Question

用矩阵除法求解以下线性方程组,写出程序过程和求解结果:2x1+3x2+x3=8;x1+3

Answer 1

问：用矩阵除法求解以下线性方程组,写出程序过程和求解结果:2x1+3x2+x3=8;x1+3

答：已知：A=[6,3,4;-2,5,7;8,-4,-3]，若AV=I，则V是A的逆矩阵：V=inv(A)ex2.10: 解下列方程组：6x1+3x2+4x3=3-2x1+5x2+7x3=-48x1-4x2-3x3=-7记为A*X=B，A=[6,3,4;-2,5,7;8,-4,-3]，X=[x1;x2;x3]B=[3;-4;-7]，则X=inv(A)*B=A\B（左除）。若X1=[x1, x2, x3], A1=[6,-2,8;3,5,-4;4,7,-3],B1=[3,-4,-7]则记为X1*A1=B1，故X1=B1*inv(A1)=B1/A1（右除）。A=[6,3,4;-2,5,7;8,-4,-3];B=[3;-4;-7]; X=A\B或：A1=[6,-2,8;3,5,-4;4,7,-3];B1=[3,-4,-7]; X1=B1/A1