Question 1204100: Consider the system of equations
x + 2y - 3z = a
2x + 6y - 11z = b
x - 2y + 7z = c
where a, b and c are three real numbers.
1 What relation must the parameters a, b and c satisfy for the system of equations to have at least
the system of equations has at least one solution?
2. Assuming that a, b and c satisfy the relation that allows the system to have at least one solution, what relation must a, b and c satisfy?
the system has at least one solution, calculate in function of a, b and c, the general solution of the system of equations using Gauss's method.
Can the above linear system have a unique solution in R3?
Found 2 solutions by MathLover1, math_tutor2020: Answer by MathLover1(20850) (Show Source): Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
I'll convert the system of equations into a matrix as shown below.
Then I'll get that matrix into row echelon form (REF)
Each matrix is presented as a table. In my opinion, the grid lines help separate things to make the entries look more cleaner.
Normally however, matrix notation will not have these helpful grid lines.
1 | 2 | -3 | a | | 2 | 6 | -11 | b | | 0 | -4 | 10 | c-a | R3 - R1 --> R3 |
Notation like R3 - R1 --> R3 means we subtract rows 3 and 1, and store the results into row 3.
1 | 2 | -3 | a | | 0 | 2 | -5 | b-2a | R2 - 2*R1 --> R2 | 0 | -4 | 10 | c-a | |
1 | 2 | -3 | a | | 0 | 1 | -5/2 | (b-2a)/2 | (1/2)*R2 --> R2 | 0 | -4 | 10 | c-a | |
1 | 2 | -3 | a | | 0 | 1 | -5/2 | (b-2a)/2 | | 0 | 0 | 0 | -5a+2b+c | R3 + 4*R2 --> R3 |
We have gone from the matrix to the matrix
The last row leads to the equation
0x+0y+0z = -5a+2b+c
or
0 = -5a+2b+c
Solve for c to get
c = 5a-2b
This will mean we have infinitely many solutions if and only if c = 5a-2b
Otherwise, -5a+2b+c is nonzero and it causes a contradiction, and hence no solutions.
The values of 'a' and b can be any two real numbers you want.
---------------------------------------------------
Let's look at an example that has infinitely many solutions.
a = 1, b = 2
c = 5a-2b = 5*1-2*2 = 1
Therefore this system
is consistent and dependent.
It has infinitely many solutions.
I'll let the student verify this claim, and the later claims mentioned below.
Another example with infinitely many solutions.
a = 5, b = 3
c = 5a-2b = 5*5-2*3 = 19
This system
is consistent and dependent.
It has infinitely many solutions.
Let's look at an example of a system that has no solutions.
a = 1, b = 2, c = 3
These a,b,c values do not satisfy the equation c = 5a-2b
Therefore this system shown below has no solutions (it is inconsistent)
As you can see, we cannot pick a trio of a,b,c values to have the system produce exactly one unique solution.
Either we have infinitely many solutions, or none at all.
|
|
|