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Solve the system Ax = b using the given LU factorization of A by first solving Ly = b for y
Solution
Given the LU factorization of matrix A, we need to solve the system Ax=b. The LU factorization expresses A as the product of a lower triangular matrix L and an upper triangular matrix U, i.e., A=LU.
The process involves two steps:
Solve the system Ly=b for y (forward substitution).
Solve the system Ux=y for x (backward substitution).
We are given:
The matrix A,
2 & -8 & 0 \\
3 & -7 & 8 \\
-1 & 4 & 2
\end{pmatrix}$$
- The LU factorization,
$$L = \begin{pmatrix}
1 & 0 & 0 \\
\frac{3}{2} & 1 & 0 \\
-\frac{1}{2} & \frac{1}{2} & 1
\end{pmatrix}, \quad
U = \begin{pmatrix}
2 & -8 & 0 \\
0 & 5 & 8 \\
0 & 0 & 2
\end{pmatrix}$$
- The vector $$b = \begin{pmatrix} 10 \\ -1 \\ -9 \end{pmatrix}$$.
### Step 1: Solve $$Ly = b$$
We need to solve:
$$L \cdot y = b$$
Where $$y = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$$ and $$L$$ is the lower triangular matrix:
$$\begin{pmatrix}
1 & 0 & 0 \\
\frac{3}{2} & 1 & 0 \\
-\frac{1}{2} & \frac{1}{2} & 1
\end{pmatrix} \cdot \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 10 \\ -1 \\ -9 \end{pmatrix}$$
This leads to the following system of equations:
1. $$y_1 = 10$$
2. $$\frac{3}{2} y_1 + y_2 = -1$$
3. $$-\frac{1}{2} y_1 + \frac{1}{2} y_2 + y_3 = -9$$
#### Solve for $$y_1$$, $$y_2$$, and $$y_3$$:
1. $$y_1 = 10$$
2. $$\frac{3}{2} (10) + y_2 = -1$$
$$15 + y_2 = -1$$
$$y_2 = -1 - 15 = -16$$
3. $$-\frac{1}{2} (10) + \frac{1}{2} (-16) + y_3 = -9$$
$$-5 - 8 + y_3 = -9$$
$$y_3 = -9 + 13 = 4$$
Thus, we have:
$$y = \begin{pmatrix} 10 \\ -16 \\ 4 \end{pmatrix}$$
### Step 2: Solve $$Ux = y$$
Next, we solve:
$$U \cdot x = y$$
Where $$x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$ and $$U$$ is the upper triangular matrix:
$$\begin{pmatrix}
2 & -8 & 0 \\
0 & 5 & 8 \\
0 & 0 & 2
\end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 10 \\ -16 \\ 4 \end{pmatrix}$$
This leads to the following system of equations:
1. $$2x_1 - 8x_2 = 10$$
2. $$5x_2 + 8x_3 = -16$$
3. $$2x_3 = 4$$
#### Solve for $$x_1$$, $$x_2$$, and $$x_3$$:
1. From equation (3):
$$2x_3 = 4$$
$$x_3 = 2$$
2. From equation (2):
$$5x_2 + 8(2) = -16$$
$$5x_2 + 16 = -16$$
$$5x_2 = -32$$
$$x_2 = -\frac{32}{5} = -6.4$$
3. From equation (1):
$$2x_1 - 8(-6.4) = 10$$
$$2x_1 + 51.2 = 10$$
$$2x_1 = 10 - 51.2 = -41.2$$
$$x_1 = \frac{-41.2}{2} = -20.6$$
Thus, the solution is:
$$x = \begin{pmatrix} -20.6 \\ -6.4 \\ 2 \end{pmatrix}$$
### Final Answer:
The solution to the system $$Ax = b$$ is:
$$x = \begin{pmatrix} -20.6 \\ -6.4 \\ 2 \end{pmatrix}$$
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Would you like more details on any step or further clarification? Here are some related questions:
1. What are the advantages of LU factorization in solving systems of equations?
2. Can LU factorization be applied to any matrix?
3. How does forward substitution work in triangular systems?
4. What are some efficient numerical methods for solving large linear systems?
5. How do you determine the LU factorization of a given matrix?
**Tip:** LU factorization is particularly useful for solving multiple systems with the same matrix but different right-hand sides, as the factorization only needs to be done once.
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Math Problem Analysis
Mathematical Concepts
Linear Algebra LU Factorization Systems of Linear Equations Forward and Backward Substitution
Formulas
LU factorization: A = L * U Forward substitution: L * y = b Backward substitution: U * x = y