Strassen-Algorithmus/2x2/Aufgabe/Lösung

Es ist

${\displaystyle {}{\begin{aligned}m_{1}+m_{4}-m_{5}+m_{7}&={\left(a_{11}+a_{22}\right)}\cdot {\left(b_{11}+b_{22}\right)}+a_{22}\cdot {\left(b_{21}-b_{11}\right)}-{\left(a_{11}+a_{12}\right)}\cdot b_{22}+{\left(a_{12}-a_{22}\right)}\cdot {\left(b_{21}+b_{22}\right)}\\&=a_{11}b_{11}+a_{22}b_{11}+a_{11}b_{22}+a_{22}b_{22}+a_{22}b_{21}-a_{22}b_{11}-a_{11}b_{22}-a_{12}b_{22}+a_{12}b_{21}-a_{22}b_{21}+a_{12}b_{22}-a_{22}b_{22}\\&=a_{11}b_{11}+a_{12}b_{21}\\&=c_{11},\,\end{aligned}}}$

${\displaystyle {}{\begin{aligned}m_{3}+m_{5}&=a_{11}\cdot {\left(b_{12}-b_{22}\right)}+{\left(a_{11}+a_{12}\right)}\cdot b_{22}\\&=a_{11}b_{12}-a_{11}b_{22}+a_{11}b_{22}+a_{12}b_{22}\\&=a_{11}b_{12}+a_{12}b_{22}\\&=c_{12},\end{aligned}}}$

${\displaystyle {}{\begin{aligned}m_{2}+m_{4}&={\left(a_{21}+a_{22}\right)}\cdot b_{11}+a_{22}\cdot {\left(b_{21}-b_{11}\right)}\\&=a_{21}b_{11}+a_{22}b_{11}+a_{22}b_{21}-a_{22}b_{11}\\&=a_{21}b_{11}+a_{22}b_{21}\\&=c_{21},\end{aligned}}}$

${\displaystyle {}{\begin{aligned}m_{1}-m_{2}+m_{3}+m_{6}&={\left(a_{11}+a_{22}\right)}\cdot {\left(b_{11}+b_{22}\right)}-{\left(a_{21}+a_{22}\right)}\cdot b_{11}+a_{11}\cdot {\left(b_{12}-b_{22}\right)}+{\left(a_{21}-a_{11}\right)}\cdot {\left(b_{11}+b_{12}\right)}\\&=a_{11}b_{11}+a_{22}b_{11}+a_{11}b_{22}+a_{22}b_{22}-a_{21}b_{11}-a_{22}b_{11}+a_{11}b_{12}-a_{11}b_{22}+a_{21}b_{11}+a_{21}b_{12}-a_{11}b_{11}-a_{11}b_{12}\\&=a_{21}b_{12}+a_{22}b_{22}\\&=c_{22},\,\end{aligned}}}$