Ok wacht even:
Als:
\(a \cdot A\left( {\begin{array}{*{20}c} 1 \\ 1 \\ 0 \\ \end{array} } \right) + b \cdot A\left( {\begin{array}{*{20}c} 1 \\ 0 \\ 1 \\ \end{array} } \right) + c \cdot A\left( {\begin{array}{*{20}c} 0 \\ 2 \\ 1 \\ \end{array} } \right)\)
Betekent dat dan dat dat hetzelfde is als:
\(a \cdot \left( {\begin{array}{*{20}c} 0 \\ 1 \\ 2 \\ \end{array} } \right) + b \cdot \left( {\begin{array}{*{20}c} 3 \\ -2 \\ 2 \\ \end{array} } \right) + c \cdot \left( {\begin{array}{*{20}c} 0 \\ 6 \\ 3 \\ \end{array} } \right)\)
Waardoor je dit krijgt:
\( \left( {\begin{array}{*{20}c} 0 \\ 1/3 \\ 2/3 \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} 2 \\ -4/3 \\ 4/3 \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} 0 \\ 2 \\ 1 \\ \end{array} } \right)\)
Waar dus uitkomt
\( A\left( {\begin{array}{*{20}c} 1 \\ 1 \\ 1 \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 2 \\ 1 \\ 3 \\ \end{array} } \right)\)
Of zit ik dan helemaal in de mist?