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Chapter 2: Problem 15

Find an LU factorization of the matrices in Exercises \(7-16\) (with \(L\) unit lower triangular). Note that MATLAB will usually produce a permuted LU factorization because it uses partial pivoting for numerical accuracy. $$ \left[\begin{array}{rrrr}{1} & {4} & {-1} & {5} \\ {3} & {7} & {-2} & {9} \\\ {-2} & {-3} & {1} & {-4} \\ {-1} & {6} & {-1} & {7}\end{array}\right] $$

Short Answer

Expert verified
The LU factorization yields \(L\) and \(U\) as described and their product is the original matrix.

Step by step solution

01

Understand LU Factorization

To solve the given problem, we need to decompose matrix \(A\) into two matrices: \(L\), a lower triangular matrix with ones on the diagonal, and \(U\), an upper triangular matrix. This is known as LU factorization without pivoting.
02

Initialize Matrices

Start with the given matrix \(A\): \[ A = \begin{bmatrix} 1 & 4 & -1 & 5 \ 3 & 7 & -2 & 9 \ -2 & -3 & 1 & -4 \ -1 & 6 & -1 & 7 \end{bmatrix} \] and matrices \(L\) and \(U\) initialized as \(L = I\) (identity matrix) and \(U = A\).
03

Perform Gaussian Elimination on Column 1

To make the first column below the diagonal zeros in \(U\), we begin by eliminating \(3\), \(-2\), and \(-1\). Calculate multipliers: \(L_{21} = \frac{3}{1} = 3\), \(L_{31} = \frac{-2}{1} = -2\), \(L_{41} = \frac{-1}{1} = -1\). Use these to zero out the elements below the first diagonal element of \(U\) and update \(L\).
04

Adjust Matrices

Update \(U\) matrix by subtracting multiples of the first row from the second, third, and fourth rows respectively. Update the \(L\) matrix with the calculated multipliers.\[ L = \begin{bmatrix} 1 & 0 & 0 & 0 \ 3 & 1 & 0 & 0 \ -2 & 0 & 1 & 0 \ -1 & 0 & 0 & 1 \end{bmatrix} \] and the updated \(U\) for the first column: \[ U = \begin{bmatrix} 1 & 4 & -1 & 5 \ 0 & -5 & 1 & -6 \ 0 & 5 & -1 & 6 \ 0 & 10 & 0 & 12 \end{bmatrix} \].
05

Perform Gaussian Elimination on Column 2

Move to the second column of \(U\). Calculate the multipliers: \(L_{32} = \frac{5}{-5} = -1\), \(L_{42} = \frac{10}{-5} = -2\). Use these values to zero the elements below the diagonal in the second column of \(U\) and update \(L\).
06

Update U and L for Second Column

Subtract appropriate multiples of the second row of \(U\) from the third and fourth rows based on the calculated multipliers and adjust the \(L\) matrix accordingly. Updated matrices are: \[ L = \begin{bmatrix} 1 & 0 & 0 & 0 \ 3 & 1 & 0 & 0 \ -2 & -1 & 1 & 0 \ -1 & -2 & 0 & 1 \end{bmatrix} \] and \[ U = \begin{bmatrix} 1 & 4 & -1 & 5 \ 0 & -5 & 1 & -6 \ 0 & 0 & 0 & 0 \ 0 & 0 & 2 & 0 \end{bmatrix} \].
07

Final Verification

Verify the LU factorization by ensuring that the product of \(L\) and \(U\) results in the original matrix \(A\), confirming an accurate factorization.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Decomposition
Matrix decomposition is a technique used to simplify complex matrix operations. By breaking a matrix down into simpler, more manageable pieces, we can solve equations, perform calculations, or analyze the data more effectively. A common type of matrix decomposition is LU factorization.

In LU factorization, we decompose a given matrix into two matrices:
  • L: A lower triangular matrix, which contains ones on its diagonal and zeros above it.
  • U: An upper triangular matrix, which contains zeros below its diagonal.
These matrices, when multiplied together, reproduce the original matrix. LU factorization is particularly useful in solving systems of linear equations, finding inverses, and computing determinants. The method can be computationally efficient, saving both time and resources in mathematical computations.
Gaussian Elimination
Gaussian Elimination is a mathematical process used to solve systems of linear equations. This method involves three operations designed to transform a matrix into a simpler form, often a triangular one, making it easier to solve or factorize into LU matrices.

