Find nature of solution for given system of equations
$2 x + y + z = 3$ ; $x + 2 y + z = 4$ ; $3 x + z = 2$

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Solution

$\begin{matrix} \Rightarrow D = ⎛ ⎜ ⎝ \begin{matrix} 2 & 1 & 1 1 & 2 & 1 3 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠ \Rightarrow D = 2 (2) - 1 (1 - 3) + 1 (- 6) \Rightarrow D = 4 + 2 - 6 = 0 \Rightarrow D_{1} = ⎛ ⎜ ⎝ \begin{matrix} 3 & 1 & 1 4 & 2 & 1 2 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠ \Rightarrow D_{1} = 3 (2) - 1 (4 - 2) + 1 (- 4) \Rightarrow D_{1} = 6 - 2 - 4 = 0 \Rightarrow D_{2} = ⎛ ⎜ ⎝ \begin{matrix} 2 & 1 & 1 1 & 2 & 1 3 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠ \Rightarrow D_{2} == 2 (2) - 3 (1 - 3) + 1 (2 + 2) \Rightarrow 4 + 6 - 10 = 0 \Rightarrow D = D_{1} = D_{2} = 0 \end{matrix}$
hence system has infinite number of solution.

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