# Homework Help: Inverse matrix with variables

1. Jul 17, 2010

### SpiffyEh

1. The problem statement, all variables and given/known data
A=
[P Q
R S]
Suppose that A and P are non-singular and show that,
A^(-1) =
[x -P^(-1)*Q*W
-W*R*P^(-1) w ]
where W= (S-R*P^(-1)*Q)^(-1) and X= P^(-1)*Q*W*R*P^(-1)

Hint: First, remember that if you are given a candidate for an inverse then you need only check that the appropriate multiplication gives you the identity. Second, you must note that we are working with a matrix whose elements are matrices and when you perform a check you are checking blocks.Thus when you perform the check [A*A^(-1)]_11 you are finding the upper-left block of the product matrix and the result should be matrix and not a scalar. What matrix should you get for this block? What about the rest?

2. Relevant equations

3. The attempt at a solution

I'm completely lost about what to do with this problem. Can someone please show or explain to me what to do

2. Jul 17, 2010

### Staff: Mentor

As it appears on the Web page it's hard to tell that this represents a 2 x 2 matrix of blocks. Here's how it would appear using LaTeX.
$$\text{Show that } A^{-1} = \left[ \begin{array}{cc} X & -P^{-1}QW\\ -WRP^{-1} & W \end{array}\right]$$
This is a good hint. Multiply A*<the matrix above>. If the product is the identity matrix, then the matrix above must be A-1 and you will have shown what needs to be shown. Keep in mind the next hint. That is, the product will be a 2 x 2 matrix of blocks.
I'm getting a block with I and another block with the zero matrix in the right places, but in the other two blocks it's not coming out. It's possible I have a mistake, but before I spend any more time on this you should make sure that you have copied the problem exactly as given. A single wrong sign or anything else wrong could foul things up.

You will need to substitute for X and W, and do some factoring.

Last edited: Jul 17, 2010
3. Jul 17, 2010

### SpiffyEh

X= P^(-1)*Q*W*R*P^(-1)
should actually be
X=P^(-1) + P^(-1)*Q*W*R*P^(-1)
sorry, I completely missed that

4. Jul 17, 2010

### SpiffyEh

I tried getting it with the LaTeX but I can't get it to show right.

A*(A^(-1)) =
[PX-QWRP^(-1) -P(^-1)QWP+QW
RX-SWRP^(-1) -RP^(-1)QW+SW]

I'm trying to subsitute in x and w and see if I can get it

Also, I think I should be getting the answer to
[ I 0
0 I] correct? this would give me an identity matrix as a whole

5. Jul 17, 2010

### SpiffyEh

I got [AA^(-1)]_11 = I by subsituting in x and using P*P^(-1) = I and I*B = B
I got [AA^(-1)]_12 to work out to 0 with the same identities.
I also got [AA^(-1)]_22 to work out to I using A*A^(-1) = I

Now, I'm having problems with [AA^(-1)]_21. I just can't see what to do with it

6. Jul 17, 2010

### Staff: Mentor

You can use my script from post 2. Just click on the expression and another window opens. Copy that and past it into your post, and change it as necessary.
Yes, that's what you should get.

That one should come out to the zero matrix. If you can't get it, post what you have for that multiplication. I don't need to see the other three multiplications because it sounds like you have them figured out.

7. Jul 17, 2010

### SpiffyEh

I have RX - SWRP^(-1)
I subsitutued X in to get RP^(-1) + RP^(-1)QWRP^(-1) - SWRP^(-1)
This is where i'm stuck. I don't see anything that could get me anywhere towards 0. Maybe i'm just looking at it wrong but I really can't come up with anything

8. Jul 17, 2010

### Staff: Mentor

Now substitute for W, which is (S - RP-1Q)-1.
RP-1 + RP-1QWRP-1 - SWRP-1
= RP-1 + RP-1Q(S - RP-1Q)-1RP-1 - S(S - RP-1Q)-1RP-1
= (I + RP-1Q(S - RP-1Q)-1 - S(S - RP-1Q)-1)RP-1
= (I + (RP-1Q - S)(S - RP-1Q)-1)RP-1

It's easy to show that the part inside the outer parentheses simplifies to the zero matrix. The key is realizing that for an invertible matrix B, B*(-B-1) = -I.

9. Jul 17, 2010

### SpiffyEh

oh! I never would have gotten the B*(-B^(-1)) = -I, thats where I thought it wasn't going anywhere. I had another one very similar to this problem and the same property helped me solve it. Thank you so much.

I also have this:
Show that if P,S,A are all non-singular matrices then (S- RP^(-1)Q)^(-1) = S^(-1) + S^(-1)RXQS^(-1).

Can I use the x and w equations I was given for this? I have two equations of X and two for W since this comes after the two similar problems.
The X and W above and I also have
X = (P-QS^(-1)R)^(-1)
W = S^(-1) + S^(-1)RXQS^(-1)

To show it do I just move things around and all that until both sides are equal?
Sorry for all the questions, I'm just really lost and I'm trying to understand it. I don't get any examples from my professor so its hard for me to pick up on things.

