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

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Solution

$Using the equations we get D = |\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2 \end{matrix}| \Rightarrow 1 (2 - 2) + 1 (4 - 1) + 1 (- 4 + 1) = 0 D_{1} = |\begin{matrix} 3 & - 1 & 1 \\ 2 & 1 & - 1 \\ 1 & - 2 & 2 \end{matrix}| \Rightarrow 3 (2 - 2) + 1 (4 + 1) + 1 (- 4 - 1) = 0 D_{2} = |\begin{matrix} 1 & 3 & 1 \\ 2 & 2 & - 1 \\ - 1 & 1 & 2 \end{matrix}| \Rightarrow 1 (4 + 1) - 3 (4 - 1) + 1 (2 + 2) = 0 D_{3} = |\begin{matrix} 1 & - 1 & 3 \\ 2 & 1 & 2 \\ - 1 & - 2 & 1 \end{matrix}| \Rightarrow 1 (1 + 4) + 1 (2 + 2) + 3 (- 4 + 1) = 0$

Here,
$D = D_{1} = D_{2} = D_{3} = 0$
Thus, the system of linear equations has infinitely many solutions.

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Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x - y - z & 2 x & 2 x 2 y & y - z - x & 2 y 2 z & 2 z & z - x - y \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣

(x + 2 y + z) + (x + y + z) - (3 x + 2 y + 2 z)

x = 1, y = 1, z = 1

x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \frac{2 y}{1 - y^{2}} \frac{2 z}{1 - z^{2}}

x + y + z = x y z

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