1. ## Simultaneous equations

Find the value of a for which there are infinitely many solutions to the equations
2x + ay - z = 0
3x + 4y - (a + 1)z = 13
10x + 8y + (a - 4)z = 26

not sure if I have to substitute in lambda or...any help would be greatly appreciated!

2. Originally Posted by scubasteve94
Find the value of a for which there are infinitely many solutions to the equations
2x + ay - z = 0
3x + 4y - (a + 1)z = 13
10x + 8y + (a - 4)z = 26

not sure if I have to substitute in lambda or...any help would be greatly appreciated!
Ok... not sure if I'm allowed to help since I'm new... but here goes...
In order for there to be infinitely many solutions to the equations, that means that all three of those equations are a multiple of each other. You can think of this as a system of equations and treat "a" as a constant, and try to solve either by addition or substitution. I would suggest multiplying the second equation by (-2). This will allow you to add this equation to the third equation and a variable will cancel out, and you also get a nice little constant on the right hand side that is very useful. Try that and see if that helps you progress a little.

3. Originally Posted by S31J41
Ok... not sure if I'm allowed to help since I'm new... but here goes...
In order for there to be infinitely many solutions to the equations, that means that all three of those equations are a multiple of each other. You can think of this as a system of equations and treat "a" as a constant, and try to solve either by addition or substitution. I would suggest multiplying the second equation by (-2). This will allow you to add this equation to the third equation and a variable will cancel out, and you also get a nice little constant on the right hand side that is very useful. Try that and see if that helps you progress a little.
Actually it really only requires one of the equations to be a multiple of another.

Because then that would equate to 2 equations in 3 unknowns. To solve this, the solution will be in terms of one of the variables, therefore infinitely many solutions...

4. Ok so I did that and ended up with
4x + (3a - 2)z = 0

5. Originally Posted by Prove It
Actually it really only requires one of the equations to be a multiple of another.

Because then that would equate to 2 equations in 3 unknowns. To solve this, the solution will be in terms of one of the variables, therefore infinitely many solutions...
Hmmm duly corrected I was thinking in simplier terms =P And that actually does not even pertain to this problem because the equations aren't a multiple of each other...

Now back to the problem

You got 4x + (3a - 2)z = 0 (which we will call equation 4) which is perfect because the first equation is 2x-z=0. Now if you multiply the first equation by (-2) and add it to equation 4. All the variables cancel out except for a, which should be easy to solve.

6. Equation 1 is 2x + ay - z = 0 though

7. Originally Posted by scubasteve94
Equation 1 is 2x + ay - z = 0 though
Alright, so

Equation 1 = 2x + ay - z = 0
Equation 4 = 4x + (3a - 2)z = 0

In order for these two equations to have infinitely many solutions, these two equations have to be a multiple of each other. which means that each coefficient must be a multiple of the corresponding coefficient in the other equation. Knowing that, this is easily solved because the corresponding coefficient for y is (0) which means that (ac), where c is any arbitrary constant is equal to (0) the only way this can happen is when a = 0.

8. 3 solutions when a <> 0:
a=1, x=-1, y=6, z=4
a=3, x=7, y=-6, z=-4
a=4, x=5, y=-3, z=-2

9. Originally Posted by Wilmer
3 solutions when a <> 0:
a=1, x=-1, y=6, z=4
a=3, x=7, y=-6, z=-4
a=4, x=5, y=-3, z=-2
I think they are asking for a value of (a) so that there will be infinitely many solutions

10. Originally Posted by Prove It
Actually it really only requires one of the equations to be a multiple of another.

Because then that would equate to 2 equations in 3 unknowns. To solve this, the solution will be in terms of one of the variables, therefore infinitely many solutions...
No, it doesn't require one of the equations to be a multiple of the other- that would be sufficient but is not necessary.

For example, if x+ y+ z= 0, x- y+ z= 0, x+ z= 0, no equation is a multiple of either of the other two, but y= 0, z= -x would be a solution for any x.

For more than two equations, it is not a matter of being a "multiple" but of being a "linear combination". Here, the third equation is 1/2 the sum of the first two.

11. Originally Posted by Wilmer
3 solutions when a <> 0:
a=1, x=-1, y=6, z=4
a=3, x=7, y=-6, z=-4
a=4, x=5, y=-3, z=-2
Were you thinking that integer solutions were required?

scubasteve94, one method doing this is to try to solve the equations and see what happens. It will eventually reduce to a fraction with a in both numerator and denominator. If the denominator is not 0, there exist a unique solution. If the denominator is 0 and the numerator is not, there exist no solution. If both numerator and denominator are 0, there exist an infinite number of solutions.

If you know about "matrices" and "determinants", the easiest way to find the denominator of that fraction is to take the determinant of the coefficient matrix. That turns out to be a fairly simple quadratic in a and it is easy to set it equal to 0 and solve for two values of a. Then you need to decide if those give no solutions or an infinite number of solutions.