几代实验1
09-10-2《几何与代数》 数学实验报告 学号：
03009427 姓名：
曾骥敏 得分：
.要求：报告中应包含实验中你所输入的所有命令及运算结果，用4A 纸打印．并在第 16 周之前交给任课老师。
实验一：
平板的稳态温度分布问题（线性方程组应用）T l =36 ,T u =16,T r =8,T d =28(1)建立可以确定平板内节点温度的线性方程组 ； ； 解：方程组如下：
4T 1-T 2-T 4 =T l +T u =52;-T 1 +4T 2-T 3-T 5 =T l =36;-T 2 +4T 3-T 6 =T l +T d =64;-T 1 +4T 4-T 5-T 7 =T u =16;-T 2-T 4 +4T 5-T 6-T 8 =0;-T 3-T 5 +4T 6-T 9 =T d =28;-T 4 +4T 7-T 8 =T u +T r =24;-T 5-T 7 +4T 8-T 9 =T r =8;-T 6-T 8 +4T 9 =T d +T r =36.(2)用 用 MATLAB 软件 的三种方法 求解该线性方程组(请输出精确解（分数形式）)； 用 方法一：利用 Cramer 法则求解； 方法二：作为 逆 矩阵 的方法 求解； 用 方法三：利用 Gauss 消元法 即通过 初等行变换 求解。
解：【方法一】 >> a=[4,-1,0,-1,0,0,0,0,0]";b=[-1,4,-1,0,-1,0,0,0,0,]";c=[0,-1,4,0,0,-1,0,0,0]";d=[-1,0,0,4,-1,0,-1,0,0]";e=[0,-1,0,-1,4,-1,0,-1,0]";f=[0,0,-1,0,-1,4,0,0,-1]";g=[0,0,0,-1,0,0,4,-1,0]";h=[0,0,0,0,-1,0,-1,4,-1]";i=[0,0,0,0,0,-1,0,-1,4]";j=[52,36,64,16,0,28,24,8,36]";>> D=det([a,b,c,d,e,f,g,h,i]), D =
100352 >> D1=det([j,b,c,d,e,f,g,h,i]), D1 = 2494464 >> D2=det([a,j,c,d,e,f,g,h,i]), D2 = 2809856 >> D3=det([a,b,j,d,e,f,g,h,i]), D3 = 2924544 >> D4=det([a,b,c,j,e,f,g,h,i]), D4 = 1949696 >> D5=det([a,b,c,d,j,f,g,h,i]), D5 = 2207744 >> D6=det([a,b,c,d,e,j,g,h,i]), D6 = 2465792 >> D7=det([a,b,c,d,e,f,j,h,i]), D7 = 1490944
>> D8=det([a,b,c,d,e,f,g,j,i]), D8 = 1605632 >> D9=det([a,b,c,d,e,f,g,h,j]), D9 = 1921024 >> T1=D1/D,T2=D2/D,T3=D3/D,T4=D4/D,T5=D5/D,T6=D6/D,T7=D7/D,T8=D8/D,T9=D9/D, T1 = 174/7 T2 = 28 T3 = 204/7 T4 = 136/7 T5 = 22 T6 = 172/7
T7 = 104/7 T8 = 16 T9 = 134/7 【方法二】 >> A=[4,-1,0,-1,0,0,0,0,0;-1,4,-1,0,-1,0,0,0,0;0,-1,4,0,0,-1,0,0,0;-1,0,0,4,-1,0,-1,0,0;0,-1,0,-1,4,-1,0,-1,0;0,0,-1,0,-1,4,0,0,-1;0,0,0,-1,0,0,4,-1,0;0,0,0,0,-1,0,-1,4,-1;0,0,0,0,0,-1,0,-1,4] A = 4-1 0-1 0 0 0 0 0-1 4-1 0-1 0 0 0 0 0-1 4 0 0-1 0 0 0-1 0 0 4-1 0-1 0 0 0-1 0-1 4-1 0-1 0 0 0-1 0-1 4 0 0-1 0 0 0-1 0 0 4-1 0 0 0 0 0-1 0-1 4-1 0 0 0 0 0-1 0-1 4 >> inv(A)ans =
67/224 11/112 1/32 11/112 1/16 3/112 1/32 3/112 3/224 11/112 37/112 11/112 1/16 1/8 1/16 3/112 5/112 3/112 1/32 11/112 67/224 3/112 1/16 11/112 3/224 3/112 1/32 11/112 1/16 3/112 37/112 1/8 5/112 11/112 1/16 3/112 1/16 1/8 1/16 1/8 3/8 1/8 1/16 1/8 1/16 3/112 1/16 11/112 5/112 1/8 37/112 3/112 1/16 11/112 1/32 3/112 3/224 11/112 1/16 3/112 67/224 11/112 1/32 3/112 5/112 3/112 1/16 1/8 1/16 11/112 37/112 11/112 3/224 3/112 1/32 3/112 1/16 11/112 1/32 11/112 67/224 >> B=[52;36;64;16;0;28;24;8;36] B = 52 36 64 16 0 28 24 8 36 >> T=inv(A)*B T = 174/7 28 204/7 136/7 22 172/7 104/7
16 134/7 【方法三】 >> format rat,% >> A=[4,-1,0,-1,0,0,0,0,0,52;-1,4,-1,0,-1,0,0,0,0,36;0,-1,4,0,0,-1,0,0,0,64;-1,0,0,4,-1,0,-1,0,0,16;0,-1,0,-1,4,-1,0,-1,0,0;0,0,-1,0,-1,4,0,0,-1,28;0,0,0,-1,0,0,4,-1,0,24;0,0,0,0,-1,0,-1,4,-1,8;0,0,0,0,0,-1,0,-1,4,36] A = 4-1 0-1 0 0 0 0 0 52-1 4-1 0-1 0 0 0 0 36 0-1 4 0 0-1 0 0 0 64-1 0 0 4-1 0-1 0 0 16 0-1 0-1 4-1 0-1 0 0 0 0-1 0-1 4 0 0-1 28 0 0 0-1 0 0 4-1 0 24 0 0 0 0-1 0-1 4-1 8 0 0 0 0 0-1 0-1 4 36 >> rref(A)ans = 1 0 0 0 0 0 0 0 0 174/7 0 1 0 0 0 0 0 0 0 28 0 0 1 0 0 0 0 0 0 204/7 0 0 0 1 0 0 0 0 0 136/7 0 0 0 0 1 0 0 0 0 22
0 0 0 0 0 1 0 0 0 172/7 0 0 0 0 0 0 1 0 0 104/7 0 0 0 0 0 0 0 1 0 16 0 0 0 0 0 0 0 0 1 134/7 所以，线性方程组的解为：T 1 = 174/7 ,T 2 = 28, T 3 =204/7, T 4 =136/7, T 5 = 22, T 6 =172/7, T 7 =104/7, T 8 =16, T 9 =134/7
