Homework Help: For which a is the mtrix nonsingular

1. Feb 1, 2010

Denver Dang

1. The problem statement, all variables and given/known data
For which a $$\in$$ C is the 3x3 matrix:

$$$\left[ \begin{matrix} a & 2 & 0 \\ 0 & -1 & 2 \\ 1 & 0 & a \\ \end{matrix} \right]$$$
nonsingular ?

2. Relevant equations
Ax = 0

3. The attempt at a solution
Well, the problem is that I think I know what to do, but I can't remember how it's done :S
If Ax = 0, then it's nonsingular, so that's all I have to show.
And then I multiply the matrix with x ([x1, x2, x3]) I just get the same matrix as I have above, but with zeros in the fourth column.
So if I put in on REF I get that x1 = -a, x2 = a2/2.
And then what ? In my head in seems so easy, but I just can't figure it out :S
What is the a's that's satisfy that the matrix is nonsingular ?
If I use RREF on AB = BA = I, I get that a $$\neq$$ (+-)2. But I can only do that if I have assumed that the matrix, A, is nonsingular, which I have not shown yet.
So, what to do ? :)

Regards

2. Feb 1, 2010

Staff: Mentor

That's not the definition of nonsingularity. The equation above is true for x = 0 for any square matrix. If the matrix is nonsingular, the equation will be true only for x = 0.

Another approach that I think you have forgotten, is that the determinant of a nonsingular (AKA invertible) matix is nonzero. Try taking the determinant and see what value(s) a needs to be so that the matrix is nonsingular.

3. Feb 2, 2010

Denver Dang

Hmmm, but I think I'm not allowed to use determinants yet. We haven't "learned" it yet, so...

4. Feb 2, 2010

HallsofIvy

Well, then, go back to your first idea but keep Mark44's caution in mind: for what values of a does
$$\begin{bmatrix} a & 2 & 0 \\ 0 & -1 & 2 \\ 1 & 0 & a \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}ax+ 2y \\ -y+ 2z \\x+ az\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\\end{bmatrix}$$
for x, y, and z NOT all 0?

5. Feb 2, 2010

Denver Dang

Well, if I'm allowed to just solve the equations but substituting some of the equations into some of the others, I get that: a = 2 or a = -2.

6. Feb 2, 2010

Staff: Mentor

Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.

7. Feb 2, 2010

Denver Dang

Ok, great, thank you :)