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Construct the operation count for solving an \(n \times n\) linear system using the Crout Factorization Algorithm.

Short Answer

Expert verified
The operation count for solving an \(n \times n\) linear system using the Crout factorization algorithm is approximately \(\frac{n^3}{3} + 2n^2\) operations. This includes the operations for matrix decomposition, forward substitution, and backward substitution.

Step by step solution

01

Matrix Decomposition into Lower and Upper Triangular Matrices

Use the Crout factorization algorithm to decompose the \(n \times n\) matrix into two triangular matrizes, one lower triangulate matrix (L) and one upper triangular matrix (U). But note, in Crout's factorization, the diagonal elements of U are 1. If we consider this decomposition process, we end up with \(n^3/3\) operations.
02

Forward Substitution

After decomposing the matrix, use forward substitution to solve the intermediate system \(Ly = b\) for \(y\). Here, y is an unknown vector. This step involves \(n^2\) operations.
03

Backward Substitution

The final step is to use backward substitution to solve the system \(Ux = y\) for the vector \(x\). This step involves \(n^2\) operations.

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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 crucial step in solving linear systems, especially using methods like the Crout Factorization Algorithm. This technique involves breaking down a matrix into simpler, more manageable components, specifically into lower (\( L \)) and upper (\( U \)) triangular matrices. In Crout's method, the matrix is decomposed such that: - The diagonal elements of matrix \( U \) are all ones.- The main computation involves forming the \( L \) matrix with non-zero elements beneath the main diagonal.By performing this decomposition, we simplify the solution process for linear systems significantly. For an \( n \times n \) matrix, this entire decomposition requires around \( \frac{n^3}{3} \) operations, which demonstrates its computational efficiency while preparing the system for further steps.
Linear Systems
Linear systems are equations involving linear algebraic expressions. They are commonly written in matrix form as \( Ax = b \), where \( A \) is a matrix, \( x \) is a vector of unknowns, and \( b \) is a vector of constants. Solving these systems efficiently is critical in mathematical modeling and engineering applications.- The Crout Factorization Algorithm becomes handy as it prepares these systems for easier solution paths.- By transforming \( A \) into triangular matrices \( L \) and \( U \), the original system is broken into two simpler systems. - First, \( Ly = b \) - Then, \( Ux = y \)Both of these systems can be solved efficiently through substitution methods, allowing us to find the unknown variables quickly.
Forward Substitution
Forward substitution is used to solve systems of equations with a lower triangular matrix, particularly after matrix decomposition. Given a system like \( Ly = b \), where \( L \) is a lower triangular matrix and \( b \) is a known vector, our objective is to find \( y \).The process is straightforward:- Start with the first row to find the first element of \( y \).- Progressively move downward, substituting found values into subsequent rows.- Each step involves simple arithmetic operations.This systematic top-down approach efficiently uses \( n^2 \) operations for an \( n \times n \) system, ensuring all values of \( y \) are determined swiftly.
Backward Substitution
Backward substitution is the complementary technique to forward substitution but is used with upper triangular matrices, such as \( Ux = y \), post-decomposition.In this scenario:- Begin from the last row, where only one unknown exists, to find its value.- Progress upward, substituting known values into previous rows.- Continue until all variables in \( x \) are determined.This method also involves \( n^2 \) operations, enabling quick calculation of unknowns in an efficient manner. Backward substitution, along with forward substitution, facilitates solving linear systems post matrix decomposition of the Crout Factorization Algorithm, making computational processes significantly streamlined.

