Find A Given A^{-1}: Systematic Solution Method

[Saladsamurai]

Homework Statement

Given

A^{-1} =<br /> \left[\begin{array} {cc}<br /> 2&amp;-1\\<br /> 3&amp;5<br /> \end{array}\right]<br />

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}<br /> \left[\begin{array} {cc}<br /> d&amp;-b\\<br /> -c&amp;a<br /> \end{array}\right]<br />

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?

 

[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.

 

[Saladsamurai]

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

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

 

[Pengwuino]

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

 

[Dick]

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

 

[Saladsamurai]

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

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.

 

[Dick]

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.

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.

 

[Saladsamurai]

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.

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 =<br /> \left[\begin{array} {cc}<br /> a&amp;b\\<br /> c&amp;d<br /> \end{array}\right]<br />

then,

<br /> A^{-1} = \frac{1}{ad-bc}<br /> \left[\begin{array} {cc}<br /> d&amp;-b\\<br /> -c&amp;a<br /> \end{array}\right]<br /> =\left[\begin{array} {cc}<br /> 2&amp;-1\\<br /> 3&amp;5<br /> \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.

 

[Hurkyl]

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

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.

 

[Saladsamurai]

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.

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

 

[Hurkyl]

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.

 

[Saladsamurai]

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)

 

[Hurkyl]

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.

 

[Hurkyl]

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.

 

[Saladsamurai]

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

easy yes. but still painfully time consuming.

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.

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?

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.

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

 

[Hurkyl]

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'.

 

[Mark44]

Re Gaussian elimination...

easy yes. but still painfully time consuming.

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

 

[Saladsamurai]

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

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
the thread.

 

[Mark44]

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].

 

[Saladsamurai]

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].

I see. The way that you snipped my quote though made it seem like when I said it was "painfully time consuming" I was speaking solving [A|I]. But I was not.


In post #12 I decided that I would like to further prod the idea of using substitution with the 4 EQs in post #1.

It gets pretty nasty as pointed out by Hurkyl. I am still curious to know if there is only one exact sequence that the EQs must be substituted in in order to reduce the EQs to 1 in 1 unknown. Though I doubt there is only one, I have yet to one. But that is because I went to bed. But now I am up at at 'em

Hopefully I won't spend all afternoon on this silly idea and will actually make some progress in the text

But I like exploring... so if I do, that's cool too.