Linear Equations: Determine the values of a in this system of linear equations

Hi,

I have a question regarding linear algebra which I am unsure of how to approach. The question says:

Determine the values of a for which the following system of equations has (a) no solutions, (b) exactly one solution, and (c) infinitely many solutions.

x + y + 7z = -7

2x + 3y + 17z = -16

x + 2y + (a^2+1)z = 3a

So I assume that we would use some form of Gaussian to figure out the solutions because thats what have have been doing in the course so far, but everything I have tried so far has failed!!!

I did get a solution of a=3 for no solutions, but I'm not actually sure if this is correct (what I did was set a^2-6 equal to 3 after using row reduction). Anyway any help I could get would be greatly appreciated or even just any clues as to how I should start this problem.

Thanks in advance!!

Re: Linear Equations: Determine the values of a in this system of linear equations

Gaussian elimination is the right way to go.

Having done that the third equation is

$\displaystyle (a^{2}-9)z=3a+9.$

You have to figure out what this implies for various values of $\displaystyle a.$

Specifically those that make the $\displaystyle z$ coefficient zero.

Re: Linear Equations: Determine the values of a in this system of linear equations

Ohhh thank you very much, i think i might be on the right track now

Re: Linear Equations: Determine the values of a in this system of linear equations

So the solutions I found are:

For no solution: a=3

For infinitely many solutions: a=-3

and

For one solution: a E {R/(-3,3)}

Sorry with that last one I know how I've written it probably isn't the correct way to write it mathematically but what i meant is that a is an element of the Real number system not including -3 and 3 (if you could show me how to write this properly it would be very appreciated as it has been a while since I have done that type of stuff). Are these the correct solutions?

Anyway thank-you very much for the help you have given me so far!