2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2

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Solution

$Using the equations we get D = |\begin{matrix} 2 & 1 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 5 & 1 \end{matrix}| \Rightarrow 2 (- 2 + 5) - 1 (1 - 5) - 2 (- 5 + 10) = 0 D_{1} = |\begin{matrix} 4 & 1 & - 2 \\ - 2 & - 2 & 1 \\ - 2 & - 5 & 1 \end{matrix}| \Rightarrow 4 (- 2 + 5) - 1 (- 2 + 2) - 2 (10 - 4) = 0 D_{2} = |\begin{matrix} 2 & 4 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 2 & 1 \end{matrix}| \Rightarrow 2 (- 2 + 2) - 4 (1 - 5) - 2 (- 2 + 10) = 0 D_{3} = |\begin{matrix} 2 & 1 & 4 \\ 1 & - 2 & - 2 \\ 5 & - 5 & - 2 \end{matrix}| \Rightarrow 2 (4 - 10) - 1 (- 2 + 10) + 4 (- 5 + 10) = 0$

$∴ D = D_{1} = D_{2} = 0$

Hence, the system of linear equations has infinitely many solutions.

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x, y, z

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} ∣ ∣ ∣ ∣

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