Chapter 3: Problem 10
Find an LU factorization of the given matrix. $$\left[\begin{array}{rrr} 2 & 2 & -1 \\ 4 & 0 & 4 \\ 3 & 4 & 4 \end{array}\right]$$
Short Answer
Expert verified
The LU factorization is L = \(\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ \frac{3}{2} & -\frac{1}{4} & 1 \end{bmatrix}\), U = \(\begin{bmatrix} 2 & 2 & -1 \\ 0 & -4 & 6 \\ 0 & 0 & \frac{13}{4} \end{bmatrix}\).
Step by step solution
01
Understand LU Factorization
LU factorization is the decomposition of a matrix into the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \). For a given square matrix \( A \), the goal is to find matrices \( L \) and \( U \) such that \( A = LU \).
02
Initialize Matrices L and U
Start by initializing the matrices \( L \) and \( U \). \( L \) is a lower triangular matrix, so it has 1s on its diagonal and zeros above. \( U \) needs to be an upper triangular matrix. Initially, set \( U \) equal to \( A \) and \( L \) as the identity matrix:
03
Transform to Upper Triangular U
Perform row operations to convert \( U \) into an upper triangular matrix. Begin with the first element in the first column that is non-zero (the pivot). Use row operations to create zeros below each pivot in the \( U \) matrix, modifying \( L \) to reflect these operations:
04
Zero Element Below First Pivot
Use the element \( u_{11} = 2 \) as the pivot, and eliminate \( u_{21} = 4 \) by subtracting 2 times the first row from the second row. This corresponds to the operation:\[R_2 = R_2 - 2R_1\]The first modification to \( U \) is:\[U = \begin{bmatrix} 2 & 2 & -1 \ 0 & -4 & 6 \ 3 & 4 & 4 \end{bmatrix}\]Update \( L \) accordingly:\( l_{21} = \frac{u_{21}}{u_{11}} = 2 \)Set \( L = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)
05
Zero Element Below Second Pivot
Proceed to eliminate the element below the new pivot \( u_{11} = 2 \) in the third row:\[R_3 = R_3 - \frac{3}{2}R_1\]After performing this operation, update \( U \):\[U = \begin{bmatrix} 2 & 2 & -1 \ 0 & -4 & 6 \ 0 & 1 & \frac{11}{2} \end{bmatrix}\]Update \( L \) similarly by using the factor \( l_{31} = \frac{3}{2} \):\( L = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ \frac{3}{2} & 0 & 1 \end{bmatrix} \)
06
Eliminate Final Element Below Pivot
The next pivot is \( u_{22} = -4 \), aim to eliminate \( u_{32} = 1 \):\[R_3 = R_3 + \frac{1}{4}R_2\]Continue with operations leading to the new \( U \):\[U = \begin{bmatrix} 2 & 2 & -1 \ 0 & -4 & 6 \ 0 & 0 & \frac{13}{4} \end{bmatrix}\]Update \( L \):\( l_{32} = \frac{-1}{4} \)\( L = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ \frac{3}{2} & -\frac{1}{4} & 1 \end{bmatrix} \)
07
Verify L and U
Finally, verify that multiplying \( L \) and \( U \) returns the original matrix \( A \). Calculate the product of \( L \) and \( U \) to ensure the decomposition is correct:\[LU = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ \frac{3}{2} & -\frac{1}{4} & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 & -1 \ 0 & -4 & 6 \ 0 & 0 & \frac{13}{4} \end{bmatrix} = \begin{bmatrix} 2 & 2 & -1 \ 4 & 0 & 4 \ 3 & 4 & 4 \end{bmatrix} = A\]This shows that \( A = LU \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lower Triangular Matrix
When you're working with matrices, a lower triangular matrix is very special. It's a matrix where all the entries above the diagonal are zero. Imagine a staircase with steps only leading downwards; this is similar to how you can think of a lower triangular matrix.
Each row starts with zero until it reaches the diagonal, then it can have non-zero numbers. If you take a 3x3 matrix as an example, its form would look something like this:
Each row starts with zero until it reaches the diagonal, then it can have non-zero numbers. If you take a 3x3 matrix as an example, its form would look something like this:
- The first row could be numbers like 1, 0, 0
- The second row would start with a zero and then numbers, like 2, 1, 0
- The third row will follow the same pattern, such as 3, 4, 1
Upper Triangular Matrix
An upper triangular matrix is another special kind of matrix. It almost works like the opposite of a lower triangular matrix. In an upper triangular matrix, all the entries below the diagonal are zero. Picture a series of steps going upwards, starting from the bottom left; this illustrates an upper triangular matrix.
The structure looks like this:
The structure looks like this:
- The first row can have all non-zero numbers, like 2, 2, -1
- The second row starts with zero and then non-zero numbers, for example, 0, -4, 6
- The last row begins with zeros, with just one non-zero number, like 0, 0, \(\frac{13}{4}\)
Matrix Decomposition
Matrix decomposition is a strategy used to break down a complex matrix into simpler, more manageable components. This process can be visualized like disassembling a large jigsaw puzzle into its pieces, only to work through them more efficiently.
A common form of matrix decomposition is LU factorization. Here, any given square matrix \(A\) is expressed as the product of two simpler matrices, \(L\) and \(U\):
This method is incredibly useful in numerical analysis and computational mathematics because:
A common form of matrix decomposition is LU factorization. Here, any given square matrix \(A\) is expressed as the product of two simpler matrices, \(L\) and \(U\):
- \(L\) is a lower triangular matrix
- \(U\) is an upper triangular matrix
This method is incredibly useful in numerical analysis and computational mathematics because:
- It simplifies solving systems of linear equations.
- It helps in efficiently finding determinants and inverses of matrices.
- It enhances operations in computer algorithms designed for matrix computations.