给以下代码每一行都添加注释：clear all; clc; A = input('请输入对阵矩阵A：'); [T,D] = quadraticForm(A); disp("T = "); disp(T); disp("D = "); disp(D); function [T,D] = quadraticForm(A) n = size(A,1); T = eye(n); D = zeros(n); for k = 1:n-1 % 判断是否需要进行变换 if A(k+1,k) ~= 0 % 计算变换矩阵 [c,s] = givens(A(k,k),A(k+1,k)); G = [c s; -s c]; % 更新A和T A([k k+1],k:n) = G'A([k k+1],k:n); T(:,[k k+1]) = T(:,[k k+1])G; end end for i = 1:n D(i,i) = sign(A(i,i)); if D(i,i) == 0 D(i,i) = 1; end end T = T'; D = Ddiag(abs(diag(A))); end function [c,s] = givens(a,b) % 计算Givens变换矩阵的元素c和s if b == 0 c = 1; s = 0; elseif abs(b) > abs(a) t = -a/b; s = 1/sqrt(1+t^2); c = st; else t = -b/a; c = 1/sqrt(1+t^2); s = c*t; end end

% 清空变量并清空命令窗口
clear all;
clc;

% 获取用户输入的对阵矩阵A
A = input('请输入对阵矩阵A：');

% 调用quadraticForm函数，计算出T和D
[T,D] = quadraticForm(A);

% 输出计算结果T和D
disp("T = ");
disp(T);
disp("D = ");
disp(D);

% 定义函数quadraticForm，用于计算矩阵的二次型
function [T,D] = quadraticForm(A)
% 获取矩阵A的大小
n = size(A,1);

% 初始化变换矩阵T和对角矩阵D
T = eye(n);
D = zeros(n);

% 进行初等变换，将A矩阵化为上三角矩阵
for k = 1:n-1
% 判断是否需要进行变换
if A(k+1,k) ~= 0
% 计算Givens变换矩阵的元素c和s
[c,s] = givens(A(k,k),A(k+1,k));
G = [c s; -s c]; % 计算变换矩阵
% 更新A和T
A([k k+1],k:n) = G'*A([k k+1],k:n);
T(:,[k k+1]) = T(:,[k k+1])*G;
end
end

% 根据A的对角线元素的正负性，构建出D矩阵
for i = 1:n
D(i,i) = sign(A(i,i));
if D(i,i) == 0
D(i,i) = 1;
end
end

% 计算出最终的T和D
T = T';
D = D*diag(abs(diag(A)));
end

% 定义函数givens，用于计算Givens变换矩阵的元素c和s
function [c,s] = givens(a,b)
% 计算Givens变换矩阵的元素c和s
if b == 0
c = 1;
s = 0;
elseif abs(b) > abs(a)
t = -a/b;
s = 1/sqrt(1+t^2);
c = s*t;
else
t = -b/a;
c = 1/sqrt(1+t^2);
s = c*t;
end
end