Question

Ma puteti ajuta?$\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

Answer 1

$\displaystyle \mathtt{2A+ \left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt1\\\end{array}\right)= \left(\begin{array}{ccc}\mathtt5&\mathtt6\\\mathtt{-1}&\mathtt3\\\end{array}\right)}\\ \\ \\\mathtt{2A=\left(\begin{array}{ccc}\mathtt5&\mathtt6\\\mathtt{-1}&\mathtt3\\\end{array}\right)- \left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt1\\\end{array}\right)}$

$\displaystyle \mathtt{2A= \left(\begin{array}{ccc}\mathtt{5-1}&\mathtt{6-2}\\\mathtt{(-1)-3}&\mathtt{3-1}\\\end{array}\right)}\\ \\ \\ \mathtt{2A= \left(\begin{array}{ccc}\mathtt4&\mathtt4\\\mathtt{-4}&\mathtt2\\\end{array}\right) }\\ \\ \\ \mathtt{A= \frac{1}{2} \left(\begin{array}{ccc}\mathtt4&\mathtt4\\\mathtt{-4}&\mathtt2\\\end{array}\right)}$

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{1}{2}\cdot 4 }&\mathtt{\displaystyle \frac{1}{2}\cdot 4 }\\\\\mathtt{\displaystyle \frac{1}{2}\cdot (-4) }&\mathtt{\displaystyle \frac{1}{2}\cdot 2 }\\\end{array}\right)}\\ \\ \\ \mathtt{A=\left(\begin{array}{ccc}\mathtt2&\mathtt2\\\mathtt{-2}&\mathtt1\\\end{array}\right)}$