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Most popular questions from this chapter

Obtain factorizations of the form \(A=P^{\prime} L U\) for the following matrices. a. \(\quad A=\left[\begin{array}{rrr}0 & 2 & 3 \\ 1 & 1 & -1 \\ 0 & -1 & 1\end{array}\right]\) b. \(\quad A=\left[\begin{array}{rrr}1 & 2 & -1 \\ 1 & 2 & 3 \\ 2 & -1 & 4\end{array}\right]\) c. \(\quad A=\left[\begin{array}{rrrr}1 & -2 & 3 & 0 \\ 3 & -6 & 9 & 3 \\ 2 & 1 & 4 & 1 \\ 1 & -2 & 2 & -2\end{array}\right]\) d. \(\quad A=\left[\begin{array}{rrrr}1 & -2 & 3 & 0 \\ 1 & -2 & 3 & 1 \\ 1 & -2 & 2 & -2 \\ 2 & 1 & 3 & -1\end{array}\right]\)

Consider the four \(3 \times 3\) linear systems having the same coefficient matrix: $$ \begin{array}{cc} 2 x_{1}-3 x_{2}+x_{3}=2, & 2 x_{1}-3 x_{2}+x_{3}=6 \\ x_{1}+x_{2}-x_{3}=-1, & x_{1}+x_{2}-x_{3}=4 \\ -x_{1}+x_{2}-3 x_{3}=0 ; & -x_{1}+x_{2}-3 x_{3}=5 \\ 2 x_{1}-3 x_{2}+x_{3}=0, & 2 x_{1}-3 x_{2}+x_{3}=-1 \\ x_{1}+x_{2}-x_{3}=1, & x_{1}+x_{2}-x_{3}=0 \\ -x_{1}+x_{2}-3 x_{3}=-3 ; & -x_{1}+x_{2}-3 x_{3}=0 \end{array} $$ a. Solve the linear systems by applying Gaussian elimination to the augmented matrix $$ \left[\begin{array}{rrrrrrrr} 2 & -3 & 1 & \vdots & 2 & 6 & 0 & -1 \\ 1 & 1 & -1 & \vdots & -1 & 4 & 1 & 0 \\ -1 & 1 & -3 & \vdots & 0 & 5 & -3 & 0 \end{array}\right] $$ b. Solve the linear systems by finding and multiplying by the inverse of $$ A=\left[\begin{array}{rrr} 2 & -3 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & -3 \end{array}\right] $$ c. Which method requires more operations?

Use the Gaussian Elimination Algorithm to solve the following linear systems, if possible, and determine whether row interchanges are necessary: a. \(\begin{aligned} x_{1}-x_{2}+3 x_{3} &=2 \\ 3 x_{1}-3 x_{2}+x_{3} &=-1 \\\ x_{1}+x_{2} &=3 \end{aligned}\) b. \(\begin{aligned} 2 x_{1}-1.5 x_{2}+3 x_{3} &=1 \\\\-x_{1}+2 x_{3} &=3 \\ 4 x_{1}-4.5 x_{2}+5 x_{3} &=1 \end{aligned}\) c. \(\begin{aligned} 2 x_{1} &=3 \\ x_{1}+1.5 x_{2} &=4.5 \\\\-3 x_{2}+0.5 x_{3} &=-6.6 \\ 2 x_{1}-2 x_{2}+x_{3}+x_{4} &=0.8 \end{aligned}\) d. \(\begin{aligned} x_{1}+x_{2}+x_{4} &=2 \\ 2 x_{1}+x_{2}-x_{3}+x_{4} &=1 \\\ 4 x_{1}-x_{2}-2 x_{3}+2 x_{4} &=0 \\ 3 x_{1}-x_{2}-x_{3}+2 x_{4} &=-3 \end{aligned}\)

