Solve $x + y + 2 z = 4; 2 x + 2 y + 4 z = 8; 3 x + 3 y + 6 z = 12$ by using determinant method.

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Solution

$x + y + 2 z = 4$
$2 x + 2 y + 4 z = 8$
$3 x + 3 y + 6 z = 12$
By determinant method:
$⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 2 2 & 2 & 4 3 & 3 & 6 \end{matrix} ⎞ ⎟ ⎠ \times ⎛ ⎜ ⎝ \begin{matrix} x y z \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} 4812 \end{matrix} ⎞ ⎟ ⎠$

$D = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 2 2 & 2 & 4 3 & 3 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Similarly, $D_{1}, D_{2}, D_{3} = 0$
As they all are 0 the set of equations will have infinite solutions.

We can consider x and y as free variables... $x = α; y = β$

Thus, the solution set is $(α, β, \frac{4 - α - β}{2})$

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