如何使用MATLAB求解二元一次矩阵方程a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0]b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0]c=[0,0,0,0,0;-0.053,0,0,0,0;3.4
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![如何使用MATLAB求解二元一次矩阵方程a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0]b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0]c=[0,0,0,0,0;-0.053,0,0,0,0;3.4](/uploads/image/z/3162057-33-7.jpg?t=%E5%A6%82%E4%BD%95%E4%BD%BF%E7%94%A8MATLAB%E6%B1%82%E8%A7%A3%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E7%9F%A9%E9%98%B5%E6%96%B9%E7%A8%8Ba%3D%5B0%2C0%2C-1%2C0%2C0%3B-0.558%2C-0.097%2C-0.903%2C-0.472%2C0%3B0%2C-1%2C0%2C0%2C0%3B0%2C0%2C0%2C0%2C0%3B0%2C0%2C0%2C0%2C0%5Db%3D%5B0%2C0%2C1%2C0%2C-0.02%3B0.7%2C0%2C1%2C0.476%2C0.0048%3B-2.95%2C-0.568%2C1.568%2C-2.758%2C0%3B0%2C0%2C0%2C0.0199%2C0.0101%3B0%2C1%2C0%2C0%2C0%5Dc%3D%5B0%2C0%2C0%2C0%2C0%3B-0.053%2C0%2C0%2C0%2C0%3B3.4)
如何使用MATLAB求解二元一次矩阵方程a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0]b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0]c=[0,0,0,0,0;-0.053,0,0,0,0;3.4
如何使用MATLAB求解二元一次矩阵方程
a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0]
b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0]
c=[0,0,0,0,0;-0.053,0,0,0,0;3.4897,0,0,2.786,0.0277;0,0,0,0,0;0,-1,0,0,0;]
求解的方程形式为ax^2-bx+c=0 希望有人能写出代码
如何使用MATLAB求解二元一次矩阵方程a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0]b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0]c=[0,0,0,0,0;-0.053,0,0,0,0;3.4
这应该是一元二次方程吧
clear all
clc
a=[0,0,-1,0,0;-0.558,-0.097,-0.903,-0.472,0;0,-1,0,0,0;0,0,0,0,0;0,0,0,0,0];
b=[0,0,1,0,-0.02;0.7,0,1,0.476,0.0048;-2.95,-0.568,1.568,-2.758,0;0,0,0,0.0199,0.0101;0,1,0,0,0];
c=[0,0,0,0,0;-0.053,0,0,0,0;3.4897,0,0,2.786,0.0277;0,0,0,0,0;0,-1,0,0,0;];
result=solve('a*x^2-b*x+c=0');
x1=eval(result(1));%x1、x2就是两个根
x2=eval(result(1));