Question 1196722: Find (if possible) values that the system of linear equations has (1) no solution, (2) exactly one solution, (3) infinitely many solution
2x-y+z = a
x+y+2z = b
3y+3z = c
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
The original system is
which is the same as
The coefficients and right hand side values form this augmented matrix
This may seem a bit strange, but I'll represent the matrix notation as a table of values.
I'll write it as When you write the steps to your teacher, you won't have the grid lines shown. Though I think the grid lines or boxes are very useful, and wish matrix notation involved it.
Let's swap the first and second row (R1 and R2)This way we can have the upper right corner as '1'
Double everything in R1 to go from 1,1,2,b to 2,2,4,2b
We'll subtract this from R2. The result replaces R2.
| 1 | 1 | 2 | b | | | 0 | -3 | -3 | a-2b | R2 - 2*R1 ---> R2 | | 0 | 3 | 3 | c | |
It might help to use a spreadsheet to line up the values.
The purpose of that previous step was to make the '2' in the second row turn into 0. The goal is to zero out everything below that '1' in the upper left corner.
Next, we'll swap R2 and R3
| 1 | 1 | 2 | b | | | 0 | 3 | 3 | c | R2 <--> R3 | | 0 | -3 | -3 | a-2b | |
We have positive 3's and negative 3's. They conveniently add to 0.
So we'll add R2 and R3 together and the result will replace R3
| 1 | 1 | 2 | b | | | 0 | 3 | 3 | c | | | 0 | 0 | 0 | a-2b+c | R2+R3 --> R3 |
We have gone from this matrix
to this one
after applying row reduction. The matrix is in REF (row echelon form)
The item in the bottom right corner is a-2b+c
Equate it with zero and solve for c.
a-2b+c = 0
c = 2b-a
If that condition is true, then the bottom right corner entry is 0 and it leads to a string of zeros along the bottom row.
This string of zeros tells us that we have infinitely many solutions since we have  along the bottom row.
The system is consistent and dependent in such a case.
If  , then the bottom right entry is nonzero and we have no solutions since we have  for some nonzero value k along the bottom row. The left side is always zero and right hand side is never zero.
The system is inconsistent in this case.
There are no cases possible where we'd get exactly one solution since either  or . Meaning that we either get a dependent system or an inconsistent one.
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Normally I'd recommend using the Linear Algebra Toolkit
http://www.math.odu.edu/~bogacki/lat/
However, such a tool doesn't work with variables as inputs. If the a,b,c were actual numbers then we can use this tool.
Instead, you can to use something like GeoGebra or WolframAlpha to confirm your answers.
The way to input matrix notation for either app is to type in something like
{{1,2},{3,4}} to mean 
So you'll type in
{{2,-1,1,a},{1,1,2,b},{0,3,3,c}}
Include the key term rref to tell WolframAlpha to apply row reduction. For GeoGebra, the command is labeled ReducedRowEchelonForm. You need to be in CAS mode to able to row reduce this matrix with a,b,c in it.
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Here is a way to do this problem without using matrices.
I'll be using the substitution method.
Solve the third equation for z
3y+3z = c
3z = c-3y
z = (c-3y)/3
Then substitute that into the second equation
x+y+2z = b
x+y+2(c - 3y)/3 = b
3x+3y+2(c-3y) = 3b
3x+3y+2c-6y = 3b
3x-3y+2c = 3b
In the third step, I multiplied both sides by 3 to clear out the fraction.
Then let's isolate y
3x-3y+2c = 3b
-3y = 3b-3x-2c
y = -(3b-3x-2c)/3
Let's plug z = (c-3y)/3 into the first equation.
2x-y+z = a
2x-y+(c-3y)/3 = a
6x-3y+c-3y = 3a
6x-6y+c = 3a
Then plug in y = -(3b-3x-2c)/3
6x-6y+c = 3a
6x-6( -(3b-3x-2c)/3 )+c = 3a
6x+2(3b-3x-2c)+c = 3a
6x+6b-6x-4c+c = 3a
6b-3c = 3a
Notice at this point all x,y,z variables are eliminated.
Let's solve for c
6b-3c = 3a
6b-3a = 3c
3c = 6b-3a
c = (6b-3a)/3
c = 2b-a
If c = 2b-a, then the system has infinitely many solutions.
If , then the system has no solutions.
There are no cases in which this system has exactly one solution.
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