For which a is the mtrix nonsingular

[Denver Dang]

Homework Statement

For which a \in C is the 3x3 matrix:

\[\left[ \begin{matrix}<br /> a &amp; 2 &amp; 0 \\<br /> 0 &amp; -1 &amp; 2 \\<br /> 1 &amp; 0 &amp; a \\<br /> \end{matrix} \right]\]<br />
nonsingular ?

Homework Equations

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]

Homework Statement

For which a \in C is the 3x3 matrix:

\[\left[ \begin{matrix}<br /> a &amp; 2 &amp; 0 \\<br /> 0 &amp; -1 &amp; 2 \\<br /> 1 &amp; 0 &amp; a \\<br /> \end{matrix} \right]\]<br />
nonsingular ?

Homework Equations

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

 

[Denver Dang]

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

 

[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 &amp; 2 &amp; 0 \\ 0 &amp; -1 &amp; 2 \\ 1 &amp; 0 &amp; 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?

 

[Denver Dang]

Well, then, go back to your first idea but keep Mark44's caution in mind: for what values of a does
\begin{bmatrix} a &amp; 2 &amp; 0 \\ 0 &amp; -1 &amp; 2 \\ 1 &amp; 0 &amp; 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]

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

 

[Denver Dang]

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

Ok, great, thank you :)