SOLUTION: 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

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

2	-1	1	a
1	1	2	b
0	3	3	c

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)

1	1	2	b	R1 <--> R2
2	-1	1	a
0	3	3	c

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.

------------------------------------------

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.

------------------------------------------

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.

RELATED QUESTIONS
Consider the equation 3x+2y=6. If possible, find a second linear equation to create a... (answered by Fombitz)

Consider the equation 3x+2y=6. If possible, find a second linear equation to create a... (answered by Fombitz)

Graph the system of linear equations. Does the system have exactly one solution no... (answered by checkley77)

The system of equations y=2x-3 and y=4x-3 has a. exactly one solution b. no solution... (answered by vleith,6785071231)

4c= 3d + 3 c= d-1 The book says use substitution to solve each system of equations.... (answered by stanbon)

Use Substitution to solve each system of equations. If the system does not have exactly... (answered by checkley77)

Find values of a, b, and c (if possible) such that the system of linear equations has a... (answered by ikleyn)

Find the value(s) of the constant k such that the system of linear equations: 9x + ky =... (answered by htmentor,ikleyn)

Determine whether the system of linear equations has one and only one solution,... (answered by stanbon)