10. Jul 17, 2010

### Staff: Mentor

Just show that (S- RP^(-1)Q) * S^(-1) + S^(-1)RXQS^(-1) = I

You'll need to substitute for X, but not W, since W doesn't appear in the equation above.

11. Jul 17, 2010

### SpiffyEh

(S-RP^(-1)Q)*(S^(-1) + S^(-1)RXQS^(-1)) = I
SS^(-1) + SS^(-1)RXQS^(-1) - S^(-1)RP^(-1)Q - RP^(-1)QS^(-1)RXQS^(-1) = I
where SS^(-1) = I. So,
I + RXQS^(-1) - S^(-1)RP^(-1)Q - RP^(-1)QS^(-1)RXQS^(-1) = I
I + RQS^(-1)(P-QS^(-1)R)^(-1) -S^(-1)RP^(-1)Q - RP^(-1)QS^(-1)RQS^(-1)(P-QS^(-1)R)^(-1) = I

I used the
X = (P-QS^(-1)R)^(-1)

From here, I'm lost again. I tried pulling things I could out but that got me no where. I tried pulling out RQS^(-1)(P-QS^(-1)R)^(-1) which didn't help. I don't know why I'm having so much trouble with this part. Should I be plugging in the other value I have for X?

12. Jul 17, 2010

### Staff: Mentor

Your comment in post 9 about two equations for X and two for W bothered me. There should not be two equations for X. If you can scan the page and post it, we should be able to figure out what to use.

Also, in your last post you have this:
You're trying to show that the expression on the left is equal to I, so don't start off with an equation that says they're equal. By doing that you are tacitly assuming the two are equal, but that's what you need to show. Work with the expression on the left until it is simplified to what's on the right.

I'm not having any success with the latest problem, but I don't want to spend any more time on it until it's clear what should be used for X.

One other thing: You're writing stuff like S^(-1) for the inverse, which is OK, but it would be better to use exponents (which is what I've been doing) to make things clearer. It's pretty easy to do this.

Click the Go Advanced button below the text input area. This opens a larger input area with a menu across the top. To make an exponent, click the X2 button and type in the exponent. There's an X2 button that you can use to make subscripts.

13. Jul 17, 2010

### SpiffyEh

I attached the problem. Between problem 2.1 and 2.2 there are some hints but I already posted the only one thats relevant to the problem earlier. It never really says which X to use or anything. Also, if you get a chance could you look at the link below please? I have all my work shown, I'm just unsure if I'm doing it correctly or not. Thank you

#### Attached Files:

• ###### problem.png
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Last edited: Jul 18, 2010
14. Jul 18, 2010

### Staff: Mentor

Problem 2.2 looks to me to be similar to 2.1, the first one you posted. This time, show that
$$A^{-1} = \left[ \begin{array}{cc} X & -XQS^{-1}\\-S^{-1}RX & W \end{array}\right]$$

Use the values given for X and W that are below the problem. That's how I see it.

15. Jul 18, 2010

### SpiffyEh

I got 2.2 to work out just like 2.1. I posted that so that you could see which X to use for 2.3 because it's a little confusing.

16. Jul 18, 2010

### Staff: Mentor

2.3 is actually pretty easy. In 2.1 and 2.1 you are finding A-1 using two different techniques. In 2.3 they are asking you to show that W in the matrix of problem 2.1 equals W in the matrix of problem 2.2. If a matrix (A) has an inverse, the inverse has to be unique, so W in the one representation must equal W in the other representation. I think that's all there is to it.

17. Jul 18, 2010

### SpiffyEh

hmm.. that seems too simple but I can't see how they got from one to another by subsituting in so I guess that will have to do. I mean it makes sense but it just seems to easy.

In 2.4 I'm supposed to check it. Is this where I would plug in the values given and try to subsitute a formula in for X?

18. Jul 18, 2010

### Staff: Mentor

In 2.4, matrix A will be just a regular 2 x 2 matrix that is invertible. If you use the representation of A-1 that is shown in 2.1, use X that is defined for that problem. If you use the representation that is shown in 2.2, use the X that is defined for that problem.

19. Jul 18, 2010

### SpiffyEh

I'm still a little confused by what you mean. Do I do this the same way i did 2.1 and 2.2 then? Could you show me an example please

20. Jul 18, 2010

### Staff: Mentor

Prob 2.1 gives you one representation for A-1, and prob. 2.2 gives you another. W and X are used in each representation, and are defined differently for each problem. If you use the representation for A-1 of 2.1, use the values of W and X defined in that problem.

If you use the representation for A-1 of 2.2, use the other values of W and X that are defined in that problem.

No, you aren't supposed to redo problems 2.1 and 2.2 again - just substitute the given values into one or the other representations of A-1, and then multiply A times A-1.