# Matrix problem - number of solutions to a linear system of equation.

• September 14th 2011, 02:25 PM
HelpMe111
Matrix problem - number of solutions to a linear system of equation.
Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions.

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

I made an augmented matrix.

I tried doing that elimination thing. I ended up with (-a^2 + 2) = a - 2 for the final line.
In order for the system to be inconsistent, -(a^2 + 2) has to equate to 0, so I solved for a and got rad 2.
So it's rad 2 for no solution.
A =/= radical 2 for unique solution (I think).

I'm not sure about infinite solutions. Any hints would be greatly appreciated.
• September 14th 2011, 04:10 PM
Ackbeet
Re: Matrix problem. Help please.
I think you'll find that you get no solutions when $a=\pm\sqrt{2}.$ For any other value of a, the determinant is nonzero, and hence you get one solution. Thus, for no values of a can you get infinite solutions.
• September 14th 2011, 06:43 PM
HelpMe111
Re: Matrix problem. Help please.
Quote:

Originally Posted by Ackbeet
I think you'll find that you get no solutions when $a=\pm\sqrt{2}.$ For any other value of a, the determinant is nonzero, and hence you get one solution. Thus, for no values of a can you get infinite solutions.

Thanks. I forgot about - rad 2.
I'm still not convinced about replacing a with 0. I'll plug it in and see what happens to the matrix.