Direct Methods for solving linear Systems
Matrices enable us to write linear systems in a compact form that makes it easier to automate the standard elimination method on an electronic computer in order to obtain solutions in a fast and efficient way.
5.1 Echelon Form of a Matrix
The main idea b...
Direct Methods for solving linear Systems
Matrices enable us to write linear systems in a compact form that makes it easier to automate the
standard elimination method on an electronic computer in order to obtain solutions in a fast and
efficient way.
5.1 Echelon Form of a Matrix
The main idea behind this method is to take the augmented matrix [𝐴 ∶ 𝑏] of a linear system
(𝐴𝑥 = 𝑏) and perform certain operations on it in order to obtain a new partitioned matrix [𝐶 ∶ 𝑑]
said to be in Row Echelon Form(REF) or in Reduced Row Echelon Form (RREF). The obtained
matrix represents a new system 𝐶𝑥 = 𝑑 that has the same solution of system 𝐴𝑥 = 𝑏 and is easier
to solve.
A matrix is said to be in RREF if it satisfies the four points below. However, if the matrix satisfies
the first three points only, it is said to be in REF.
1. All zero rows, if there are any, appear at the bottom of the matrix.
2. The first non-zero entry from the left of a non-zero row is a 1. This is called the
leading one.
3. For each non-zero row, the leading one appears to the right and below any leading
ones in preceding rows.
4. If a column contains a leading one, then all other entries in that column are zero.
Example:
Matrix A is in RREF and Matrix B is in REF.
1 0 0 0 1 2 3 4
0 1 0 0=A 0 1 2 5= B
0 0 1 0 0 0 1 2
0 0 0 1 0 0 0 0
Any matrix can be put in REF and RREF by:
Interchange of rows.
Row multiplication by a non-zero number.
Addition of the multiple of one row to another.
Example 1:
Lab Numerical Analysis
, Solve the above linear system by (a) Gaussian Elimination method (matrix in REF), (b)
GaussJordan Reduction method (matrix in RREF).
𝑥 + 2𝑦 + 3𝑧 = 𝟔
{2𝑥 − 3𝑦 + 2𝑧 = 𝟏𝟒
3𝑥 + 𝑦 − 𝑧 = −𝟐
Solution:
a. Gaussian Elimination method The
Augmented Matrix is:
Add (-2) times the first row to the second one and (-3) times the first row to the third one:
Multiply the third row by and interchange the second and third rows:
Add 7 times the second row to the third row then multiply the third row by :
Ag =
We can compute the solution of the system by Gaussian Elimination from this matrix which is now
in REF:
𝑥 = 1; 𝑦 = −2; 𝑧 = 3
b. Gauss-Jordan Reduction method
To solve the given linear system by Gauss-Jordan reduction, we transform the last obtained matrix
to RREF through the below steps:
Add (-2) times the third row to the second row and get Ag1
Add (-3) times the third row to the first row and get Ag2
Finally add (-2) times the second row to the first row and get final Ag matrix in RREF.
Lab Numerical Analysis
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