Question

Buna te rog cum rezolv eu ex

Answer 1

$\mathtt{ \left(\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt2\\\mathtt2&\mathtt1&\mathtt2\\\mathtt1&\mathtt2&\mathtt3\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-1}&\mathtt{cos0}\\ \mathtt2&\mathtt{-1}&\mathtt{sin \frac{\pi}{2}} \\\mathtt0&\mathtt1&\mathtt{tg \frac{\pi}{4}} \end{array}\right)=}\\ \\ =\left(\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt2\\ \mathtt2&\mathtt1&\mathtt2\\\mathtt1&\mathtt2&\mathtt3\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-1}&\mathtt1\\ \mathtt2&\mathtt{-1}&\mathtt1\\ \mathtt0&\mathtt1&\mathtt1\end{array}\right)=$
$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{3 \cdot (-1)+1 \cdot 2+2 \cdot 0}&\mathtt{3 \cdot (-1)+1 \cdot (-1)+2 \cdot 1}&\mathtt{3 \cdot 1+1 \cdot 1+2 \cdot 1}\\\mathtt{2 \cdot (-1)+1 \cdot 2+2 \cdot 0}&\mathtt{2 \cdot (-1)+1 \cdot (-1)+2 \cdot 1}&\mathtt{2 \cdot 1+1 \cdot 1+2 \cdot 1}\\\mathtt{1 \cdot (-1)+2 \cdot 2+3 \cdot 0}&\mathtt{1 \cdot (-1)+2 \cdot (-1)+3 \cdot 1}&\mathtt{1 \cdot 1+2 \cdot 1+3 \cdot 1}\end{array}\right)=}$
$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{-3+2+0}&\mathtt{-3-1+2}&\mathtt{3+1+2}\\\mathtt{-2+2+0}&\mathtt{-2-1+2}&\mathtt{2+1+2}\\\mathtt{-1+4+0}&\mathtt{-1-2+3}&\mathtt{1+2+3}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt6\\\mathtt0&\mathtt{-1}&\mathtt{5}\\\mathtt{3}&\mathtt{0}&\mathtt6\end{array}\right)}$