Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system

arenceabigns

arenceabigns

Answered question

2021-06-03

Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system by back substitution. Assume that the variables are named x1,x2,from left to right. [105320124700113]

Answer & Explanation

oppturf

oppturf

Skilled2021-06-04Added 94 answers

The linear system corresponding to the augmented matrix is
x1+5x3+3x4=2

x22x3+4x4=7
x3+x4=3
Solve for the leading variables. x3+x4=3
x2=7+2x34x4
x3=3x4
Substitute x3=3x4 into the first two equations.
x1=13+2x4
x2=16x4
x3=3x4
Assign arbitrary value to free variable x4, say x4=t. Then the solution is described by the parametric equations. x1=13+2t, x2=16t, x3=3t, x4=t

RizerMix

RizerMix

Expert2023-05-27Added 656 answers

Given the row echelon form:
[105320124700113]
We can see that the third row corresponds to the equation:
0x1+0x2+1x3+1x4=3
This simplifies to:
x3+x4=3
Now, moving to the second row, we have:
0x1+1x22x3+4x4=7
Substituting the value of x3 from the third row into this equation, we get:
x22(1x4)+4x4=7x2+2x4=7
Finally, considering the first row, we have:
1x1+0x2+5x3+3x4=2
Substituting the values of x2 and x3 from the second and third rows, respectively, we get:
x1+5(1x4)+3x4=2x1+8x4=2
Therefore, the solution to the system of linear equations is:
x1=28x4x2=72x4x3=3x4 where x4 is a free variable.
nick1337

nick1337

Expert2023-05-27Added 777 answers

To solve the system of linear equations using back substitution, we start from the last equation and work our way up. Let's go step by step.
The given row echelon form is:
[105320124700113]
From the third equation, we have:
x3+x4=3
Let's solve for x3 in terms of x4:
x3=3x4
Now, substituting this value of x3 in the second equation, we get:
x22x3+4x4=7
Replacing x3 with 3x4, we have:
x22(3x4)+4x4=7
Simplifying this equation gives us:
x26+2x4+4x4=7
Combining like terms:
x2+6x4=1
From the first equation, we have:
x1+5x3+3x4=2
Substituting the value of x3 as 3x4:
x1+5(3x4)+3x4=2
Expanding and simplifying:
x1+155x4+3x4=2
Combining like terms:
x12x4=13
Finally, we have the equations:
x12x4=13x2+6x4=1x3=3x4
These are the solutions to the system of linear equations in terms of x1, x2, and x4.
Don Sumner

Don Sumner

Skilled2023-05-27Added 184 answers

Result:
x1=72x3, x2=19+6x3, x3, x4=3x3
Solution:
From the given row echelon form:
x3+x4=3
x22x3+4x4=7
x1+5x3+3x4=2
Solving the first equation for x4:
x4=3x3
Substituting this value of x4 into the second equation:
x22x3+4(3x3)=7
Simplifying:
x22x3+124x3=7
x26x3=19
Solving the equation for x2:
x2=19+6x3
Substituting the values of x2 and x4 into the third equation:
x1+5x3+3(3x3)=2
Simplifying:
x1+5x3+93x3=2
x1+2x3=7
Solving the equation for x1:
x1=72x3
Thus, the solution to the system of linear equations is:
x1=72x3
x2=19+6x3
x3 is a free variable (it can take any real value)
x4=3x3

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