Determinant Properties of 3x3 Matrices | Linear Algebra Homework

[nicknaq]

Homework Statement

Let A be a 3x3 matrix with determinant 5. Then det(adj(A^T))=____, det(adj(A^−1))=____ and det(adj(7A))=____.

Homework Equations

Well, I know that the the adjoint is the transpose of the matrix of cofactors.
Also, these may be useful:
A^-1=adj(A)/det(A)
Aadj(A)=det(A)I
adj(A^T)=(adj(A))^T
A*adj(A)=det(A)*I

The Attempt at a Solution

There's not much process involved in the questions, so I haven't really had an attempt. Thanks for the help.

 

[CompuChip]

Why don't you try it out?
Just construct some (easy) 3 x 3 matrices with determinant 5, like
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{pmatrix}
or
\begin{pmatrix} 3 & -1 & 2 \\ 0 & 5 & 0 \\ 1 & 0 & 1 \end{pmatrix}

(By the way, by "adj(A)" do you mean the adjoint, i.e. conjugate transpose?)

More general hint: det(AB) = det(A) det(B) - this combines nicely with some of the identities you quoted.

 

[nicknaq]

Why don't you try it out?
Just construct some (easy) 3 x 3 matrices with determinant 5, like
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{pmatrix}
or
\begin{pmatrix} 3 & -1 & 2 \\ 0 & 5 & 0 \\ 1 & 0 & 1 \end{pmatrix}

(By the way, by "adj(A)" do you mean the adjoint, i.e. conjugate transpose?)

More general hint: det(AB) = det(A) det(B) - this combines nicely with some of the identities you quoted.

Yes, that's what I mean by adj.

Just wondering, how did you come up with the second matrix? The first one is obvious, of course.

Now I'll start trying to solve this. I'll ask again if I don't get it. Thanks!

 

[nicknaq]

Hey CompuChip,

Just to follow up, I got them all right.
Thanks!

 

[CompuChip]

Great.
It may be useful to simply remember some identities for determinants, like
det(A^-1) = 1/det(A)
det(A^T) = det(A)
det(A*) = det(A)* [with x* the complex conjugate of x]

Then you can easily work out things like det(adj(A^T)): adj(A^T) is ((A^T)^T)^* = A^* so the determinant is det(A)* = 5,
etc.