# 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

Staff Emeritus
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 :)