[tex]Fie,A= \left[\begin{array}{ccc}x&y\\z&t\\\end{array}\right] [/tex], Matricea transpusa este: [tex] A^{t}= \left[\begin{array}{ccc}x&z\\y&t\\\end{array}\right] [/tex], Suma lor: [tex]A+ A^{t}= \left[\begin{array}{ccc}2x&y+z&\\z+y&2t&\end{array}\right]= [/tex][tex] \left[\begin{array}{ccc}0&0\\0&0\end{array}\right] [/tex]. Rezulta, egalind elementele ce ocupa aceeas pozitie : 2x=2t=0 deci x=t=0, si z+y=y+z=0, adica y=-z, deci solutiile sunt matricele de forma : [tex]A= \left[\begin{array}{ccc}0&a\\-a&0\end{array}\right] [/tex], cu a∈R.
