2x − 3z + w = 1
x − y + 2w = 1
− 3y + z + w = 1
x + y + z = 1

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Solution

$D = |\begin{matrix} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| 2 |\begin{matrix} - 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (0 - 1) - 0 (0 - 1) + 2 (- 3 - 1)] - 3 [1 (0 - 1) + 1 (0 - 1) + 2 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 0 (0 + 3)] = - 21 D_{1} = |\begin{matrix} 1 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 1 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| 1 |\begin{matrix} - 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 2 \\ 1 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 1 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 1 [- 1 (0 - 1) - 0 (0 - 1) + 2 (- 3 - 1)] - 3 [1 (0 - 1) + 1 (0 - 1) + 2 (1 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 2 (1 + 3)] = - 21 D_{2} = |\begin{matrix} 2 & 1 & - 3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| = 2 |\begin{matrix} 1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| + (- 3) |\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}| 2 [1 (0 - 1) + 2 (1 - 1)] - 1 [1 (0 - 1) + 2 (0 - 1)] - 3 [1 (0 - 1) - 1 (0 - 1) + 2 (0 - 1)] - 1 [1 (1 - 1) - 1 (0 - 1)] = 6 D_{3} = |\begin{matrix} 2 & 0 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| = 2 |\begin{matrix} - 1 & 1 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 + 1 |\begin{matrix} 1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (0 - 1) - 1 (0 - 1) + 2 (- 3 - 1)] + 1 [1 (0 - 1) + 1 (0 - 1) + 2 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 1 (0 + 3)] = - 6 D_{4} = |\begin{matrix} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix}| = 2 |\begin{matrix} - 1 & 0 & 1 \\ - 3 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (1 - 1) + 1 (- 3 - 1)] - 3 [1 (- 3 - 1) + 1 (0 - 1) + 1 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1)] = 3 So, by Cramer' s rule, we obtain x = \frac{D_{1}}{D} = \frac{21}{21} = 1 y = \frac{D_{2}}{D} = \frac{6}{- 21} = - \frac{2}{7} z = \frac{D_{3}}{D} = \frac{- 6}{- 21} = \frac{2}{7} w = \frac{D_{4}}{D} = \frac{3}{- 21} = - \frac{1}{7} Hence, x = 1, y = - \frac{2}{7}, z = \frac{2}{7}, w = - \frac{1}{7}$

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