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Cramer's Rule

The solution ...

Question

The solution set of the system of equations $x + y - z = 6; 3 x - 2 y + z = - 5; x + 3 y - 2 z = 14$ is $(x, y, z)$ then $x + y + z$ is equal to

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Solution

Using Cramer's rule,
$D = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & - 1 3 & - 2 & 1 1 & 3 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = - 3$

$D_{x} = ∣ ∣ ∣ ∣ \begin{matrix} 6 & 1 & - 1 - 5 & - 2 & 1 14 & 3 & - 2 \end{matrix} ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} + R_{2}$
$R_{3} \to R_{3} + 2 R_{2}$
$\Rightarrow D_{x} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & 0 - 5 & - 2 & 1 4 & - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ = - 3$

$D_{y} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 6 & - 1 3 & - 5 & 1 1 & 14 & - 2 \end{matrix} ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} + R_{2}$
$R_{3} \to R_{3} + 2 R_{2}$
$\Rightarrow D_{y} = ∣ ∣ ∣ ∣ \begin{matrix} 4 & 1 & 0 3 & - 5 & 1 7 & 4 & 0 \end{matrix} ∣ ∣ ∣ ∣ = - 9$

$D_{z} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 6 3 & - 2 & - 5 1 & 3 & 14 \end{matrix} ∣ ∣ ∣ ∣ = 6$

Then, $x = \frac{D_{x}}{D} = \frac{- 3}{- 3} = 1$

$y = \frac{D_{y}}{D} = \frac{- 9}{- 3} = 3$

$z = \frac{D_{z}}{D} = \frac{6}{- 3} = - 2$
$∴ x + y + z = 2$

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