This example shows how to solve a simple system of linear equations $\mathit{Ax}=\mathit{b}$, using QR decomposition.

In this example, define A as a 5-by-3 matrix with a large condition number. To solve a system of linear equations involving ill-conditioned (large condition number) non-square matrices, you must use QR decomposition.

rngdefault; A = gallery('randsvd', [5,3], 1000000); b = [1; 1; 1; 1; 1]; x = fixed.qrMatrixSolve(A,b)

x =3×1104× -2.3777 7.0686 -2.2703

Compare the result of thefixed.qrMatrixSolvefunction with the result of themldivideor\function.

x = A\b

x =3×1104× -2.3777 7.0686 -2.2703

This example shows the effect of a regularization parameter when solving an overdetermined system. In this example, a quantityyis measured at several different values of timetto produce the following observations.

t = [0 .3 .8 1.1 1.6 2.3]'; y = [.82 .72 .63 .60 .55 .50]';

Model the data with a decaying exponential function

$y(t)=c_1+c_2{e}^{-t}$.

The preceding equation says that the vectoryshould be approximated by a linear combination of two other vectors. One is a constant vector containing all ones and the other is the vector with componentsexp(-t). The unknown coefficients, $$c_1$$and $$c_2$$, can be computed by doing a least-squares fit, which minimizes the sum of the squares of the deviations of the data from the model. There are six equations and two unknowns, represented by a 6-by-2 matrix.

E = [ones(size(t)) exp(-t)]

E =6×21.0000 1.0000 1.0000 0.7408 1.0000 0.4493 1.0000 0.3329 1.0000 0.2019 1.0000 0.1003

Use thefixed.qrMatrixSolvefunction to get the least-squares solution.

c = fixed.qrMatrixSolve(E, y)

c =2×10.4760 0.3413

In other words, the least-squares fit to the data is

The following statements evaluate the model at regularly spaced increments int, and then plot the result together with the original data:

T = (0:0.1:2.5)'; Y = [ones(size(T)) exp(-T)]*c; plot(T,Y,“- - -”,t,y,'o')

In cases where the input matrices are ill-conditioned, small positive values of a regularization parameter can improve the conditioning of the least squares problem, and reduce the variance of the estimates. Explore the effect of the regularization parameter on the least squares solution for this data.

figure; lambda = [0:0.1:0.5]; plot(t,y,'o','DisplayName','Original Data');fori = 1:长度(λ)c =fixed.qrMatrixSolve(E, y, numerictype('double'), lambda(i)); Y = [ones(size(T)) exp(-T)]*c; holdonplot(T,Y,“- - -”,'DisplayName', ['lambda =', num2str(lambda(i))])endlegend('Original Data','lambda = 0','lambda = 0.1','lambda = 0.2','lambda = 0.3','lambda = 0.4','lambda = 0.5')