Question

Problema matrici
Daca se poate,multumesc :)

Answer 1

$\displaystyle A= \left(\begin{array}{ccc}2&1\\-1&3\end{array}\right)~~~~~~~~~~~~~~~~~~~~B= \left(\begin{array}{ccc}5&-2\\4&1\end{array}\right)\\ \\ \\AB= \left(\begin{array}{ccc}2&1\\-1&3\end{array}\right)\left(\begin{array}{ccc}5&-2\\4&1\end{array}\right)=\\ \\ \left(\begin{array}{ccc}2\cdot5+1\cdot4&2\cdot(-2)+1\cdot1\\(-1)\cdot5+3\cdot4&(-1)\cdot(-2)+3\cdot1\end{array}\right)=\left(\begin{array}{ccc}10+4&(-4)+1\\(-5)+12&2+3\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}14&-3\\7&5\end{array}\right)$

$AB=\left(\begin{array}{ccc}14&-3\\7&5\end{array}\right)$

$\displaystyle BA=\left(\begin{array}{ccc}5&-2\\4&1\end{array}\right)\left(\begin{array}{ccc}2&1\\-1&3\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}5\cdot2+(-2)\cdot(-1)&5\cdot1+(-2)\cdot3\\4\cdot2+1\cdot(-1)&4\cdot1+1\cdot3\end{array}\right)=\left(\begin{array}{ccc}10+2&5-6\\8-1&4+3\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}12&-1\\7&7\end{array}\right)\\ \\ \\ BA=\left(\begin{array}{ccc}12&-1\\7&7\end{array}\right)$

$\displaystyle A^2=A\cdot A=\left(\begin{array}{ccc}2&1\\-1&3\end{array}\right)\left(\begin{array}{ccc}2&1\\-1&3\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}2\cdot2+1\cdot(-1)&2\cdot1+1\cdot3\\(-1)\cdot2+3\cdot(-1)&(-1)\cdot1+3\cdot3\end{array}\right)=\\ \\=\left(\begin{array}{ccc}4-1&2+3\\(-2)-3&(-1)+9\end{array}\right)=\left(\begin{array}{ccc}3&5\\-5&8\end{array}\right)\\ \\ \\ A^2=\left(\begin{array}{ccc}3&5\\-5&8\end{array}\right)$

$\displaystyle B^2=B\cdot B=\left(\begin{array}{ccc}5&-2\\4&1\end{array}\right)\left(\begin{array}{ccc}5&-2\\4&1\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}5\cdot5+(-2)\cdot4&5\cdot(-2)+(-2)\cdot1\\4\cdot5+1\cdot4&4\cdot(-2)+1\cdot1\end{array}\right)=\left(\begin{array}{ccc}25-8&(-10)-2\\20+4&(-8)+1\end{array}\right)=\\ \\ =\left(\begin{array}{ccc}17&-12\\24&-7\end{array}\right)\\ \\ \\ B^2=\left(\begin{array}{ccc}17&-12\\24&-7\end{array}\right)$