solve-inside-R-3-the-system-2x-y-z-1-x-2y-z-2-x-y-2z-3-

Question Number 67189 by mathmax by abdo last updated on 23/Aug/19

s o l v e i n s i d e R^{3} t h e s y s t e m {\begin{cases} 2 x + y + z = 1 \\ x + 2 y + z = 2 \end{cases}

{x + y + 2 z = 3

Answered by MJS last updated on 23/Aug/19

D = | \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} | = 4

D_{x} = | \begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 1 \\ 3 & 1 & 2 \end{matrix} | = - 2 x = \frac{D_{x}}{D} = - \frac{1}{2}

D_{y} = | \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 2 \end{matrix} | = 2 y = \frac{D_{y}}{D} = \frac{1}{2}

D_{z} = | \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{matrix} | = 6 z = \frac{D_{z}}{D} = \frac{3}{2}

Commented by turbo msup by abdo last updated on 24/Aug/19

t h a n k y o u s i r m j s .