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Question 1006275: Solve the system of equations by converting the matrix in reduced echelon form.
2x-2y+2z=1
3y-z =5
-2x+3y-z=1
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! This system
2x-2y+2z=1
3y-z =5
-2x+3y-z=1
Is the same as this system
2x-2y+2z=1
0x+3y-1z=5
-2x+3y-1z=1
Convert to matrix form
Add row 1 to row 3. The result will replace row 3
R1+R3 --> R3
Swap row 2 and row 3
R2 <--> R3
Subtract 3 times row 3 from row 2. This result will replace row 3
R3 - 3*R2 ---> R3
Multiply every term of row 3 by -0.25
-0.25*R3 ---> R3
Subtract 1*Row 3 from row 2. Then replace all of row 2 with the result
R2 - 1*R3 --> R2
Subtract 2*Row 3 from row 1. Then replace all of row 1 with the result
R1 - 2*R3 --> R1
Subtract -2*Row 2 from row 1. Then replace all of row 1 with the result
R1 - (-2)*R2 --> R1
Multiply all of row 1 by 0.5
0.5*R1 ---> R1
The last matrix above is in reduced row echelon form. So we're done.
The last thing to do is simply read off the values in the last column. They are: 2, 1.75, 0.25
So
x = 2
y = 1.75
z = 0.25
In fraction form
x = 2
y = 7/4
z = 1/4
The solution written as an ordered triple is (x,y,z) = (2,1.75,0.25)
In fraction form, the solution as an ordered triple is (x,y,z) = (2,7/4,1/4)
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Check:
Let's check the answer (x,y,z) = (2,1.75,0.25). To do this, we replace the x,y,z with the values we found above (2, 1.75 and 0.25 respectively) into each equation. The x,y,z values must make ALL of the equations of the system true.
Let's check equation 1
2x-2y+2z=1
2*2-2*1.75+2*0.25=1 ... plug in x = 2,y = 1.75,z = 0.25
1=1 ... true
Now let's check equation 2
3y-z =5
3*1.75-0.25 =5 ... plug in x = 2,y = 1.75,z = 0.25
5=5 ... true
and finally equation 3
-2x+3y-z=1
-2*2+3*1.75-0.25=1 ... plug in x = 2,y = 1.75,z = 0.25
1=1 ... true
All three equations are true when (x,y,z) = (2,1.75,0.25). So the solution is confirmed.
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