$$\begin{array}{rcl}
  {
    \left(\!\!\!
    \begin{array}{rrr}
     4&1&8\\
	 7&4&-4\\
    \end{array}
    \!\!\right)
  }

    &
  \xymatrix@C=15ex{
    \ar[r]^-{\small
      \begin{array}{r}
        \frac{1}{4}\mathbf{r_1} \to \mathbf{r_1} \\
      \end{array}
    } &
  }

  &

  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 7&4&-4\\
    \end{array}
    \!\!\right)
  }

  \end{array}$$

$$\begin{array}{rcl}
  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 7&4&-4\\
    \end{array}
    \!\!\right)
  }

    &
  \xymatrix@C=15ex{
    \ar[r]^-{\small
      \begin{array}{r}
        -7\mathbf{r_1} + \mathbf{r_2} \to \mathbf{r_2} \\
      \end{array}
    } &
  }

  &

  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 0&\frac{9}{4}&-18\\
    \end{array}
    \!\!\right)
  }

  \end{array}$$

$$\begin{array}{rcl}
  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 0&\frac{9}{4}&-18\\
    \end{array}
    \!\!\right)
  }

    &
  \xymatrix@C=15ex{
    \ar[r]^-{\small
      \begin{array}{r}
        \frac{4}{9}\mathbf{r_2} \to \mathbf{r_2} \\
      \end{array}
    } &
  }

  &

  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 0&1&-8\\
    \end{array}
    \!\!\right)
  }

  \end{array}$$

$$\begin{array}{rcl}
  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&\frac{1}{4}&2\\
	 0&1&-8\\
    \end{array}
    \!\!\right)
  }

    &
  \xymatrix@C=15ex{
    \ar[r]^-{\small
      \begin{array}{r}
        \frac{-1}{4}\mathbf{r_2} + \mathbf{r_1} \to \mathbf{r_1} \\
      \end{array}
    } &
  }

  &

  {
    \left(\!\!\!
    \begin{array}{rrr}
     1&0&4\\
	 0&1&-8\\
    \end{array}
    \!\!\right)
  }
  = \mathbf{(A^T)^*}
  \end{array}$$

For at løse ligningssystemet sætter jeg $x_3 = t$.\\

$$x_1 = -4t$$
$$x_2 = 8t$$
$$x_3 = t$$

En basis for $\mathcal{U}^\perp =$ null $\mathbf{A}$ er

$$\begin{array}{rcl}
  {
    \left\{\left(
    \begin{array}{r}
     -4\\
	 8\\
	 1\\
    \end{array}
    \right)\right\}
  }
  \end{array}$$