Solve the given system of equations, or else show that

dublattm 2022-01-22 Answered
Solve the given system of equations, or else show that there is no solution.
a) x1+2x2x3=1
2x1+x2+x3=1
x1x2+2x3=1
b) x1+2x2x3=2
2x14x2+2x3=4
2x1+4x22x3=4
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Expert Answer

coolbananas03ok
Answered 2022-01-23 Author has 20 answers

Step 1
Consider the following system:
x1+2x2x3=1
2x1+x2+x3=1
x1x2+2x3=1
Solve the system
Write augmented matrix
[121121111121]
R3R1R22R1
[121103310330]
R3R2
[121103310001]
From the last row, we got 0=1 which is absqrt.
Hence the system has no solution
Step 2
Consider the following syste:
x1+2x2x3=2
2x14x2+2x3=4
2x1+4x22x3=4
Solve the system
Write the augmented matrix
[121224242424]
R2+2R1R32R1
[121200000000]
From this, we get x1+2x2x3=2
Here, x2, x3 are free variables.
So choose x2=s, x3=t, where s, t are any parameters.
Then x1=22s+t
Hence the solution to the given system is
(x1x2x3)=(22s+tst), where s, t are any parameters

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ocretz56
Answered 2022-01-24 Author has 16 answers
Step 1
Part A
The coefficient matrix is :
A=[121211112]
Now
det(A)=[121211112]=1×(2+1)2×(41)1×(21)
det(A)=36+3=0
Since det(A)=0, the matrix is not invertible and thus, has no solution.
Step 2
Part B:
The coefficient matrix is:
A=[121242242]
Now
det(A)==[121242242]=1×(88)2×(44)1×(8+8)
det(A)=0+0+0=0
Since det(A)=0, the matrix is not invertible and thus, has no solution.
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