Solve the system 3x1+2x2−x3=4 x1−2x2+2x3=1 11x1+2x2+x3=14

Caroline Elliott

Caroline Elliott

Answered

2022-01-21

Solve the system
3x1+2x2x3=4
x12x2+2x3=1
11x1+2x2+x3=14

Answer & Explanation

Damian Roberts

Damian Roberts

Expert

2022-01-22Added 14 answers

Step 1
Given:
3x1+2x2x3=4
x12x2+2x3=1
11x1+2x2+x3=14
Step 2
[3211221121][x1x2x3]=[4114]
Therefore augmented matrix is
[32141221112114]
Step 3
Row transformations
R23R2R1, R33R311R1
[32140871016142]
R3R32R2
[321408770000]
Step 4
Thus,
x3 is free variable
x2=18(17x3)
x1=14(5x3)
for x3=1
x2=1
x1=1
Therefore, (111) is one of the solution of given system.
enguinhispi

enguinhispi

Expert

2022-01-23Added 15 answers

Step 1
Consider the provided question,
The given system of equation is,
x12x2+2x3=1 (I)
3x1+2x2x3=4 (II)
11x1+2x2+x3=14 (III)
Operate (II)((I)×3) and (III)((I)×11) we obtain,
8x27x3=1 (IV)
And, 24x221x3=38x27x3=1 (V)
From (IV) and (V),
0x20x3=0
Thus, the system of equation is written as,
x12x2+2x3=1
8x27x3=1
0x20x3=0
And, the corresponding coefficient matrix is,
[112087000]
Step 2
Since, the corresponding coefficient matrix in row echelon form,
And, the required equivalent system is,
x12x2+2x3=1
8x27x3=1
0x20x3=0
Augumented matrix:
[122108710000]
Since
Rank=number of non=zero row in row echelon form of matrix
So, rank of coefficient matrix is 2 and, that of augumented matrix is also 2. NKS Therefore, the given system of equation is consistent.
Since, number of free variable
=number of unknown variablerank of coefficient matrix
=32
=1
So, the system has one free variable.
let it be x3 and x3=a; aR
Step 3
x12x2+2a=1
8x27a=1
0x20a=0
x2=1+7a8
Now,

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