\title	{División por la derecha 2x2}
\author	{XIAO Gang}
\email	{xiao@unice.fr}
\format	{tex}

\integer{range=20}
\integer{sign1=random(1,-1)}
\integer{sign2=random(1,-1)}
\integer{a0=random(0..\range)}
\integer{b0=random(1..\range)}
\integer{c0=random(1..\range)}
\integer{d0=random(0..\range)}
\integer{a2=\a0*(\sign1)}
\integer{b2=\b0*(\sign1)}
\integer{c2=-(\c0)*(\sign2)}
\integer{d2=\d0*(\sign2)}
\integer{a1=random(-\range..\range)}
\integer{b1=random(-\range..\range)}
\integer{c1=random(-\range..\range)}
\integer{d1=random(-\range..\range)}
\integer{a3=(\a1)*(\a2)+(\b1)*(\c2)}
\integer{b3=(\a1)*(\b2)+(\b1)*(\d2)}
\integer{c3=(\c1)*(\a2)+(\d1)*(\c2)}
\integer{d3=(\c1)*(\b2)+(\d1)*(\d2)}

\statement{Determina la matriz $A=\pmatrix{a&b\cr c&d}$ tal que
 $$ \pmatrix{a&b\cr c&d} \pmatrix{\a2&\b2\cr\c2&\d2}
 = \pmatrix{\a3&\b3\cr\c3&\d3} .$$}
\answer{a=\a1}
\answer{b=\b1}
\answer{c=\c1}
\answer{d=\d1}
