# Homework Help: Linear Algebra: given adj(A) find A

1. Dec 17, 2007

### mbrmbrg

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

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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 $$det\mathbb A$$

Last edited: Dec 17, 2007
2. Dec 17, 2007

### Hurkyl

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

3. Dec 18, 2007

### 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.

4. Dec 18, 2007

### 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!

5. Dec 19, 2007

### 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.

6. Dec 19, 2007

### mbrmbrg

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

7. Dec 31, 2007

### 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!

8. Dec 31, 2007

### Hurkyl

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

9. Jan 3, 2008

### 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!

Last edited: Jan 3, 2008