Using Cramer's rule or otherwise, the solution of equations:
x + y + z = 6
x - y + z = 2 and
3x + 2y - 4z = -5
is .

x = 2, y = 1, z = 3

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x = 2, y = 3, z = 1

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x = 1, y = 3, z = 2

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x = 1, y = 2, z = 3

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Solution

The correct option is D x = 1, y = 2, z = 3
$D = ⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 1 & - 1 & 1 3 & 2 & - 4 \end{matrix} ∣ ∣ ∣ ∣ Applying C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} = ⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 1 & - 2 & 0 3 & - 1 & - 7 \end{matrix} ∣ ∣ ∣ ∣ = 14 D_{1} = ⎛ ⎜ ⎝ \begin{matrix} 6 & 1 & 1 2 & - 1 & 1 - 5 & 2 & - 4 \end{matrix} ∣ ∣ ∣ ∣ Applying R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} + 4 R_{1} = ⎛ ⎜ ⎝ \begin{matrix} 6 & 1 & 1 - 4 & - 2 & 0 19 & 6 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 14 D_{2} = ⎛ ⎜ ⎝ \begin{matrix} 1 & 6 & 1 1 & 2 & 1 3 & - 5 & - 4 \end{matrix} ∣ ∣ ∣ ∣ Applying R_{2} \to R_{2} - R_{1} = ⎛ ⎜ ⎝ \begin{matrix} 1 & 6 & 1 0 & - 4 & 0 3 & - 5 & - 4 \end{matrix} ∣ ∣ ∣ ∣ = 28 D_{3} = ⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 6 1 & - 1 & 2 3 & 2 & - 5 \end{matrix} ∣ ∣ ∣ ∣ Applying C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - 3 C_{1} = ⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 6 0 & - 2 & - 4 0 & - 1 & - 23 \end{matrix} ∣ ∣ ∣ ∣ = 42 Hence by cramer' s rule : x = \frac{D_{1}}{D} = \frac{14}{14} = 1 y = \frac{D_{2}}{D} = \frac{28}{14} = 2 z = \frac{D_{3}}{D} = \frac{42}{14} = 3$

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Similar questions

x + y + z = 6

x - y + z = 2

3 x + 2 y - 4 z = - 5

2 x - y + 3 z = 9, x + y + z = 6, x - y + z = 2

x + y + 4 z = 6

3 x + 2 y - 2 z = 9

5 x + y + 2 z = 13

x + 2 y + z = 1

3 x + y + z = 6

x + 2 y = 0

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