Solve by crouts methods the system of equations 3x+2y+7z=4 2x+3y+z=5 3x+4y+z=7.

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$\begin{bmatrix} 3 & 2 & 7 \ 2 & 3 & 1 \ 3 & 4 & 1 \end{bmatrix}$$\ \ \begin{bmatrix} x \ y \ z \end{bmatrix}$ = $\begin{bmatrix} 4 \ 5 \ 7 \end{bmatrix}$ Writing [A] as [L][U] we have $\begin{bmatrix} 3 & 2 & 7 \ 2 & 3 & 1 \ 3 & 4 & 1 \end{bmatrix}$ = $\begin{bmatrix} u_{11} & u_{12} & u_{13} \ l_{21}u_{11} & l_{21}u_{12} + u_{22} & l_{21}u_{13} + u_{23} \ l_{31}u_{11} & l_{31}u_{12}+l_{32}u_{22} & l_{31}u_{13} + l_{32}u_{23} + u_{33} \end{bmatrix}$ * Hence on comparison of LHS and RHS matrices we get, $u_{11} = 3 $ $u_{12}=2$ $u_{13}=7$ $l_{21}u_{11} = 2$ $l_{21}u_{12}+u_{22}=3$ $l_{21}u_{13}+u_{23} = 1$ $l_{21}=\dfrac{2}{3}$ $u_{22}=3-\dfrac{4}{3} = \dfrac{5}{3}$ $u_{23}=1-\dfrac{14}{3}=\dfrac{-11}{3}$ $l_{31}u_{11} =3$ $l_{31}u_{12}+l_{32}u_{22} =4$ $l_{31}u_{13}+l_{32}u_{23}+u_{33} =1$ $l_{31}=\dfrac{3}{3}=1$ $l_{32}=\dfrac{6}{5}$ $u_{33}=\dfrac{-24}{15} $ then [A] = $\begin{bmatrix} 1 & 0 & 0 \ \dfrac{2}{3} & 1 & 0\ 1 & \dfrac{6}{5} & 1 \end{bmatrix}\ \ \ \ \begin{bmatrix} 3 & 2 & 7 \ 0 & \dfrac{5}{3} & \dfrac{-11}{3}\ 0 & 0 & \dfrac{-24}{15} \end{bmatrix}$ writing LY =B the system becomes $\begin{bmatrix} 1 & 0 & 0 \ \frac{2}{3} & 1 & 0 \ 1 & \frac{6}{5} & 1 \end{bmatrix}$$\ \ \ \begin{bmatrix} y1\ y2 \ y3 \end{bmatrix}$ = $\begin{bmatrix} 4 \ 5 \ 7 \end{bmatrix}$ Solving we get From R1, y1 = 4 From R2, $\dfrac{2}{3}$y1 + y2 = 5 thus y2 = $\dfrac73$ From R3, y1 + $\dfrac65y2 + y3 = 7$ thus y3 = $\dfrac15$ * Solving UX = Y we get, $\begin{bmatrix} 3 & 2 & 7 \ 0 & \dfrac53 & \dfrac{-11}{3} \ 0 & 0 & \dfrac{24}{15} \end{bmatrix}\ \ \ \ \begin{bmatrix}x\ y \ z \end{bmatrix}\ \ \ \ \begin{bmatrix}4 \\dfrac73 \\dfrac15 \end{bmatrix}$ From R3, $\dfrac{-24}{15}z = \dfrac15$ z = $\dfrac{-1}{8}$ From R2, $\dfrac53y-\dfrac{11}{3}z = \dfrac73$ $y = \dfrac{15}{8}$ From r3, 3x + 2y + 7z = 4 x = $\dfrac{-3}{32}$

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