步驟 1

A = {{1., 2.}, {5., 6.}, {4.5, 6.}};
B = {5., 6., 8.};
LinearSolve[A, B]

輸出

LinearSolve[{{1., 2.}, {5., 6.}, {4.5, 6.}}, {5., 6., 8.}]

步驟 2

A = {{1., 2.}, {2., 4.}, {3., 6.}};
B = {5., 6., 8.};
MatrixRank[A]

輸出

步驟 3

X = PseudoInverse[A].B

輸出

{0.5857142857142857, 1.1714285714285715}

步驟 4

A.X

輸出

{2.928571428571429, 5.857142857142858, 8.785714285714285}

步驟 5

data = {{0.2, 4.59}, {0.4, 5.05}, {0.6, 6.2}, {0.8, 6.6}, {1., 7.4}};

沒有輸出——這個步驟是在為下一步做準備。

步驟 6

plot1 = ListPlot[ data, PlotStyle -> PointSize[0.02]]

輸出

-Graphics-

步驟 7

xData = data[[All, 1]];
yData = data[[All, 2]];

沒有輸出——這個步驟是在為下一步做準備。

步驟 8

mat = Transpose[ {Table[1, {Length[xData]}], xData}];
mat//MatrixForm

輸出

MatrixForm[{{1, 0.2}, {1, 0.4}, {1, 0.6}, {1, 0.8}, {1, 1.}}]

步驟 9

PseudoInverse[mat].yData

輸出

{3.8170000000000015, 3.585}

步驟 10

sol = FindFit[data, α + β x, {α, β}, x]

輸出

{α -> 3.8169999997513093, β -> 3.5849999944044773}

步驟 11

Plot[ (α + β x) /. sol, {x, 0, 1}, Epilog -> First[plot1]]

輸出

-Graphics-

使用到的函式

Linear algebra matrix computations least squares solutions