
    • $\begin{bmatrix}

1 & 0 & 0\\ ? & ? & ?\\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix}$

    • The first and last rows remain unchanged - hence we have $\begin{bmatrix}

1\\ ?\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0\\ ?\\ 1 \end{bmatrix}$

    • what should we put instead of $[? \ ? \ ?]$ so multiplication takes us from one matrix to another?
    • $\begin{bmatrix}

1 & 0 & 0\\ \boxed{-3} & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

    • we want to subtract 3 times row 1 from row 2:
    • $\begin{bmatrix}

1 & 0 & 0\\ \boxed{-3} & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 1\\ 3 & 8 & 1\\ 0 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 1\\ 0 & 2 & -2\\ 0 & 4 & 1 \end{bmatrix}$

  • let $E_{21}$ be the matrix that's used for step 2,1 of the elimination
    • $E_{21} A = \begin{bmatrix}

1 & 2 & 1\\ 0 & 2 & -2\\ 0 & 4 & 1 \end{bmatrix}$

  • we continue this way until we transform $A$ to $U$
    • so we have $E_{21} (E_{21} A) = U$
    • or $\underbrace{(E_{21} E_{21})}_E A = EA = U $
    • This a part of $LU$ Factorization


What if we need to exchange rows?


See Also

Sources

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