# 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

Staff Emeritus
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

Staff Emeritus
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