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Solve by Crout's method the following system of equations: 3x+2y+7z=4, 2x+3y+z=5, 3x+4y+z=7

Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 8 M

Year : May 2015

1 Answer
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I$_{11}=3, \; I_{21}=2, \; I_{31}=3 \\ \; \\ \; \\ u_{12}= \dfrac{a_{12}}{l_{11}} =\dfrac{2}{3} \; , \; u_{13}= \dfrac{a_{13}}{l_{11}} =\dfrac{7}{3} \\ \; \\ \; \\ I^{22} \;=\; a^{22} - I_{21}u_{12} \;=\; 3-2\times \dfrac{2}{3} \;=\; \dfrac{5}{3} \\ \; \\ \; \\ I^{32} \;=\; a^{32} - I_{31}u_{12} \;=\; 4-2\times \dfrac{2}{3} \;=\; 2 \\ \; \\ \; \\ u_{23}= \dfrac{a_{23} - I_{23}u_{13} }{I_{22}} =\dfrac{1-2 \times \dfrac{7}{3}}{ \dfrac{5}{3} } \;=\; \dfrac{-11}{5} \\ \; \\ \; \\ \; \\ I^{33} \;=\; a^{33} - I_{31}u_{13} - I_{32}u_{23} \;=\; 1-3\times \dfrac{7}{3} -2\times \dfrac{-11}{5} \;=\; \dfrac{-8}{5} \\ \; \\ \; \\ \left[ \begin{array}{ccc} 3 & 0 & 0 \\ 2 & \dfrac{5}{3} & 0 \\ 3 & 2 & \dfrac{-8}{5} \end{array}\right] \left[ \begin{array}{ccc} 1 & \dfrac{2}{3} & \dfrac{7}{3} \\ 0 & 1 & \dfrac{-11}{5} \\ 0 & 0 & 1 \end{array}\right] $

If we multiply this,we'll get the original coefficient matrix.

Now, $ \left[ \begin{array}{ccc} 3 & 0 & 0 \\ 2 & \dfrac{5}{3} & 0 \\ 3 & 2 & \dfrac{-8}{5} \end{array}\right] \left[ \begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 4 \\ 5 \\ 7 \end{array}\right] \\ \; \\ \; \\ \; \\ \therefore d_1 \;=\; \dfrac{4}{3} \; , \; d_2 \;=\; \dfrac{7}{5} \; , \; d_3 \;=\; \dfrac{-1}{8} \\ \; \\ \left[ \begin{array}{ccc} 1 & \dfrac{2}{3} & \dfrac{7}{3} \\ 0 & 1 & \dfrac{-11}{5} \\ 0 & 0 & 1 \end{array}\right] \left[ \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right] \\ \; \\ \; \\ \left[ \begin{array}{ccc} 1 & \dfrac{2}{3} & \dfrac{7}{3} \\ 0 & 1 & \dfrac{-11}{5} \\ 0 & 0 & 1 \end{array}\right] \left[ \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} \dfrac{4}{3} \\ \dfrac{7}{5} \\ \dfrac{-1}{8} \end{array}\right] \\ \; \\ \; \\ \therefore x_3 \;=\; \dfrac{-1}{8} \\ \; \\ x_2 -\dfrac{11}{5} \times \dfrac{-1}{8} \;=\; \dfrac{7}{5} \; \; \; \ \therefore x_2 \;=\; \dfrac{9}{8} \\ \; \\ x_1 + \dfrac{2}{3} \times \dfrac{9}{8} - \dfrac{7}{3} \times \dfrac{1}{8} \;=\; \dfrac{4}{3} \; \; \; \; \therefore x_1 \;=\; \dfrac{7}{8} $

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