# Homework Help: Inverse matrix problem

1. Mar 4, 2010

### Precursor

The problem statement, all variables and given/known data

http://img199.imageshack.us/img199/9336/matho.jpg [Broken]

The attempt at a solution
I don't really know how to go about solving this problem, since it's a partitioned matrix. If I write it out in its complete 4 * 4 form, it will take a long time to reduce it, and I won't be able to get an answer in the form required by the question. So what method of inverse should I use?

Last edited by a moderator: May 4, 2017
2. Mar 4, 2010

### Staff: Mentor

I'm not sure this would work, but I would give it a shot. For a 2 x 2 matrix A =
[a b]
[c d]
, A-1 =
(1/det(A)) *
[d -b]
[-c a]

Maybe you can extend this idea, with |A| being |B|*0 - 2|I|2.

3. Mar 4, 2010

### Hurkyl

Staff Emeritus
Does that matter? Where does this cause problems?

4. Mar 4, 2010

### Precursor

According to this method, the inverse turns out to be:

$$\begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & -2I \\-I & B\end{bmatrix} = \begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\ 0 & -2 & 3 & 0\end{bmatrix}$$

Now how do I get the inverse in terms of $$B, B^{-1}, 0, I$$, as the problem asks?

5. Mar 4, 2010

### Hurkyl

Staff Emeritus
Why did you do arithmetic with the individual components instead of with the blocks?

6. Mar 4, 2010

### Precursor

$$A^{-1} = (AD- BC)^{-1}\begin{bmatrix}D & -B \\ -C & A\end{bmatrix}$$

where $$A= B, B= 2I, C= I, D= 0$$.

I don't know how to continue from this without using the individual components. But why should that matter anyway? If the inverse I posted above is correct, then shouldn't I be able to put it back into partitioned form?

7. Mar 4, 2010

### Hurkyl

Staff Emeritus
I don't understand why you have a problem turning it back into partitioned form and relabeling the blocks. Could you explain more?

But I don't understand why you took it out of partitioned form and wrote everything in terms of their components either.

8. Mar 4, 2010

### Precursor

If I turn it back into partitioned form I get

$$\begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\ 0 & -2 & 3 & 0\end{bmatrix} = \begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & -2I \\-I & B\end{bmatrix}$$

which is not in terms of $$B, B^{-1}, 0, I$$. The one that's missing is $$B^{-1}$$.

9. Mar 4, 2010

### Staff: Mentor

You don't want (AD - BC)-1, just 1/(|A||D| - |B||C|). This works out to be an ordinary common fraction. I calculated A-1 as I described. As a check, I calculated AA-1 and got a block matrix with I blocks on the main diagonal and 0 blocks elsewhere. There's no need to calculate B-1.

10. Mar 4, 2010

### Precursor

So was your final answer a partitioned matrix in terms of $$B, B^{-1}, 0, I$$, or was it like mine?

11. Mar 4, 2010

### Hurkyl

Staff Emeritus
You know how to check if you have the inverse right?

A priori, there's no reason that B-1 must appear in your answer. "In terms of" means that you should limit yourself only* to those things -- it does not imply all of those things should appear.

P.S. you should probably multiply out your two matrices....

*: There are usually some other things implicitly included -- in this case probably things like scalars, products, and so forth

12. Mar 4, 2010

### Hurkyl

Staff Emeritus
For the record, this problem has a special form -- all of the blocks commute with each other. If it weren't for that fact, the quoted formula for 2x2 inverses would not work here.

If it works, that is -- I haven't checked that it actually does. And I think you are right that you want (AD-BC)-1 rather than the determinant.

13. Mar 4, 2010

### Staff: Mentor

I've checked it out in two ways: working with the blocks, and substituting the blocks back in and multiplying the expanded forms of A and A^(-1). I get the identity matrix both ways.

I realize that the inverse formula I used works only for 2 x 2 matrices. Apparently it also works if you have 2 x 2 block matrices where the blocks are each 2 x 2.

My formula for the determinant is, I believe, correct. It lead to the answer I expected, but I haven't proven that it is correct. For the block matrix A, the "determinant" I came up with was 1/(|A||D| - |B||C|), which works out to -1/2.

14. Mar 4, 2010

### Hurkyl

Staff Emeritus
And those blocks commute with each other!

Commuting is a key point -- it means that the block matrix arithmetic behaves sufficiently similar to ordinary arithmetic that you wind up getting the same formulas when all is said and done.

When the blocks don't commute, things get messier. You can still find the inverse via (block) Gaussian elimination, of course.