[tex] \left[\begin{array}{ccc}2&-2&-4\\-1&3&4\\1&-2&-3\end{array}\right] + \left[\begin{array}{ccc}2&-3&-5\\-1&4&5\\1&-3&-4\end{array}\right]= \left[\begin{array}{ccc}4&-5&-9\\-2&7&9\\2&-5&-7\end{array}\right] \\ \\ [/tex]
[tex] \left[\begin{array}{ccc}2&-2&-4\\-1&3&4\\1&-2&-3\end{array}\right] - \left[\begin{array}{ccc}2&-3&-5\\-1&4&5\\1&-3&-4\end{array}\right]= \left[\begin{array}{ccc}0&1&1\\0&-1&-1\\0&1&1\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}2&-2&-4\\-1&3&4\\1&-2&-3\end{array}\right] \cdot \left[\begin{array}{ccc}2&-3&-5\\-1&4&5\\1&-3&-4\end{array}\right]=\left[\begin{array}{ccc}2&-2&-4\\-1&3&4\\1&-2&-3\end{array}\right][/tex]
la adunare se aduna elementul 1 din prima matrice cu elementul 1 din a doua matrice , apoi se aduna elementul 2 din prima matrice cu elementul 2 din a doua matrice
3.
4x+y+z=1⇒x=(1-y-z)/4
(1-y-z)/4+4y+z=2 si (1-y-z)/4+y+4z=
(1-y-z)/{4+4y+z=2⇒
y=(-3z+7)/15
(1-(-3z+7)/15-z)/4}+(-3z+7)/15+4z=3⇒z=2/3
dar y=(-3z+7)/15⇒y=1/3
iar x=0
deci x = 0, y = 1/3 si z = 2/3
