Question

Solve the following
0.6x + 0.8y + 0.1z = 1
1.1 x + 0.4y + 0.3z= 0.2
x + y + 2z =0.5
by LU decomposition method and find the inverse of the coefficient matrix

Answer 1

Answer on Question #81954 — Math — Linear Algebra

Question

Solve the following

$$0.6x + 0.8y + 0.1z = 1$$ $$1.1x + 0.4y + 0.3z = 0.2$$ $$x + y + 2z = 0.5$$

by LU decomposition method and find the inverse of the coefficient matrix

Solution

$$A = \begin{pmatrix} 0.6 & 0.8 & 0.1 \\ 1.1 & 0.4 & 0.3 \\ 1 & 1 & 2 \end{pmatrix}; \quad b = \begin{pmatrix} 1 \\ 0.2 \\ 0.5 \end{pmatrix}$$

Swap some rows in matrices.

$$A = \begin{pmatrix} 1 & 1 & 2 \\ 1.1 & 0.4 & 0.3 \\ 0.6 & 0.8 & 0.1 \end{pmatrix}; \quad b = \begin{pmatrix} 0.5 \\ 0.2 \\ 1 \end{pmatrix}$$

LU = A

$$L = \begin{pmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{pmatrix}; \quad U = \begin{pmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{pmatrix};$$ $$\begin{pmatrix} 1 & 1 & 2 \\ 1.1 & 0.4 & 0.3 \\ 0.6 & 0.8 & 0.1 \end{pmatrix} \begin{pmatrix} 0.5 \\ 0.2 \\ 1 \end{pmatrix} \to R_2^* = R_2 + (-1.1) * R_1; \quad R_3^* = R_3 + (-0.6) * R_1; \quad \to$$ $$\begin{pmatrix} 1 & 1 & 2 \\ 0 & -0.7 & -1.9 \\ 0 & 0.2 & -1.1 \end{pmatrix} \begin{pmatrix} 0.5 \\ -0.35 \\ 0.7 \end{pmatrix};$$ $$U = \begin{pmatrix} 1 & 1 & 2 \\ 0 & -0.7 & -1.9 \\ 0 & 0 & -1.643 \end{pmatrix}; \quad b = \begin{pmatrix} 0.5 \\ -0.35 \\ 0.6 \end{pmatrix}$$ $$L = \begin{pmatrix} 1 & 0 & 0 \\ 1.1 & 1 & 0 \\ 0.6 & * & 1 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 1 & 2 \\ 0 & -0.7 & -1.9 \\ 0 & 0.2 & -1.1 \end{pmatrix} \begin{pmatrix} 0.5 \\ -0.35 \\ 0.7 \end{pmatrix} \to R_3^* = R_3 + \begin{pmatrix} 0.2 \\ 0.7 \end{pmatrix} * R_2 \to \begin{pmatrix} 1 & 1 & 2 \\ 0 & -0.7 & -1.9 \\ 0 & 0 & -1.643 \end{pmatrix} \begin{pmatrix} 0.5 \\ -0.35 \\ 0.6 \end{pmatrix}$$ $$L = \begin{pmatrix} 1 & 0 & 0 \\ 1.1 & 1 & 0 \\ 0.6 & -0.2857 & 1 \end{pmatrix}$$ $$U = \begin{pmatrix} 1 & 1 & 2 \\ 0 & -0.7 & -1.9 \\ 0 & 0 & -1.643 \end{pmatrix}; \quad b = \begin{pmatrix} 0.5 \\ -0.35 \\ 0.6 \end{pmatrix}$$ $$\begin{cases} x + y + 2z = 0.5 \\ -0.7y - 1.9z = -0.35 \\ -1.643z = 0.6 \end{cases}$$

So, we have: $Z = 0.6 / (-1.643) = -0.365$

$$Y = \frac {- 0.35 + 1.9 * (- 0.365)}{- 0.7} = 1.4907$$ $$X = 0.5 - 2 * (-0.365) - 1.4907 = -0.2607$$

Find the inverse of the coefficient matrix

$$\left( \begin{array}{cccc} 0.6 & 0.8 & 0.1 & 1 & 0 & 0 \\ 1.1 & 0.4 & 0.3 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0 & 0 & 1 \end{array} \right);$$

1. $R_{1}^{*} = R_{1} / 0.6$

$$\left( \begin{array}{cccc} 1 & 1.33 & 0.167 & 1.67 & 0 & 0 \\ 1.1 & 0.4 & 0.3 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0 & 0 & 1 \end{array} \right);$$

2. $R_{2}^{*} = R_{2} - 1.1 * R_{1}; R_{3}^{*} = R_{3} - R_{1}$

$$\left( \begin{array}{cccc} 1 & 1.33 & 0.167 & 1.67 & 0 & 0 \\ 0 & -1.067 & 0.1167 & -1.83 & 1 & 0 \\ 0 & -0.33 & 1.83 & -1.67 & 0 & 1 \end{array} \right);$$

3. $R_{2}^{*} = R_{2} / (-1.067)$

$$\left( \begin{array}{cccc} 1 & 1.33 & 0.167 & 1.67 & 0 & 0 \\ 0 & 1 & -0.109 & 1.719 & -0.9375 & 0 \\ 0 & -0.33 & 1.83 & -1.67 & 0 & 1 \end{array} \right);$$

4. $R_{1}^{*} = R_{1} - 1.33 * R_{2}; R_{3}^{*} = R_{3} + 0.33 * R_{2}$

$$\left( \begin{array}{cccc} 1 & 0 & 0.3125 & -0.625 & 1.25 & 0 \\ 0 & 1 & -0.109 & 1.719 & -0.9375 & 0 \\ 0 & 0 & 1.797 & -1.094 & -0.3125 & 1 \end{array} \right);$$

5. $R_{3}^{*} = \frac{R_{3}}{1.797}$

$$\left( \begin{array}{cccc} 1 & 0 & 0.3125 & -0.625 & 1.25 & 0 \\ 0 & 1 & -0.109 & 1.719 & -0.9375 & 0 \\ 0 & 0 & 1 & -0.609 & -0.174 & 0.556 \end{array} \right);$$

6. $R_{1}^{*} = R_{1} - 0.3125 * R_{3}; R_{2}^{*} = R_{2} + 0.109 * R_{3}$

$$\left( \begin{array}{cccc} 1 & 0 & 0 & -0.435 & 1.304 & -0.174 \\ 0 & 1 & 0 & 1.65 & -0.957 & 0.061 \\ 0 & 0 & 1 & -0.609 & -0.174 & 0.556 \end{array} \right);$$

Answer: $Z = -0.365$; $Y = 1.4907$; $X = -0.2607$;

$$\left( \begin{array}{cccc} -0.435 & 1.304 & -0.174 \\ 1.65 & -0.957 & 0.061 \\ -0.609 & -0.174 & 0.556 \end{array} \right).$$

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