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Pivot matrix function help

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Luis Gonzalez
Luis Gonzalez 2022 年 3 月 7 日
回答済み: Siraj 2023 年 12 月 4 日
hi i wanna make a code that does the Pivot matrix thing from the Gauss method please
i need help with a pivot matrix gauss method code
function [A,b]=pivotea(A,b)
n = size(A, 1);
r = zeros(n, 1);
for i = 1 : 1 : n
r(i) = i;
end
x = zeros(n, 1);
for k = 1 : 1 : n
max = abs(A(r(k), r(k)));
max_pos = k;
for l = k : 1 : n
if abs(A(r(l), r(k))) > max
max = abs(A(r(l), r(k)));
max_pos = l;
end
end
temp_r = r;
r(k) = temp_r(max_pos);
r(max_pos) = temp_r(k);
for i = 1 : 1 : n
if i ~= k
zeta = A(r(i), k) / A(r(k), k);
for j = k : 1 : n
A(r(i), j) = A(r(i), j) - A(r(k), j) * zeta;
end
b(r(i)) = b(r(i)) - b(r(k)) * zeta;
end
end
end
for i = 1 : 1 : n
x(i) = b(r(i)) / A(r(i), i);
end
end
look i have this now problem is whenever i try pivot a matrix some spaces are left on 0 like here
Anueva =
0 -0.3636 0 0
-11.0000 0 0 -0.0000
0 0 0 -3.0486
0 0 61.7500 0
when it should be like
  2 件のコメント
Matt J
Matt J 2022 年 3 月 7 日
It does not appear that you have begun anything. That makes it difficult for us to help you finish.
Luis Gonzalez
Luis Gonzalez 2022 年 3 月 9 日
i need help with a pivot matrix gauss method code
function [A,b]=pivotea(A,b)
n = size(A, 1);
r = zeros(n, 1);
for i = 1 : 1 : n
r(i) = i;
end
x = zeros(n, 1);
for k = 1 : 1 : n
max = abs(A(r(k), r(k)));
max_pos = k;
for l = k : 1 : n
if abs(A(r(l), r(k))) > max
max = abs(A(r(l), r(k)));
max_pos = l;
end
end
temp_r = r;
r(k) = temp_r(max_pos);
r(max_pos) = temp_r(k);
for i = 1 : 1 : n
if i ~= k
zeta = A(r(i), k) / A(r(k), k);
for j = k : 1 : n
A(r(i), j) = A(r(i), j) - A(r(k), j) * zeta;
end
b(r(i)) = b(r(i)) - b(r(k)) * zeta;
end
end
end
for i = 1 : 1 : n
x(i) = b(r(i)) / A(r(i), i);
end
end
look i have this now problem is whenever i try pivot a matrix some spaces are left on 0 like here
Anueva =
0 -0.3636 0 0
-11.0000 0 0 -0.0000
0 0 0 -3.0486
0 0 61.7500 0
when it should be like

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回答 (1 件)

Siraj
Siraj 2023 年 12 月 4 日
Hi!
It seems that you are attempting to solve a system of linear equations using Gaussian elimination with partial pivoting. However, there are a couple of issues in the code. Firstly, all row swapping and elimination operations necessary to achieve the echelon form should be applied to the augmented matrix instead of the matrix "A". Secondly, to calculate the value of "x", the process should begin with the last row, a method known as back propagation or back substitution.
Please refer to the modified code below which uses an Augmented matrix "Aug" which serves as the main structure for pivoting operations (Row swapping and Gaussian Elimination).
A = [0.02 0.01 0 0; 1 2 1 0; 0 1 2 1; 0 0 100 200];
b = [0.02;1;4;800];
[reducedA, reduced_b] = pivot(A,b)
A = 4×4
0.0200 0.0100 0 0 1.0000 2.0000 1.0000 0 0 1.0000 2.0000 1.0000 0 0 100.0000 200.0000
b = 4×1
0.0200 1.0000 4.0000 800.0000
Aug = 4×5
0.0200 0.0100 0 0 0.0200 1.0000 2.0000 1.0000 0 1.0000 0 1.0000 2.0000 1.0000 4.0000 0 0 100.0000 200.0000 800.0000
x = 4×1
1 0 0 4
reducedA = 4×4
1.0000 2.0000 1.0000 0 0 1.0000 2.0000 1.0000 0 0 100.0000 200.0000 0 0 0 -0.0500
reduced_b = 4×1
1.0000 4.0000 800.0000 -0.2000
function [A,b]=pivot(A,b)
% Get the length of b
n = length(b);
% Initialize x with zeros
x = zeros(n,1);
% Create the augmented matrix
Aug = [A,b];
%Display
A
b
Aug
% Loop for partial pivoting and echelon form
for j = 1:n-1
% partial pivoting
% Find the maximum value and its position for partial pivoting
[maxi, max_pos] = max(abs(Aug(j:n,j)));
% Swap the rows for partial pivoting
C = Aug(j,:);
Aug(j,:) = Aug(max_pos+j-1,:);
Aug(max_pos+j-1,:) = C;
% echelon form
for i = j+1:n
% Calculate zeta for echelon form
zeta = Aug(i,j)/Aug(j,j);
% Perform row operations for echelon form
Aug(i,:) = Aug(i,:) - zeta*Aug(j,:);
end
end
% Back substitution
% Calculate x(n) using back substitution
x(n) = Aug(n,n+1)/Aug(n,n);
% Loop for back substitution
for k = n-1:-1:1
% Calculate x(k) using back substitution
x(k) = (Aug(k,n+1)-Aug(k,k+1:n)*x(k+1:n))/Aug(k,k);
end
% Display the result
x
% Separate A and b from the augmented matrix
A = Aug(:,1:end-1 );
b = Aug(:,end);
end
You can see the algorithm in action by visiting the following link:
You can explore various methods for solving a system of linear equations in MATLAB by referring to the following link:
To obtain the reduced row echelon form of a matrix in MATLAB, you can utilize the "rref" function. For more information about this function, please visit the following link:
Hope this helps

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