A) [tex]X+2Y= \left[\begin{array}{ccc}1&2\\3&4\\\end{array}\right] [/tex]
[tex]X-2Y= \left[\begin{array}{ccc}1&3\\5&7&\\\end{array}\right] [/tex]
2Y, -2Y se reduc asa ca avem: [tex]2X= \left[\begin{array}{ccc}1&2&\\3&4\\\end{array}\right] + \left[\begin{array}{ccc}1&3\\5&7\\\end{array}\right] [/tex]
[tex]2X= \left[\begin{array}{ccc}2&5\\8&11\\\end{array}\right] [/tex]
[tex]X= \left[\begin{array}{ccc}1& \frac{5}{2} \\4& \frac{11}{2} \\\end{array}\right] [/tex]
Apoi inlocuim in prima sau a doua ecuatie pentru a-l afla pe Y, eu zic in prima:
[tex]2Y= \left[\begin{array}{ccc}1&2\\3&4\\\end{array}\right] - \left[\begin{array}{ccc}1& \frac{5}{2} \\4& \frac{11}{2} \\\end{array}\right] [/tex]
[tex] 2Y= \left[\begin{array}{ccc}0& \\ \frac{-1}{2} \\ -1&\frac{-3}{2} \\\end{array}\right]
[/tex]
[tex]Y= \left[\begin{array}{ccc}0& -1\\ \frac{-1}{2} &-3\\\end{array}\right] [/tex]
