Question

Gauss Elimination Method

0 = 0.13D1 + 0.03D2
0 = 0.3D1 + 0.13D2 - 0.1D4
2 = 0.13D3 - 0.03D4
-4= -0.1D2 - 0.03D3 + 0.13D4

Answer 1

Answer on Question#54307 – Math – Linear Algebra

$$D _ {1} = - 3 2 \frac {8 4}{2 1 3}; \quad D _ {2} = 1 4 0 \frac {8 0}{2 1 3}; \quad D _ {3} = 3 5 \frac {1 5}{2 1 3}; \quad D _ {4} = 8 5 \frac {6 5}{2 1 3}.$$

Question

Gauss Elimination Method

$$0 = 0. 1 3 D _ {1} + 0. 0 3 D _ {2}$$ $$0 = 0. 3 D _ {1} + 0. 1 3 D _ {2} - 0. 1 D _ {4}$$ $$2 = 0. 1 3 D _ {3} - 0. 0 3 D _ {4}$$ $$- 4 = - 0. 1 D _ {2} - 0. 0 3 D _ {3} + 0. 1 3 D _ {4}$$

Solution

Normalize notation

$$\left\{ \begin{array}{l} 0. 1 3 D _ {1} + 0. 0 3 D _ {2} + 0. 0 D _ {3} + 0. 0 D _ {4} = 0 \\ 0. 3 D _ {1} + 0. 1 3 D _ {2} + 0. 0 D _ {3} - 0. 1 D _ {4} = 0 \\ 0. 0 D _ {1} + 0. 0 D _ {2} + 0. 1 3 D _ {3} - 0. 0 3 D _ {4} = 2 \\ 0. 0 D _ {1} - 0. 1 D _ {2} - 0. 0 3 D _ {3} + 0. 1 3 D _ {4} = - 4 \end{array} \right.$$

Eliminate $D_{1}$ in the second row, using first one

$$\left(0. 3 - 0. 1 3 * \frac {0 . 3}{0 . 1 3}\right) D _ {1} + \left(0. 1 3 - 0. 0 3 * \frac {0 . 3}{0 . 1 3}\right) D _ {2} + 0. 0 D _ {3} - 0. 1 D _ {4} = 0$$ $$0. 0 D _ {1} + \frac {0 . 7 9}{1 3} D _ {2} + 0. 0 D _ {3} - 0. 1 D _ {4} = 0$$

Last three rows after operation (omit $D_{1}$ term). Rearrange order of third and fourth rows

$$\left\{ \begin{array}{c} \frac {0 . 7 9}{1 3} D _ {2} + 0. 0 D _ {3} - 0. 1 D _ {4} = 0 \\ - 0. 1 D _ {2} - 0. 0 3 D _ {3} + 0. 1 3 D _ {4} = - 4 \\ 0. 0 D _ {2} + 0. 1 3 D _ {3} - 0. 0 3 D _ {4} = 2 \end{array} \right.$$

Eliminate $D_{2}$ in the second row, using first one

$$\left(- 0. 1 + \frac {0 . 7 9}{1 3} * \frac {1 . 3}{0 . 7 9}\right) D _ {2} + (- 0. 0 3) D _ {3} + \left(0. 1 3 - 0. 1 * \frac {1 . 3}{0 . 7 9}\right) D _ {4} = - 4$$ $$0. 0 D _ {2} - 0. 0 3 D _ {3} - 0. 1 3 * \frac {0 . 2 1}{0 . 7 9} D _ {4} = - 4$$

Last two rows after operation (omit $D_{2}$ term)

$$\left\{ \begin{array}{c} - 0. 0 3 D _ {3} - 0. 1 3 * \frac {0 . 2 1}{0 . 7 9} D _ {4} = - 4 \\ 0. 1 3 D _ {3} - 0. 0 3 D _ {4} = 2 \end{array} \right.$$

Eliminate $D_{3}$ in the second row, using first one

$$\left(0. 1 3 - 0. 0 3 * \frac {0 . 1 3}{0 . 0 3}\right) D _ {3} + \left(- 0. 0 3 - 0. 1 3 * \frac {0 . 2 1}{0 . 7 9} * \frac {0 . 1 3}{0 . 0 3}\right) D _ {4} = \left(2 - 4 * \frac {0 . 1 3}{0 . 0 3}\right)$$ $$0.0D_3 - \left(0.03 + 7 * \frac{0.13^2}{0.79}\right) D_4 = \left(2 - \frac{0.52}{0.03}\right)$$ $$0.0D_3 - \frac{0.142}{0.79} D_4 = -\frac{0.46}{0.03}$$ $$D_4 = \frac{0.46}{0.03} * \frac{0.79}{0.142} = \frac{46}{3} * \frac{790}{142} = \frac{23}{3} * \frac{790}{71} = \frac{18170}{213} = 85 \frac{65}{213}$$

Eliminate $D_4$ term from the equations backward ($3^{\text{rd}}$ and $2^{\text{nd}}$ equations respectively):

$$-0.03D_3 - 0.13 * \frac{0.21}{0.79} * \left(\frac{0.46}{0.03} * \frac{0.79}{0.142}\right) = -4$$ $$0.03D_3 = 4 - 7 * 0.13 * \frac{0.23}{0.071} = \frac{0.0747}{0.071} = \frac{747}{710}$$ $$D_3 = \frac{2490}{71} = 35 \frac{15}{213}$$ $$\frac{0.79}{13} D_2 - 0.1 * \left(\frac{0.46}{0.03} * \frac{0.79}{0.142}\right) = 0$$ $$D_2 = \frac{13}{0.79} * 0.1 * \left(\frac{0.46}{0.03} * \frac{0.79}{0.142}\right)$$ $$D_2 = \frac{13}{0.142} * \frac{0.046}{0.03} = \frac{13}{0.142} * \frac{23}{15} = \frac{299}{2.13} = \frac{29900}{213} = 140 \frac{80}{213}$$

Note: due to particular form of our equations, we have lower number of steps in algorithm.

Eliminate $D_2$ from the $1^{\text{st}}$ equation

$$0.13D_1 + 0.03 * \left(\frac{13}{0.142} * \frac{0.046}{0.03}\right) = 0$$ $$D_1 = -\frac{0.03}{0.13} * \frac{13}{0.142} * \frac{0.046}{0.03} = -\frac{0.13}{0.13} * \frac{4.6}{0.142} = -\frac{6.9}{0.213} = -\frac{6900}{213} = -32 \frac{84}{213}$$

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