3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11

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Solution

Given: 3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11

$D = |\begin{matrix} 3 & 1 & 1 \\ 2 & - 4 & 3 \\ 4 & 1 & - 3 \end{matrix}| = 3 (12 - 3) - 2 (- 3 - 1) + 4 (3 + 4) = 27 + 8 + 28 = 63 D_{1} = |\begin{matrix} 2 & 1 & 1 \\ - 1 & - 4 & 3 \\ - 11 & 1 & - 3 \end{matrix}| = 2 (12 - 3) + 1 (- 3 - 1) - 11 (3 + 4) = 18 - 4 - 77 = - 63 D_{2} = |\begin{matrix} 3 & 2 & 1 \\ 2 & - 1 & 3 \\ 4 & - 11 & - 3 \end{matrix}| = 3 (3 + 33) - 2 (- 6 + 11) + 4 (6 + 1) = 108 - 10 + 28 = 126 D_{3} = |\begin{matrix} 3 & 1 & 2 \\ 2 & - 4 & - 1 \\ 4 & 1 & - 11 \end{matrix}| = 3 (44 + 1) - 2 (- 11 - 2) + 4 (- 1 + 8) = 135 + 26 + 28 = 189 Now, x = \frac{D_{1}}{D} = \frac{- 63}{63} = - 1 y = \frac{D_{2}}{D} = \frac{126}{63} = 2 z = \frac{D_{3}}{D} = \frac{189}{63} = 3 ∴ x = - 1, y = 2 and z = 3$

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