Răspuns :
[tex]\dislaystyle\bf\\\begin{cases}3x+y-z=4\\2x+2y-z=4\\x+y+z=2\end{cases}\\\\\\Matricea~extinsa~a~sistemului\\\\\left|\begin{array}{ccc}3&1&-1\\2&2&-1\\1&1&1\end{array}\right|\left|\begin{array}{c}4\\4\\2\end{array}\right|\\\\Aplicam~metoda~lui~Gauss~de~triangularizare~a~mareicii~coeficientilor.\\Pasul~1:~~L_1~\leftrightarrow~L_3\\\\\left|\begin{array}{ccc}1&1&1\\2&2&-1\\3&1&-1\end{array}\right|\left|\begin{array}{c}2\\4\\4\end{array}\right|\\\\[/tex]
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[tex]\dislaystyle\bf\\Pasul~2:~~L2+(-2)\times L1\\\\\left|\begin{array}{ccc}1&1&1\\0&0&-3\\3&1&-1\end{array}\right|\left|\begin{array}{c}2\\0\\4\end{array}\right|\\\\Pasul~3:~~L_2~\leftrightarrow~L_3\\\\\left|\begin{array}{ccc}1&1&1\\3&1&-1\\0&0&-3\end{array}\right|\left|\begin{array}{c}2\\4\\0\end{array}\right|\\\\Pasul~4:~~L2+(-3)\times L1\\\\\left|\begin{array}{ccc}1&1&1\\0&-2&-4\\0&0&-3\end{array}\right|\left|\begin{array}{c}2\\-2\\0\end{array}\right|\\\\Pasul~5:~~Rezolvam~sistemul~de~jos~in~sus.[/tex]
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[tex]\dislaystyle\bf\\\left|\begin{array}{ccc}1&1&1\\0&-2&-4\\0&0&-3\end{array}\right|\left|\begin{array}{c}2\\-2\\0\end{array}\right|\\\\\\\begin{cases}x+y+z=2\\~~-2y-4z=-2\\~~~~~~~~~-3z=0\\\end{cases}\\\\\\Din~E_3~\implies~~-3z=0~\implies~\boxed{\bf z=0}\\\\Din~E_2~\implies~-2y-4\times0=-2\implies~-2y=-2\implies~\boxed{\bf y=1}\\\\Din~E_1~\implies~x+1+0=2~~\implies~x+1=2~\implies~\boxed{\bf x=1}[/tex]