Linear Algebra: given adj(A) find A

[mbrmbrg]

[SOLVED] Linear Algebra: given adj(A) find A

Homework Statement

If adj\mathbb A = \left(\begin{array}{ccc}1&0&1\\1&-1&0\\0&2&1\end{array}\right), find A. Briefly justify your algorithm.

Homework Equations

adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})

The Attempt at a Solution

adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})
invert both sides to get:
(adj\mathbb A)^{-1} = [(det\mathbb A)(\mathbb A^{-1})]^{-1}
(adj\mathbb A)^{-1} = (\mathbb A^{-1})^{-1}(det\mathbb A)^{-1}
(adj\mathbb A)^{-1} = (\mathbb A)(\frac{1}{det\mathbb A})
\mathbb A = (det\mathbb A)(adj\mathbb A)^{-1}

My, isn't that nice.

I computed (adj\mathbb A)^{-1}, and found it to be
\left(\begin{array}{ccc}-1&2&1\\-1&1&1\\2&-2&-1\end{array}\right)
But I have no clue how to find detA, so I'm stuck with one equation:
\mathbb A = (det\mathbb A)(adj\mathbb A)^{-1}
and two unknowns:
\mathbb A and det\mathbb A

 

[Hurkyl]

I'm stuck with one equation:
\mathbb A = (det\mathbb A)(adj\mathbb A)^{-1}
and two unknowns:
\mathbb A and det\mathbb A

You have an equation involving A, and you need an equation involving det A... that should suggest something...

 

[HallsofIvy]

Are you aware that "adjoint" is a "dual" property?
that is, that the "adjoint of the adjoint of A" is A itself. You are given the adjoint of A and are asked to find A- just find the adjoint of the matrix you are given.

 

[mbrmbrg]

Hurkyl, did you mean for me to use HOI's formula, or did you have something else in mind? (and no, I'm not trying to mooch the answer )

Thanks, HallsofIvy! No, I had no idea that adj(adjA)=A. Did I ever know that...? Nope; can I derive it?
:much kerfuffle, then gives up and looks through notes:
Well, fancy that! I have played with adj(adjA) before, and found that it equals (det\mathbb A)^{n-2}\mathbb A (if you'd like to see the whole derivation, I'll type it up, but I did it a very long way and then we proved it differently in class, and got the same result)
Playing with it again (using adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})), I can't get adj(adjA)=A. Can it be derived using only that formula?
Thanks!

 

[HallsofIvy]

What exactly is your definition of adj(A)? (There are several equivalent definitions.) How you would prove that depends strongly on your definition of adjoint.

 

[mbrmbrg]

We defined adj(A) as the transpose of the cofactor matrix of A.

 

[mbrmbrg]

OK, I asked my professor about adj(adjA), and he said it only equals A in special cases. The general formula is adj(adj\mathbb A)=(det\mathbb A)^{n-2}\mathbb A, so you can be sure that it equals A when A is a 2x2 matrix.

Right, so basically, what I'm asking is: is there another method of finding A given adjA?

Thanks!

 

[Hurkyl]

OK, I asked my professor about adj(adjA), and he said it only equals A in special cases. The general formula is adj(adj\mathbb A)=(det\mathbb A)^{n-2}\mathbb A, so you can be sure that it equals A when A is a 2x2 matrix.

There are other times you can solve that equation for A...

 

[mbrmbrg]

Hurkyl, I so have not been ignoring your help.
I just spent a couple of days randomly interrupting conversations with, "Hang on! I think I got it!" only to find that I hadn't got it, after all.
But now... da dum!
I was playing with my calculator, and I accidentally found the numeric value of det(adjA).
One thing led to another, and now: BEHOLD! (where's that emoticon with a brass brand when you need one?)

Given that for my matrix, det(adjA)=1.

I happen to know (well, my notes know it, but I could theoretically re-derive it, right?) that det(adj\mathbb A)=(det\mathbb A)^{n-1}.

So... 1=(det\mathbb A)^{3-1}, or more simply det A=\pm 1

At last, I can use that equation that I'm in love with, and say
\mathbb A^{-1}=\frac{adj\mathbb A}{det\mathbb A}

\mathbb A=(adj\mathbb A)^{-1}det\mathbb A

\mathbb A=(adj\mathbb A)^{-1}(\pm 1)

\mathbb A=(\pm1)\left(\begin{array}{ccc}-1&2&1\\-1&1&1\\2&-2&-1\end{array}\right)

Which would explain why my professor gave me so much credit on the exam for finding the inverse of adjA and mumbling stuff.

Yay!

Thanks, people!