This example shows how to compute the upper-triangular factor $\mathit{R}$, and $\mathit{C}={\mathit{Q}}^{\prime }\mathit{b}$.

Define the input matrices,A, andb.

rng('default'); m = 6; n = 3; p = 1; A = randn(m,n)

A =6×30.5377 -0.4336 0.7254 1.8339 0.3426 -0.0631 -2.2588 3.5784 0.7147 0.8622 2.7694 -0.2050 0.3188 -1.3499 -0.1241 -1.3077 3.0349 1.4897

b = randn(m,p)

b =6×11.4090 1.4172 0.6715 -1.2075 0.7172 1.6302

Thefixed.qrABfunction returns the upper-triangular factor, $\mathit{R}$, and $\mathit{C}={\mathit{Q}}^{\prime }\mathit{b}$.

[C, R] = fixed.qrAB(A,b)

C =3×1-0.3284 0.4055 2.5481

R =3×33.3630 -2.8841 -1.0421 0 4.8472 0.6885 0 0 1.3258

This example shows how to solve a system of linear equations, $\mathit{Ax}=\mathit{b}$通过计算upper-triangular factor $\mathit{R}$, and $\mathit{C}={\mathit{Q}}^{\prime }\mathit{b}$. A regularization parameter can improve the conditioning of least squares problems, and reduce the variance of the estimates when solving linear systems of equations.

Define input matrices,A, andb.

rng('default'); m = 50; n = 5; p = 1; A = randn(m,n); b = randn(m,p);

Use thefixed.qrABfunction to compute the upper-triangular factor, $\mathit{R}$, and $\mathit{C}={\mathit{Q}}^{\prime }\mathit{b}$.

[C, R] = fixed.qrAB(A, b, 0.01)

C =5×1-0.6361 1.7663 1.5892 -2.0638 -0.1327

R =5×59.0631 0.7471 0.4126 -0.3606 0.1883 0 7.2515 -1.1145 0.6011 -0.7544 0 0 7.6132 -0.9460 -0.7062 0 0 0 6.3065 -2.3238 0 0 0 0 5.9297

Use this result to solve $\mathit{Ax}=\mathit{b}$usingx = R\C. Computex = R\Cusing thefixed.qrMatrixSolvefunction.

x = fixed.qrMatrixSolve(R,C)

x =5×1-0.1148 0.2944 0.1650 -0.3355 -0.0224

Compare the result to computingx = A\bdirectly.

x = A\b

x =5×1-0.1148 0.2944 0.1650 -0.3355 -0.0224