Suppose \(m\) linear systems $$A \mathbf{x}^{(p)}=\mathbf{b}^{(p)}, \quad p=1,2, \ldots, m$$ are to be solved, each with the \(n \times n\) coefficient matrix \(A\). a. Show that Gaussian elimination with backward substitution applied to the aug- mented matrix $$\left[\begin{array}{ll} A: & \mathbf{b}^{(1)} \mathbf{b}^{(2)} \cdots \mathbf{b}^{(m)} \end{array}\right]$$ requires $$\frac{1}{3} n^{3}+m n^{2}-\frac{1}{3} n \quad \text { multiplications/ divisions }$$ $$\frac{1}{3} n^{3}+m n^{2}-\frac{1}{2} n^{2}-m n+\frac{1}{6} n \quad \text { additions/subtractions. }$$ b. Show that the Gauss-Jordan method (see Exercise 12 , Section \(6.1\) ) applied to the augmented matrix $$\left[\begin{array}{ll} A: & \left.{\mathbf{b}^{(1)} \mathbf{b}^{(2)}} \cdots \mathbf{b}^{(m)}\right] \end{array}\right.$$ requires $$\frac{1}{2} n^{3}+m n^{2}-\frac{1}{2} n \quad \text { multiplications/divisions }$$ and $$\frac{1}{2} n^{3}+(m-1) n^{2}+\left(\frac{1}{2}-m\right) n \quad \text { additions/subtractions. }$$ c. For the special case $$\mathbf{b}^{(p)}=\left[\begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ \vdots \\ 0 \end{array}\right] \leftarrow p \text { th row }$$ for each \(p=1, \ldots, m\), with \(m=n\), the solution \(x^{(p)}\) is the \(p\) th column of \(A^{-1}\). Show that Gaussian elimination with backward substitution requires \(\frac{4}{3} n^{3}-\frac{1}{3} n \quad\) multiplications/divisions and $$\frac{4}{3} n^{3}-\frac{3}{2} n^{2}+\frac{1}{6} n \quad \text { additions/subtractions }$$ for this application, and that the Gauss-Jordan method requires $$\frac{3}{2} n^{3}-\frac{1}{2} n \quad \text { multiplications/divisions }$$ and $$\frac{3}{2} n^{3}-2 n^{2}+\frac{1}{2} n \quad \text { additions/subtractions. }$$ d. Construct an algorithm using Gaussian elimination to find \(A^{-1}\), but do not per- form multiplications when one of the multipliers is known to be 1 , and do not per- form additions/subtractions when one of the elements involved is known to be \(0 .\) Show that the required computations are reduced to \(n^{3}\) multiplications/divisions and \(n^{3}-2 n^{2}+n\) additions/subtractions. e. Show that solving the linear system \(A x=\mathbf{b}\), when \(A^{-1}\) is known, still requires \(n^{2}\) multiplications/divisions and \(n^{2}-n\) additions/subtractions. f. Show that solving \(m\) linear systems \(A x^{(p)}=\mathbf{b}^{(p)}\), for \(p=1,2, \ldots, m\), by the method \(x^{(p)}=\) \(A^{-1} \mathbf{b}(p)\) requires \(m n^{2}\) multiplications and \(m\left(n^{2}-n\right)\) additions, if \(A^{-1}\) is known. g. Let \(\mathrm{A}\) be an \(n \times n\) matrix. Compare the number of operations required to solve \(n\) linear systems involving \(A\) by Gaussian elimination with backward substitution and by first inverting \(A\) and then multiplying \(A x=\mathbf{b}\) by \(A^{-1}\), for \(n=3,10,50,100 .\) Is it ever advantageous to compute \(A^{-1}\) for the purpose of solving linear systems?

Use Gaussian elimination with backward substitution and two-digit rounding arithmetic to solve the following linear systems. Do not reorder the equations. (The exact solution to each system is \(\left.x_{1}=-1, x_{2}=1, x_{3}=3 .\right)\) a. \(\quad-x_{1}+4 x_{2}+x_{3}=8\) \(\frac{5}{3} x_{1}+\frac{2}{3} x_{2}+\frac{2}{3} x_{3}=1\) \(2 x_{1}+x_{2}+4 x_{3}=11\) b. \(\quad \begin{aligned} 4 x_{1}+2 x_{2}-x_{3} &=-5 \\ \frac{1}{9} x_{1}+\frac{1}{9} x_{2}-\frac{1}{3} x_{3} &=-1 \\ x_{1}+4 x_{2}+2 x_{3} &=9 \end{aligned}\)

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