Therefore, the area of the parallelogram determined by x 1 ( a 11 a 21 ) {\displaystyle x_{1}{\binom {a_{11}}{a_{21}}}} and ( a 12 a 22 ) {\displaystyle {\binom {a_{12}}{a_{22}}}} has to be x 1 {\displaystyle x_{1}} times the area of the first one since one of the sides has been multiplied by this factor. Now, this last parallelogram, by Cavalieri's principle, has the same area as the parallelogram determined by ( b 1 b 2 ) = x 1 ( a 11 a 21 ) + x 2 ( a 12 a 22 ) {\displaystyle {\binom {b_{1}}{b_{2}}}=x_{1}{\binom {a_{11}}{a_{21}}}+x_{2}{\binom {a_{12}}{a_{22}}}} and ( a 12 a 22 ) {\displaystyle {\binom {a_{12}}{a_{22}}}} .