Solving Inhomogeneous Linear Equations: Proving Solvability for All a-Values

[Mathman23]

Hi Guys,

I got this linear algebra question I hope You Guys can assist me with

Below there is a inhomogeneous system of linear equations which I solve:

x_{1} + 2x_{2} + x_{3} = a

3x_{1} + 4x_{2} + 2x_{3} = a - 3

-4x_{1} + 2x_{2} + x_{3} = 3

I then end up the following matrix:

1 0 0 0 -3a/5

0 1 1/2 (2a+3)/5

0 0 0 (-11a/10)

I´m then supose to prove if its possible to solve the system for every a-value.

By solving the equation (2a+3)/5) = 0. I then get an a-value a = -1/2 .

If I insert this a into the above matrix and row-reduce that matrix I get:

1 0 0 0
0 1 1/2 0
0 0 0 1

I then do some tests and then conclude if I choose an a-value in the interval
[-1000000,1000000]. I still end up the same matrix above.

Is it then correct to assume if I chose an a-value in the interval [-infty,infty]. I would then still end up with the same matrix?

If my assumption is correct, is it then safe to assume that its possible to solve the system of linear-equations for every a-value??

sincerely
Fred

 

[Nenad]

why did you set your second line equal to 0. You should set the last row equal to 0, for to solve the system with real solutions, you need to have 0 = 0. It also epends on what kind of solutions you are looking for, do you want infinite solutions, one real solution or No solution to the system.

Regards,

Nenad

 

[Galileo]

If you write it out in matrix form:

<br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 1 \\<br /> 3 &amp; 4 &amp;2 \\<br /> -4 &amp; 2 &amp; 1\\ \end{array}\right)<br /> \left(<br /> \begin{array}{c}<br /> x_1 \\ x_2 \\ x_3\\ \end{array}\right)<br /> =\left(<br /> \begin{array}{c}<br /> a \\ a-3 \\ 3 \\ \end{array}\right)

which is solvable iff the vector on the right is in the span of the column vectors in the matrix. Do the column vectors of the matrix span the entire space?

 

[xanthym]

If you write it out in matrix form:

<br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 1 \\<br /> 3 &amp; 4 &amp;2 \\<br /> -4 &amp; 2 &amp; 1\\ \end{array}\right)<br /> \left(<br /> \begin{array}{c}<br /> x_1 \\ x_2 \\ x_3\\ \end{array}\right)<br /> =\left(<br /> \begin{array}{c}<br /> a \\ a-3 \\ 3 \\ \end{array}\right)

which is solvable iff the vector on the right is in the span of the column vectors in the matrix. Do the column vectors of the matrix span the entire space?

The Determinant of the 3x3 matrix is ZERO (0). Therefore, there does NOT exist a SINGLE UNIQUE solution. Either there are no solutions or an infinite number, depending on the value of "a".


~~

 

[HallsofIvy]

1 0 0 0 -3a/5

0 1 1/2 (2a+3)/5

0 0 0 (-11a/10)

I´m then supose to prove if its possible to solve the system for every a-value.

"prove if"? Do you mean "determine whether or not it is possible to solve the system for each a-value"?

The obvious point is that the last row of your final matrix: 0 0 0 (-11a/10)
is only valid if (-11a/10)= 0 which reduces to a= 0.

However, I don't get that. I get:

1 2 1 a
0 -2 -1 -2a-3
0 0 0 -6a-12

Which means a should be -2 in order for this system to have a solution. (And it will, in that case, have an infinite number of solutions.)

 

[Mathman23]

Hello Hall,

Yes what I'm attempting to determain if the system of non-homogeneous linear equations can be solved for any a !

I guess then my matrix operations are done incorrectly.

With Your result in mind I will retry them

Sincerley and thanks for Your answer.

/Fred

 

[Mathman23]

"prove if"? Do you mean "determine whether or not it is possible to solve the system for each a-value"?

The obvious point is that the last row of your final matrix: 0 0 0 (-11a/10)
is only valid if (-11a/10)= 0 which reduces to a= 0.

However, I don't get that. I get:

1 2 1 a
0 -2 -1 -2a-3
0 0 0 -6a-12

Which means a should be -2 in order for this system to have a solution. (And it will, in that case, have an infinite number of solutions.)

Hi Hall,

I have rerun my matrix operations and I now get the same result as You

How come its possible to conclude by solving -6a-12=0 that "a" must equal
-2 in order to the system of linear equations to have a solution?

And do this contradict the claim "determine whether or not it is possible to solve the system for each and any a-value?"

Sincerley

Fred

 

[HallsofIvy]

Isn't it obvious? What you are saying in doing that "row-reduction', is that the original three equations are equivalent to
x+ 2y+ z= a
-2y- z= -2a-3
0= -6a-12

In order for all those equations to be true- certainly the third one has to be true!

Since there are no x, y, or z in it we MUST have -6a- 12= 0 (therefore a= -2) in order for it to be true.