6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

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Solution

Given: 6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

$D = |\begin{matrix} 6 & 1 & - 3 \\ 1 & 3 & - 2 \\ 2 & 1 & 4 \end{matrix}| = 6 (12 + 2) - 1 (4 + 4) - 3 (1 - 6) = 6 (14) - 1 (8) - 3 (- 5) = 91 D_{1} = |\begin{matrix} 5 & 1 & - 3 \\ 5 & 3 & - 2 \\ 8 & 1 & 4 \end{matrix}| = 5 (12 + 2) - 1 (20 + 16) - 3 (5 - 24) = 5 (14) - 1 (36) - 3 (- 19) = 91 D_{2} = |\begin{matrix} 6 & 5 & - 3 \\ 1 & 5 & - 2 \\ 2 & 8 & 4 \end{matrix}| = 6 (20 + 16) - 5 (4 + 4) - 3 (8 - 10) = 6 (36) - 5 (8) - 3 (- 2) = 182 D_{3} = |\begin{matrix} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{matrix}| = 6 (24 - 5) - 1 (8 - 10) + 5 (1 - 6) = 6 (19) - 1 (- 2) + 5 (- 5) = 91 Now, x = \frac{D_{1}}{D} = \frac{91}{91} = 1 y = \frac{D_{2}}{D} = \frac{182}{91} = 2 z = \frac{D_{3}}{D} = \frac{91}{91} = 1 ∴ x = 1, y = 2 and z = 1$

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