Inverse matrix problem

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March 4th 2010, 10:02 AM
temaire

Inverse matrix problem

http://img199.imageshack.us/img199/9336/matho.jpg

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?
March 4th 2010, 12:19 PM
qmech

For simple 2 x 2 matrices, the inverse of
a b
c d

is
d -b
-c a

divided by the determinant: ad-bc.

Why don't you treat your 2x2 partitioned matrix as though it were a simple 2x2? You will be surprised how easily the inverse of A appears.
March 4th 2010, 12:27 PM
HallsofIvy

Quote:

Originally Posted by temaire

http://img199.imageshack.us/img199/9336/matho.jpg

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?

The inverse of the 2 by 2 matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is $\frac{1}{ad- bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$ , as long as $ad- bc\ne 0$ of course.

The whole point of "block matrices" is that the same thing is true of them: the inverse of $\begin{bmatrix}A & B \\ C & D\end{bmatrix}$ is $(AD- BC)^{-1}\begin{bmatrix}D & -B \\ -C & A\end{bmatrix}$

Here, A= B (sorry about that), B= 2I, C= I, and D= 0. AD- BC= 0- 2I so $(AD- BC)^{-1}= -\frac{1}{2}I$ .

The inverse of this matrix is $\begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & -2I \\-I & B\end{bmatrix}$ = $\begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\ 0 & -2 & 3 & 0\end{bmatrix}$ .
March 4th 2010, 02:02 PM
temaire

This really helped guys. But I can't seem to get $\begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\ 0 & -2 & 3 & 0\end{bmatrix}$ in terms of $B, B^{-1}, 0, I$ , as the question requires. Any ideas?

Thanks for the help by the way.