Solving 4x3 Matrix: [[1 2 3][2 -1 1][0 1 1][1 -1 0]]*[[x][y][z]]=[[-3][4][-2][a]]

[Mrmatheco]

Hi,

This matrix-system is given:



I am asked to explain how the number of solutions depends on the parameter a.
I'm really clueless as to how I find the solution(s).
If I just use the three first equations in the matrix and use the gauss elimination I got the solution:
x= -1, Y=-4, z=2 (in this solution a=3)

Apparently there are infinite solutions... How do I find that solution?

I would really appreciate some help! Thanks

 

[tkhunny]

Hi,

This matrix-system is given:

View attachment 10447

I am asked to explain how the number of solutions depends on the parameter a.
I'm really clueless as to how I find the solution(s).
If I just use the three first equations in the matrix and use the gauss elimination I got the solution:
x= -1, Y=-4, z=2 (in this solution a=3)

Apparently there are infinite solutions... How do I find that solution?

I would really appreciate some help! Thanks

Do EXACTLY what you would do with a 3x3 situation. What do you get and how do you interpret it.

 

[Mrmatheco]

Do EXACTLY what you would do with a 3x3 situation. What do you get and how do you interpret it.

I'm not sure I understand... You can only use Gauss-elimination if the matrix is squared right?

The solution (x, y z) = (-1, -4, 2) for a=3 is the only solution I am able to find that satisfy all four equations...
I've been trying to wrap my head around this the past couple of days
I don't understand how I can solve it as I would with a 3x3 situation?

 

[Deleted member 4993]

Hi,

This matrix-system is given:

View attachment 10447

I am asked to explain how the number of solutions depends on the parameter a.
I'm really clueless as to how I find the solution(s).
If I just use the three first equations in the matrix and use the gauss elimination I got the solution:
x= -1, Y=-4, z=2 (in this solution a=3)

Apparently there are infinite solutions... How do I find that solution?

I would really appreciate some help! Thanks

Why do you think so?

 

[Mrmatheco]

Why do you think so?

Because of how the questions are formulated.

These are the questions:
a) Explain how the number of solutions of the equations depend on the parameter ?.
b) Find the complete solution for the case where there are infinite solutions.
c) Which of the solutions in b) is closest to origo?

 

[topsquark]

I'm not sure I understand... You can only use Gauss-elimination if the matrix is squared right?

So don't do Gaussian elimination! Just go ahead and multiply it out. You'll get four equations, the first of which is x + 2y + 3z = -3.

-Dan

 

[tkhunny]

I'm not sure I understand... You can only use Gauss-elimination if the matrix is squared right?

The solution (x, y z) = (-1, -4, 2) for a=3 is the only solution I am able to find that satisfy all four equations...
I've been trying to wrap my head around this the past couple of days
I don't understand how I can solve it as I would with a 3x3 situation?

What's stopping you from performing the exercise? No one thinks I will do EXACTLY what you might expect. Just do it and interpret what it does do.

 

[HallsofIvy]

Hi,

This matrix-system is given:

View attachment 10447

I am asked to explain how the number of solutions depends on the parameter a.
I'm really clueless as to how I find the solution(s).
If I just use the three first equations in the matrix and use the gauss elimination I got the solution:
x= -1, Y=-4, z=2 (in this solution a=3)

This is equivalent to the four equations
x+ 2y+ 3z= -3
2x- y+ z= 4
y+ z= -2
x- y= a

The third equation gives z= -2- y. Putting that into the first equation, x+ 2y- 6- 3y= x- y- 6= -3 so x- y= 3. Putting z= -2- y into the second equation, 2x- y- 2- y= 2x- 2y- 2= 4 so x- y= 3 again! There are an infinite number of solutions to those 3 equations. If we put y= x- 3 into the fourth equation we get x- y= x- (x- 3)= 3= a. If a is any number other than 3, there are no values of x, y, z that satisfy all four equations. If a= 3, y= x- 3, z= -2- y= -2- (x- 3)= 1- x satisfy all four equations for any x.

For example, taking x= 0, y= -3 and z= 1 satisfy all four equations:
x+ 2y+ 3z= 0- 6+ 3= -3
2x- y+ z= 0+ 3+ 1= 4
y+ z= -3+ 1= -2
x- y= 0+ 3= 3= a

But if x= 1, y= -2, z= 0 also satisfy all four equations:
x+ 2y+ 3z= 1- 4+ 0= -3
2x- y+ z= 2+ 2+ 0= 4
y+ z= -2+ 0= -2
x- y= 1+ 2= 3= a

etc.

Apparently there are infinite solutions... How do I find that solution?

I would really appreciate some help! Thanks

 

[Mrmatheco]

This is equivalent to the four equations
x+ 2y+ 3z= -3
2x- y+ z= 4
y+ z= -2
x- y= a

The third equation gives z= -2- y. Putting that into the first equation, x+ 2y- 6- 3y= x- y- 6= -3 so x- y= 3. Putting z= -2- y into the second equation, 2x- y- 2- y= 2x- 2y- 2= 4 so x- y= 3 again! There are an infinite number of solutions to those 3 equations. If we put y= x- 3 into the fourth equation we get x- y= x- (x- 3)= 3= a. If a is any number other than 3, there are no values of x, y, z that satisfy all four equations. If a= 3, y= x- 3, z= -2- y= -2- (x- 3)= 1- x satisfy all four equations for any x.

For example, taking x= 0, y= -3 and z= 1 satisfy all four equations:
x+ 2y+ 3z= 0- 6+ 3= -3
2x- y+ z= 0+ 3+ 1= 4
y+ z= -3+ 1= -2
x- y= 0+ 3= 3= a

But if x= 1, y= -2, z= 0 also satisfy all four equations:
x+ 2y+ 3z= 1- 4+ 0= -3
2x- y+ z= 2+ 2+ 0= 4
y+ z= -2+ 0= -2
x- y= 1+ 2= 3= a

etc.

Thank you so much! It all make sense now. Finally it is solved.