Which systems of linear equations have no solution?
x + y + z = 1,100
x - 2y - z = -500
2x + 3y + 2z = 2,600
x + y + z = 1,500
x - y -z = -500
2x + y + z = 2,000
x + y + z = 1,400
X-2y - Z = -500
2x + 2y + 2z = 2,700
x + y + z = 1,400
-0.5x - 0.5y - 0.5z = -900
2x + 3y + 2z = 3,000
x + y + z = 1,900
x - y - 2z = -2,000
2x + 2y + z = 1,100
x + y + z = 2,400
2x - 2y + 2z = 700
x + 3y + z = 2,400
Answers
Answer:
Step-by-step explanation:
The systems of linear equations that have no solution are:
a)
x + y + z = 1,100
x - 2y - z = -500
2x + 3y + 2z = 2,600
b)
x + y + z = 2,400
2x - 2y + 2z = 700
x + 3y + z = 2,400
What is a solution?Solutions are the values of an equation where the values are substituted in the variables of the equation and make the equality in the equation true.
We also find the solution in a system of equations using the substitution or elimination method.
Example:
2x + 4 = 8
The solution is x = 2.
We have,
To determine which systems of linear equations have no solution, we need to check whether the system is inconsistent, i.e., there are no values of x, y, and z that satisfy all the equations simultaneously.
One way to check this is to use row reduction to put the system into row echelon form or reduced row echelon form and see if we obtain any contradictory equations.
Another way is to use determinants, but we'll use row reduction here.
a)
| 1 1 1 | 1,100 |
| 1 -2 -1 | -500 |
| 2 3 2 | 2,600 |
R2 = R2 - R1 and R3 = R3 - 2R1:
| 1 1 1 | 1,100 |
| 0 -3 -2 | -1,600 |
| 0 1 -1 | -1,800 |
R2 = -1/3 R2 and R3 = R3 + R2:
| 1 1 1 | 1,100 |
| 0 1/3 2/3 | 533.33|
| 0 0 1/3 | -933.33|
Now, we can solve for z in the last equation:
z = -2800
Substituting into the second equation gives:
y = 2666.67
But substituting into the first equation gives:
x = -566.67
So, this system has no solution.
b)
| 1 1 1 | 1,500 |
| 1 -1 -1 | -500 |
| 2 1 1 | 2,000 |
R2 = R2 - R1 and R3 = R3 - 2R1:
| 1 1 1 | 1,500 |
| 0 -2 -2 | -2,000 |
| 0 -1 -1 | -1,000 |
R2 = -1/2 R2 and R3 = R3 - 1/2 R2:
| 1 1 1 | 1,500 |
| 0 1 1 | 1,000 |
| 0 0 -1 | -500 |
Now, we can solve for z in the last equation:
z = 500
Substituting into the second equation gives:
y = 500
But substituting into the first equation gives:
x = 500
So, this system has a unique solution.
c)
| 1 1 1 | 1,400 |
| 1 -2 -1 | -500 |
| 2 2 2 | 2,700 |
R2 = R2 - R1 and R3 = R3 - 2R1:
| 1 1 1 | 1,400 |
| 0 -3 -2 | -900 |
| 0 0 0 | 0 |
The last row of the augmented matrix corresponds to the equation 0x + 0y + 0z = 0, which is always true.
Therefore, this equation is not contradictory and doesn't add any new information to the system.
We can eliminate it and rewrite the system as:
| 1 1 1 | 1,400 |
| 0 -3 -2 | -900 |
R2 = -1/3 R2:
| 1 1 1 | 1,400 |
| 0 1 2/3 | 300 |
R1 = R1 - R2:
| 1 0 1/3 | 1,100 |
| 0 1 2/3 | 300 |
Now, we can solve for z in the first equation:
z = 3,300
Substituting into the second equation gives:
y = -1,200
But substituting into the first equation gives:
x = -2,200
So, this system has a unique solution.
d)
| 1 1 1 | 1,400 |
|-0.5 -0.5 -0.5 | -900 |
| 2 3 2 | 3,000 |
R2 = 2R1 + R2 and R3 = -4R1 + R3:
| 1 1 1 | 1,400 |
| 0 1/2 1/2 | -100 |
| 0 -1 -2 | -3,600 |
R3 = -2R2 + R3:
| 1 1 1 | 1,400 |
| 0 1/2 1/2 | -100 |
| 0 0 -3 | -3,400 |
Now, we can solve for z in the last equation:
z = 1,133.33
Substituting into the second equation gives:
y = -1,066.67
But substituting into the first equation gives:
x = 2,333.33
So, this system has a unique solution.
e)
| 1 1 1 | 1,900 |
| 1 -1 -2 | -2,000 |
| 2 2 1 | 1,100 |
R2 = R2 - R1 and R3 = R3 - 2R1:
| 1 1 1 | 1,900 |
| 0 -2 -3 | -3,900 |
| 0 0 -3 | -4,100 |
Now, we can solve for z in the last equation:
z = 1,366.67
Substituting into the second equation gives:
y = -1,033.33
But substituting into the first equation gives:
x = 1,667
So, this system has a unique solution.
f)
| 1 1 1 | 2,400 |
| 2 -2 2 | 700 |
| 1 3 1 | 2,400 |
R2 = R2 - 2R1 and R3 = R3 - R1:
| 1 1 1 | 2,400 |
| 0 -4 0 | -1,100 |
| 0 2 0 | 1,000 |
R2 = -1/4 R2 and R3 = 1/2 R3:
| 1 1 1 | 2,400 |
| 0 1 0 | 275 |
| 0 1 0 | 500 |
R3 = R3 - R2:
| 1 1 1 | 2,400 |
| 0 1 0 | 275 |
| 0 0 0 | 225 |
The last equation shows that 0x + 0y + 0z = 225, which is impossible. Therefore, this system has no solution.
Thus,
The systems of linear equations that have no solution are:
a)
x + y + z = 1,100
x - 2y - z = -500
2x + 3y + 2z = 2,600
b)
x + y + z = 2,400
2x - 2y + 2z = 700
x + 3y + z = 2,400
Learn more about solutions of equations here:
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