An nxn homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions.
What does that tell you?
Thanks for the tips on trivial solutions and non trivial solutions... I guess I just didnt understand the difference. However I still am unsure how to go about solving this. can someone make a step by step. I would really really appreciate the extra help on this one.
Please look at this webpage. As you can see, $6$ is the correct to your problem.
Now see why that is the case. Look Here. I replaced $a$ with $6$. Then row reduced the the result. Now we can see that with $a=6$ we have:
$x=0,~y=1,\&,z=1$ or $(0,1,1)$ is a non-trivial solution to the question.
But as pointed out, there are infinitely many non-trivial solutions.
Thanks Plato! So I took the determinant of the matrix and set it equal to zero, then solved for a and got 6. So from the comments above, this would indicate that 6 is a non-trivial solution that will not have its factors equaling all zero. As you stated, there are infinietely many non-trivial solutions, but what could be some others. I have gone to that website and typed many other numbers for a and they all make the factors equal zero except 6.
Close. What it means is that 6 is the only value for a that allows the system of linear equations to have a non-trivial solutions. But 6 is not the non-trivial solution. Plato gave you an example of a non-trivial solution that could be found if a=6. Namely, x=0, y=1, z=1. 6 is the value for a that allows the system to have non-trivial solutions. Do you see the distinction? A solution to the system would be values for x,y, and z that make the system true. A value for $a$ is not the same as a solution to the system of linear equations, trivial, non-trivial, or otherwise. a=6 is the solution to the problem that was asked, though.
One obvious method should be to try to solve the equations and see what happens! The three equations are 3x+ 2y- 2z= 0, 9x+ ay- 6z= 0, 4x+ 2y- 2z= 0. Seeing that "2y- 2z" in both first and third equations, I would immediately subtract the first equation from the third to get x= 0. With x= 0, we have 2y- 2z= 0 and ay- 6z= 0. We can subtract 3 times the first of those equations from the other to eliminate get (a- 6)y= 0. As long as a is not 6, we can divide both sides by a- 6 to get y= 0 and then -2z= 0 and -6z= 0 so z= 0. That, x= y= z= 0, is the "trivial solution". In order that we not have that trivial solution, we have to have a= 6 so that we cannot solve for y: with a= 6, (a- 6)y= 0y= 0 for all y so that z= y is true for all z. Any (0, y, y) is a solution.