Consider the following 4 × 4 system. y1 1.5y2 + y3 + 2.5y4 = 1.5 y2 y4 = 1 0.5y1

Question

Consider the following 4 × 4 system.

y1 1.5y2 + y3 + 2.5y4 = 1.5

y2 y4 = 1

0.5y1 + 5.75y2 0.5y3 5.25y4 = 3.75

2.5y2 4y3 3.5y4 = 1.5

First, write down the coefficient matrix A and the right-hand side b for this system. Next, solve this system as follows. Write a Matlab script which enters in A and b, forms the corresponding augmented matrix (A|b), and then reduces the matrix to reduced row echelon form using the rref command. Run your script, and use the results to write down the linear system corresponding to the reduced row echelon form of (A|b).

Explanation / Answer

clc;
clear all;

coeff_A = [1 -1.5 1 2.5;
0 1 0 -1;
0.5 5.75 -0.5 -5.25;
0 2.5 -4 -3.5]
b = [1.5; -1; -3.75; -1.5]

augmented_matrix = [coeff_A, b]
Reduced_Matrix_row_echlon = rref(augmented_matrix)

OUTPUT

b =

1.5000
-1.0000
-3.7500
-1.5000

augmented_matrix =

1.0000 -1.5000 1.0000 2.5000 1.5000
0 1.0000 0 -1.0000 -1.0000
0.5000 5.7500 -0.5000 -5.2500 -3.7500
0 2.5000 -4.0000 -3.5000 -1.5000

Reduced_Matrix_row_echlon =

1 0 0 0 -5
0 1 0 0 6
0 0 1 0 -2
0 0 0 1 7

>>

Hence from the reduced matrix following equation can be derived.

y1 = -5

y2 = 6

y3 = -2

y4 = 7

clc;
clear all;

coeff_A = [1 -1.5 1 2.5;
0 1 0 -1;
0.5 5.75 -0.5 -5.25;
0 2.5 -4 -3.5]
b = [1.5; -1; -3.75; -1.5]

augmented_matrix = [coeff_A, b]
Reduced_Matrix_row_echlon = rref(augmented_matrix)

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