For which a is the mtrix nonsingular

  • #1
148
1

Homework Statement


For which a [tex]\in[/tex] C is the 3x3 matrix:

[tex]\[\left[ \begin{matrix}
a & 2 & 0 \\
0 & -1 & 2 \\
1 & 0 & a \\
\end{matrix} \right]\]
[/tex]
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 [tex]\neq[/tex] (+-)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
 

Answers and Replies

  • #2
35,125
6,872

Homework Statement


For which a [tex]\in[/tex] C is the 3x3 matrix:

[tex]\[\left[ \begin{matrix}
a & 2 & 0 \\
0 & -1 & 2 \\
1 & 0 & a \\
\end{matrix} \right]\]
[/tex]
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 [tex]\neq[/tex] (+-)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
 
  • #3
148
1
Hmmm, but I think I'm not allowed to use determinants yet. We haven't "learned" it yet, so...
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
964
Well, then, go back to your first idea but keep Mark44's caution in mind: for what values of a does
[tex] \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}[/tex]
for x, y, and z NOT all 0?
 
  • #5
148
1
Well, then, go back to your first idea but keep Mark44's caution in mind: for what values of a does
[tex] \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}[/tex]
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.
 
  • #6
35,125
6,872
Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.
 
  • #7
148
1
Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.

Ok, great, thank you :)
 

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