LAST The process:1. Forward elimination: reduction to row echelon form. Using it one can tell whetherthere are no solutions, or unique solution, or infinitely many solutions.2. Back substitution: further reduction to reduced row echelon form.Algorithm:1. Partial pivoting: Find the kth pivot by swapping rows, to move the entry with thelargest absolute value to the pivot position. This imparts computational stability tothe algorithm.2. For each row below the pivot, calculate the factor f which makes the kth entryzero,and for every element in the row subtract the fth multiple of the correspondingelement in the kth row.3. Repeat above steps for each unknown. We will be left with a partial R.E.F. matrix. 1 2 300 130 0 0 1Figure 1.Row echelon form (see figure 1): Matrix is said to be in R.E.F. if the following conditionshold:1. The first non-zero element in each row, called the leading coefficient, is 1.2 Each leading coefficient is in a column to the right of the previous rowleadingcoefficient.3. Rows with all zeros are below rows with at least one non-zero element.1 2000 0 100 0 0 1Figure 2.Reduced row echelon form (see figure 2): Matrix is said to be in R.R.E.F. if the followingconditions hold:1. All the conditions for R.E.F.2 The leading coefficient in each row is the only non-zero entry in its column.The algorithm is majorly about performing a sequence of operations on the rows of thematrix. What we would like to keep in mind while performing these operations is that wewant to convert the matrix into an upper triangular matrix in row echelon form. Theoperations can be:1. Swapping two rows2 Multiplying a row by a non-zero scalar3. Adding to one row a multiple of another Programming Project 1 - Gaussian Elimination to Solve Linear EquationsThe project focuses on using an algorithm for solving a system of linear equations. We will deal with thematrix of coefficients. Gaussian Elimination does not work on singular matrices (they lead to division byzero).Input: For N unknowns, input is an augmentedmatrix of size N x (N+1). One extracolumn is for Right Hand Side (RHS)mat[N][N+1] = {{3.0, 2.0, –4.0, 3.0},{2.0, 3.0, 3.0, 15.0},{5.0, –3, 1.0, 14.0)Output: Solution to equations is:3.0000001.0000002.000000Explanation:Given matrix represents following equations3.0X1 + 2.0X2 – 4.0X3 = 3.02.0x1 + 3.0X2 + 3.0X3 = 15.05.0X1 — 3.0Х2 +X3 = 14.0There is a unique solution for given equations,solutions is, X1 = 3.0, X2 = 1.0, X3 = 2.0,

Answered: Given matrix represents following…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

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LAST

The process:
1. Forward elimination: reduction to row echelon form. Using it one can tell whether
there are no solutions, or unique solution, or infinitely many solutions.
2. Back substitution: further reduction to reduced row echelon form.
Algorithm:
1. Partial pivoting: Find the kth pivot by swapping rows, to move the entry with the
largest absolute value to the pivot position. This imparts computational stability to
the algorithm.
2. For each row below the pivot, calculate the factor f which makes the kth entryzero,
and for every element in the row subtract the fth multiple of the corresponding
element in the kth row.
3. Repeat above steps for each unknown. We will be left with a partial R.E.F. matrix.

1 2 3
00 13
0 0 0 1
Figure 1.
Row echelon form (see figure 1): Matrix is said to be in R.E.F. if the following conditions
hold:
1. The first non-zero element in each row, called the leading coefficient, is 1.
2 Each leading coefficient is in a column to the right of the previous rowleading
coefficient.
3. Rows with all zeros are below rows with at least one non-zero element.
1 200
0 0 10
0 0 0 1
Figure 2.
Reduced row echelon form (see figure 2): Matrix is said to be in R.R.E.F. if the following
conditions hold:
1. All the conditions for R.E.F.
2 The leading coefficient in each row is the only non-zero entry in its column.
The algorithm is majorly about performing a sequence of operations on the rows of the
matrix. What we would like to keep in mind while performing these operations is that we
want to convert the matrix into an upper triangular matrix in row echelon form. The
operations can be:
1. Swapping two rows
2 Multiplying a row by a non-zero scalar
3. Adding to one row a multiple of another

Programming Project 1 - Gaussian Elimination to Solve Linear Equations
The project focuses on using an algorithm for solving a system of linear equations. We will deal with the
matrix of coefficients. Gaussian Elimination does not work on singular matrices (they lead to division by
zero).
Input: For N unknowns, input is an augmented
matrix of size N x (N+1). One extra
column is for Right Hand Side (RHS)
mat[N][N+1] = {{3.0, 2.0, –4.0, 3.0},
{2.0, 3.0, 3.0, 15.0},
{5.0, –3, 1.0, 14.0)
Output: Solution to equations is:
3.000000
1.000000
2.000000
Explanation:
Given matrix represents following equations
3.0X1 + 2.0X2 – 4.0X3 = 3.0
2.0x1 + 3.0X2 + 3.0X3 = 15.0
5.0X1 — 3.0Х2 +
X3 = 14.0
There is a unique solution for given equations,
solutions is, X1 = 3.0, X2 = 1.0, X3 = 2.0,

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