For which a is the mtrix nonsingular

  • Thread starter Denver Dang
  • Start date
In summary, the 3x3 matrix is nonsingular for a = 2 or a = -2. This can be shown by taking the determinant or by substituting values of a into the matrix equation and solving for x, y, and z.
  • #1
Denver Dang
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
 
Physics news on Phys.org
  • #2
Denver Dang said:

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.
Denver Dang said:
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
Hmmm, but I think I'm not allowed to use determinants yet. We haven't "learned" it yet, so...
 
  • #4
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
HallsofIvy said:
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
Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.
 
  • #7
Mark44 said:
Yes, those are correct. And you can check these by substituting into your original matrix to show that Ax = 0.

Ok, great, thank you :)
 

FAQ: For which a is the mtrix nonsingular

What does it mean for a matrix to be nonsingular?

A nonsingular matrix is a square matrix whose determinant is non-zero. This indicates that the matrix is invertible, meaning it has an inverse matrix that can be multiplied with the original matrix to get the identity matrix.

How can I determine if a matrix is nonsingular?

There are a few methods to determine if a matrix is nonsingular. One way is to calculate the determinant of the matrix. If the determinant is non-zero, then the matrix is nonsingular. Another way is to perform row reduction on the matrix. If the resulting matrix has a row of zeros, then the original matrix is singular.

Why is it important for a matrix to be nonsingular?

Nonsingular matrices are useful in solving systems of linear equations and in many other applications in mathematics, physics, and engineering. They also have an inverse matrix, which allows for efficient computation of solutions to equations involving the matrix.

Can a matrix be both singular and nonsingular?

No, a matrix cannot be both singular and nonsingular. A matrix is either one or the other.

How can I make a singular matrix into a nonsingular matrix?

A singular matrix cannot be made into a nonsingular matrix. However, you can perform operations on the matrix, such as adding or subtracting rows or columns, to transform it into a nonsingular matrix.

Similar threads

Back
Top