代码搜索：线性分析

找到约 10,000 项符合「线性分析」的源代码

代码结果 10,000
热门代码： main.c GPIO I2C SPI UART timer interrupt ADC PWM LED
www.eeworm.com/read/197958/7960488

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/196814/8058542

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/244945/12829400

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/329331/12960295

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/147529/5728597

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/147529/5728807

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/263879/11338086

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/337307/12377273

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/394381/8227492

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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www.eeworm.com/read/334860/12567968

m rooteg1.m

%方程求根例1：线性方程组 clear; A=[-1 1;-2 2]; B=[1;2]; x1=A\B,x2=pinv(A)*B A(3,2)=0; B(3)=0; x1=A\B C=[-1 1;-1 1]; D=[1;0]; y1=C\D,y2=pinv(C)*D C(3,2)=0; D(3)=0; y1=C\D
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