SÄTT IN DATAPUNKTERNA I MODELEN.
MAN FÅR OÄ

\left\{ \begin{array}{l} 1 = -A + B \\ -1 = 0 \\ -1 = A + B \\ 3 = 2A + 4B \end{array} \right.

Lösningen skiljer sig uppenbarligen.
PÅ MATRISFORM:

\left[ \begin{array}{cc} -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \end{array} \right] \left[ \begin{array}{c} A \\ B \end{array} \right] = \left[ \begin{array}{c} 1 \\ -1 \\ -1 \\ 3 \end{array} \right]

MOTSVARANDE NORMALKLUBGEN:

\left[ \begin{array}{ccc} -1 & 0 & 1 & 2 \\ 1 & 0 & 1 & 4 \end{array} \right] \left[ \begin{array}{c} -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \end{array} \right] \left[ \begin{array}{c} A \\ B \end{array} \right] = \left[ \begin{array}{ccc} -1 & 0 & 1 & 2 \\ 1 & 0 & 1 & 4 \end{array} \right] \left[ \begin{array}{c} 1 \\ -1 \\ -1 \\ 3 \end{array} \right] \left[ \begin{array}{c} 6 & 8 \\ 8 & 18 \end{array} \right] \sim \left[ \begin{array}{c} 3 & 4 \\ 4 & 9 \end{array} \right] \sim \left[ \begin{array}{c} 12 & 16 \\ 12 & 27 \end{array} \right] \sim \left[ \begin{array}{c} 12 & 16 \\ 0 & 11 \end{array} \right]

Högerled:

\left[ \begin{array}{c} 8 \\ 12 \end{array} \right] \sim \left[ \begin{array}{c} 8 \\ 18 \end{array} \right] \sim \left[ \begin{array}{c} 8 \\ 10 \end{array} \right] \therefore B = 10/11 A = \frac{1}{12} \left(8 - \frac{160}{11}\right) = \frac{-72}{132} = \frac{-6}{11} y = -\frac{6}{11} x + \frac{10}{11} x^2