# For which a is the mtrix nonsingular

## Homework Statement

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 ?

Ax = 0

## 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

Mark44
Mentor

## Homework Statement

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 ?

Ax = 0

## 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.
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.
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

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

HallsofIvy
Homework Helper
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?

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?
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.

Mark44
Mentor
Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.

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

Ok, great, thank you :)