Question

2x+y+z=3
x+y+z=1
x-y-z=1

Answer 1

$\begin{cases}2x+y+z=3\\x+y+z=1\\x-y-z=1\end{cases}$



$A= \left[\begin{array}{ccc}2&1&1\\1&1&1\\1&-1&-1\end{array}\right]$




$D = \left[\begin{array}{ccccc}2&1&1&2&1\\1&1&1&1&1\\1&-1&-1&1&-1 \end{array} \right] \\\\\\ D=\{[2.1.(-1)]+[1.1.1]+[1.1.(-1)]\}-\{[1.1.1]+[-1.1.2]+[-1.1.1]\}\\\\ D = \{[-2]+[1]+[-1]\}-\{[1]+[-2]+[-1]\}\to\\\\ D=\{-2+1-1\}-\{1-2-1\}\to\\\\ D=\{-2\}-\{-2\}\to\\ \\ D=-2+2\to\\\\ \large\boxed{D=0}$




$D_x = \left[\begin{array}{ccccc}3&1&1&3&1\\1&1&1&1&1\\1&-1&-1&1&-1 \end{array} \right] \\\\\\ D_x=\{[3.1.(-1)]+[1.1.1]+[1.1.(-1)]\}-\{[1.1.1]+[-1.1.3]+[-1.1.1]\}\\\\ D_x = \{[-3]+[1]+[-1]\}-\{[1]+[-3]+[-1]\}\to\\\\ D_x=\{-3+1-1\}-\{1-3-1\}\to\\\\ D_x=\{-3\}-\{-3\}\to\\ \\ D=-3+3\to\\\\ \large\boxed{D_x=0}$




$D_y = \left[\begin{array}{ccccc}2&3&1&2&3\\1&1&1&1&1\\1&1&-1&1&1 \end{array} \right] \\\\\\ D_y=\{[2.1.(-1)]+[3.1.1]+[1.1.1]\}-\{[1.1.1]+[1.1.2]+[-1.1.3]\}\\\\ D_y = \{[-2]+[3]+[1]\}-\{[1]+[2]+[-3]\}\to\\\\ D=\{-2+3+1\}-\{1+2-3\}\to\\\\ D_y=\{2\}-\{0\}\to\\ \\ D_y=2-0\to\\\\ \large\boxed{D_y=2}$




$D_z = \left[\begin{array}{ccccc}2&1&3&2&1\\1&1&1&1&1\\1&-1&1&1&-1 \end{array} \right] \\\\\\ D_z=\{[2.1.1]+[1.1.1]+[3.1.(-1)]\}-\{[1.1.3]+[-1.1.2]+[1.1.1]\}\\\\ D_z = \{[2]+[1]+[-3]\}-\{[3]+[-2]+[1]\}\to\\\\ D_z=\{2+1-3\}-\{3-2+1\}\to\\\\ D=\{0\}-\{2\}\to\\ \\ D_z=0-2\to\\\\ \large\boxed{D_z=-2}$




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