# Homework Help: Given A^-1 find A

1. Oct 11, 2009

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

Given

$$A^{-1} = \left[\begin{array} {cc} 2&-1\\ 3&5 \end{array}\right]$$

find A.

The only thing that we have learned regarding inverses so far is that for a 2x2 matrix, the inverse is given by

$$A^{-1} = \frac{1}{ad-bc} \left[\begin{array} {cc} d&-b\\ -c&a \end{array}\right]$$

So I set corresponding entries equal to each other and got the following 4 EQs:

$$a = \frac12-\frac{bc}{d}$$

$$c = \frac{ad}{b}-1$$

$$b = \frac13+\frac{ad}{b}$$

$$d = \frac15 +\frac{ad}{c}$$

So my question is, what is the best way to solve these? I have been trying substitution, but I guess I really don't know how to use that method with so many EQs?

I feel like I am just running in circles here! I took the 1st EQ and plugged the 2nd in for 'c' then I plugged the 3rd in for 'b' and the 4th for 'd.'

But that did not eliminate anything. I still have all 4 vars in the final EQ.

I need something more systematic. I am not sure which moves are legal here. How can eliminate something here?

2. Oct 11, 2009

### Pengwuino

No no, for 2x2, a,b,c,d are the entries in the actual matrix A. a = 2, b = -1, c = 3, d = 5.

3. Oct 11, 2009

Sorry. I don't follow. I know that a,b,c,d are the entries in A. But I am given the entries in A-1. So I need to work backwards, right?

Thanks.

EDIT: I think that I see what you mean. I did that; I am just showing the final forms of my EQS.

For example the 1st EQ comes from d/(ad-bc) = 2

4. Oct 11, 2009

### Pengwuino

Oh wait! I didnt see you were given the inverse and need to find A. Hmm let me think about this.

5. Oct 11, 2009

### Dick

Uh, (A^(-1))^(-1)=A, right? Just invert A^(-1) to find A.

6. Oct 11, 2009

Oh. I guess that would be a lot easier!

But while you are here: do you see what I mean about the endless loop of substitutions if I were to use that method?

Is there a better way to solve this system? I thought about using Gaussian Elim, but for some reason my gut is against it.

7. Oct 11, 2009

### Dick

No, I don't see any problems with endless loops of substitution. You seem to be using the same a,b,c,d notation for the entries of A as for A^(-1), if A^(-1)_11 is 'a' then A_11 probably shouldn't also be denoted by 'a'. Just invert A^(-1) to get A. Period. That's an order. In the world of invertible matrices there is no distinction between "matrices" and "inverse matrices". They all use the same formulas.

Last edited: Oct 11, 2009
8. Oct 11, 2009

Okay!! I will! BUT ( I don't expect you to read this, 'cause I know how that goes) I don't think that I abused any notation here.

If

$$A = \left[\begin{array} {cc} a&b\\ c&d \end{array}\right]$$

then,

$$A^{-1} = \frac{1}{ad-bc} \left[\begin{array} {cc} d&-b\\ -c&a \end{array}\right] =\left[\begin{array} {cc} 2&-1\\ 3&5 \end{array}\right]$$

Set corresponding entries equal and you get for example

$$\frac{1}{ad-bc}*d=2\Rightarrow a = \frac12-\frac{bc}{d}$$

which is what I got in the first post so and so forth.

So I believe that my EQs are correct and therefore I do see a problem with nasty substitutions.

Maybe you are a substitution wizard. I am not.

9. Oct 11, 2009

### Hurkyl

Staff Emeritus
The problem is that you never eliminated any of the variables. You eliminated the 'c' from the first equation by using the second equation to make a substitution. But because you did not eliminate 'c' from the third and fourth equations, it reappeared later in your calculation.

10. Oct 11, 2009

Exactly! SO can I use Gaussian elimination? I just feel weird about it since I have a's times b's and so and so forth.

That's like having x1*x2 in GE and I have yet to come upon that. Is GE still a valid solution for problems involving variables with other variables as coefficients?

Thanks

11. Oct 11, 2009

### Hurkyl

Staff Emeritus
It's not Gaussian elimination.

But yes, you can solve this system of equations by systematically eliminating variables through substitution.

This is a slow and painful way of finding the inverse of A-1, but you can do it.

12. Oct 11, 2009

Painful. Yes, I have noticed. So would it be safe to say that if I choose to start with the 1st EQ:

$$a = \frac12-\frac{bc}{d}$$

and then start substituting EQS 2,3 & 4, I can use 2,3, & 4 as many times as needed to eliminate all of the variables?

I just cannot reuse EQ 1 again.

Or is that not a safe assumption?

(I know that Dick's way is the "best" way. But sometimes I like to do it the 'hard' way. I learn more that way)

Last edited: Oct 11, 2009
13. Oct 11, 2009

### Hurkyl

Staff Emeritus
Well, I suspect you have the right idea -- but can't be sure. (it's easy enough for you to test yourself )

The "can't reuse" bit isn't that it's mathematically unsound to do so -- it's that it merely acts contrary to the way you're simplifying the system of equations.

14. Oct 11, 2009

### Hurkyl

Staff Emeritus
Incidentally, Gaussian elimination can be used to find the inverse of a matrix T. Presumably you've seen the algorithm in your book? It involves creating the augmented matrix
[T | I]​
and then row reducing.

15. Oct 11, 2009

easy yes. but still painfully time consuming.

Yeah. I just know that in order to eliminate a variable, the equations have to be linearly independent of one another.

So if the whole reusing of EQs gets really confusing for me.

My original approach was the following:

1. Use EQ 1 which is solved for 'a'
2. plug EQ 2 into 1 where ever I see 'c'
3. plug EQ 3 into the resultant of the above where ever I see 'b'
4. plug EQ 4 into the resultant of the above where ever I see 'd'

rinse and repeat steps 2 - 4 until all variables are gone except for 'a.'

It just got to the point, though, that I wasn't sure if that would actually ever happen!

Will it? Or could it be that there is a unique sequence in which I would have to substitute the EQs in order to successfully solve for 'a' alone?

Not there yet. But I peaked ahead a little and that does look familiar.
Thanks!!!

16. Oct 11, 2009

### Hurkyl

Staff Emeritus
And the problem you saw is that when you eliminated 'b' from equation 1, you used a substitution that introduced 'c', which undid your earlier work to eliminate 'b' from equation 1.

What you need, then, is a substitution that eliminates 'b', but doesn't have any occurrences of 'c'.

17. Oct 12, 2009

### Staff: Mentor

Re Gaussian elimination...
C'mon, this is just a 2 x 2 matrix! How painfully time consuming can that be?

18. Oct 12, 2009

This comment was regarding using substitution
method using the 4 EQs in post #1, not GE.

So if you would really like to know, give it a try

As an aside, even if GE were employed, it still wouldn't be
a 2x2 matrix as you say. I think maybe you misunderstood

19. Oct 12, 2009

### Staff: Mentor

My comment pertained to the 2x2 matix in post #1, and using GE to solve the 2x4 matrix system [A-1|I] to get [I|A].

20. Oct 12, 2009