Solve the following system by using MATLAB a) Solve the following system by usin

Question

Solve the following system by using MATLAB

a) Solve the following system by using MATLAB with Back Substitution method.

4X1 – X2 +2X3+2X4 –X5 = 4

-2X2+6X3+2X4+7X5= 0

   X3 –X4-2X5 = 3

   -2X4 –X5 =10

   3X5=6

b) If A= [3 2 ; 1 -1] and B= [-1 ; 1] then find

I. Determent of A

II. Inverse

III. x=B

IIII. Trace

V. Rank

VI. Row reduced echelon

c). Make a program with for loop by using Cramer’s rule A= [2 1 1; 1 -1 -1; 1 2 1 ] and B= [3; 0 ; 0]

NOTE : breifly explain the Following , thank You so much !

Explanation / Answer

problem (A)

matlab code:

clc;
clear;
x5=2
x4=(10+x5)/-2
x3=3+2*x5+x4
x2=((7*x5)+(2*x4)+(6*x3))/2
x1=(4+x5-(2*x4)-(2*x3)+x2)/4

problem (B) matlab code:

clc;
clear;
A=[3 2;1 -1];
B=[-1;1]
('Determent of A')
det(A)
('Inverse of A')
inv(A)
('solution of AX=B')
X=inv(A)*B
('trace of A')
trace(A)
('Row reduced echelon')
rref(A)
('rank of A')
rank(A)

cramer rule matlab code with for loop:

clc;
clear;
A=[2 1 1;1 -1 -1;1 2 1]
B=[3;0;0;]
D=zeros(1,3)
C=det(A);
E=A;
F=A;
G=A;
('cramer rule')

for i=1:3
G(:,i)=B(:,1)

D(i)=det(G);
G=A;

end
D
('solution is')
x=D(1)/C
y=D(2)/C
z=D(3)/C

cramer rule with out for loop:

clc;
clear;
A=[2 1 1;1 -1 -1;1 2 1]
B=[3;0;0;]
D=zeros(1,3)
C=det(A);
E=A;
F=A;
G=A;
('cramer rule')

for i=1:3
G(:,1)=B(:,1)
E(:,2)=B(:,1)
F(:,3)=B(:,1)

D(1)=det(A)
D(2)=det(E)
D(3)=det(F)
('solution is')
x=D(1)/C
y=D(2)/C
z=D(3)/C

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