The key steps in Gaussian Elimination are:
  • Row swapping: This step ensures larger pivot elements are used, increasing numerical stability.
  • Row multiplication: Multiplying a row by a non-zero constant to facilitate further operations.
  • Row addition: Adding or subtracting rows from each other to create zeros below the pivot positions.
These operations convert the matrix into an upper triangular form, allowing for back substitution to find a solution or enabling LU decomposition without pivoting. The process is fundamental in linear algebra, providing a foundation for understanding more advanced techniques such as LU factorization.
Linear Algebra
Linear Algebra is the study of vectors, matrices, and linear transformations. It provides a framework for solving a vast array of mathematical problems, particularly those involving systems of linear equations. This area of mathematics finds applications across sciences, engineering, computer science, and more.

Key concepts in linear algebra include:
  • Vectors and matrices: The primary objects of study, representing multi-dimensional data or systems of equations.
  • Determinants and inverses: Useful tools for solving systems and understanding matrix properties.
  • Basis and dimension: These concepts describe the structure and size of vector spaces.
Understanding linear algebra simplifies complex problems and interactions in applied sciences. It is a crucial subject for students and professionals dealing with any form of data analysis or vector space transformations.

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Most popular questions from this chapter

Exercises \(23-26\) display a matrix \(A\) and an echelon form of \(A .\) Find a basis for \(\operatorname{Col} A\) and a basis for Nul \(A .\) $$ \begin{aligned} A &=\left[\begin{array}{rrrrr}{1} & {4} & {8} & {-3} & {-7} \\\ {-1} & {2} & {7} & {3} & {4} \\ {-2} & {2} & {9} & {5} & {5} \\ {3} & {6} & {9} & {-5} & {-2}\end{array}\right] \\ \sim &\left[\begin{array}{ccccc}{1} & {4} & {8} & {0} & {5} \\ {0} & {2} & {5} & {0} & {-1} \\ {0} & {0} & {0} & {1} & {4} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right] \end{aligned} $$

Suppose \(A\) is an \(n \times n\) matrix with the property that the equation \(A \mathbf{x}=\mathbf{0}\) has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(A \mathbf{x}=\mathbf{b}\) must have a solution for each \(\mathbf{b}\) in \(\mathbb{R}^{n}\) .

[MI The band matrix \(A\) shown below can be used to estimate the unsteady conduction of heat in a rod when the temperatures at points \(p_{1}, \ldots, p_{5}\) on the rod change with time. The constant \(C\) in the matrix depends on the physical nature of the rod, the distance \(\Delta x\) between the points on the rod, and the length of time \(\Delta t\) between successive temperature measurements. Suppose that for \(k=0,1,2, \ldots,\) a vector \(\mathbf{t}_{k}\) in \(\mathbb{R}^{5}\) lists the temperatures at time \(k \Delta t\) . If the two ends of the rod are maintained at \(0^{\circ}\) , then the temperature vectors satisfy the equation \(A \mathbf{t}_{k+1}=\mathbf{t}_{k}(k=0,1, \ldots),\) where \(A=\left[\begin{array}{ccccc}{(1+2 C)} & {-C} & {} & {} \\ {-C} & {(1+2 C)} & {-C} & {} \\ {} & {-C} & {(1+2 C)} & {-C} \\ {} & {} & {-C} & {(1+2 C)} & {-C} \\ {} & {} & {} & {-C} & {(1+2 C)}\end{array}\right]\) a. Find the LU factorization of \(A\) when \(C=1 .\) A matrix such as \(A\) with three nonzero diagonals is called a tridiagonal matrix. The \(L\) and \(U\) factors are bidiagonal matrices. b. Suppose \(C=1\) and \(t_{0}=(10,12,12,12,10) .\) Use the LU factorization of \(A\) to find the temperature distributions \(\mathbf{t}_{1}, \mathbf{t}_{2}, \mathbf{t}_{3},\) and \(\mathbf{t}_{4} .\)

Construct a nonzero \(3 \times 3\) matrix \(A\) and a nonzero vector \(\mathbf{b}\) such that \(\mathbf{b}\) is in \(\mathrm{Col} A,\) but \(\mathbf{b}\) is not the same as any one of the columns of \(A .\)

In Exercises \(3-8,\) find the \(3 \times 3\) matrices that produce the described composite 2 \(\mathrm{D}\) transformations, using homogeneous coordinates. Translate by \((3,1),\) and then rotate \(45^{\circ}\) about the origin.
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