# Using block multiplication to find the inverse of a matrix?

I'm working through the exercises in my linear algebra book. Normally they show the answers to odd-numbered exercises, however for some reason not for these, so I have no idea what to do.

They want us to verify by block multiplication that the inverse of a matrix, if partitioned as shown, is as claimed (assume that all inverses exist as needed).

Then after this, partition the matrix below, so that you can apply the formula acquired from the above exercise to calculate the inverse.

I honestly have no idea where to start, because there aren't any examples in the book showing how it's done. I found someone else asking a similar question here, but I still don't understand how to solve it.

• Well, you have the formula for the inverse posted above, but in order to use it, you must first partition your matrix into the desired blocks. So, you need one identity matrix in the upper left corner, one in the lower right, and these should cover all of the main diagonal too. The remaining parts of the matrix will then be the "B" and "C" parts in the formula. Can you identify the two necessary identity blocks? (They are both of the same size) – A.Sh Feb 23 '16 at 9:57
• To verify that a matrix $S$ is the inverse of a matrix $T$, all you need to do is to verify that $ST=I$. For the second part, you can easily read off $B$ and $C$ from the matrix, and form the inverse using the formula you just verified. – Mårten W Feb 23 '16 at 9:57
• Regarding solutions to odd exercises btw: In most of the more advanced math books (at the masters level and beyond), you are not even given solutions to begin with. This is intended to provoke development of independency of the student. – A.Sh Feb 23 '16 at 10:02
• Okay, I got the second part. However, I still don't understand how exactly they would want me to verify it in the first part? – amosbastian Feb 23 '16 at 10:18

In your problem you are given a candidate for the inverse. All you need to do is to verify that $$\begin{pmatrix} I & B \\ C & I \end{pmatrix} \begin{pmatrix} (I - BC)^{-1} & -(I - BC)^{-1} B \\ -C (I-BC)^{-1} & I + C(I-BC)^{-1} B \end{pmatrix} = \begin{pmatrix} I & 0\\ 0 & I \end{pmatrix}.$$ There are four blocks to compute. I will do the block $(2,2)$ and leave the three others to you. We have $$C(-(I-BC)^{-1}B) + I(I+C(I-BC)^{-1}B) = I - C(I-BC)^{-1}B + C(I-BC)^{-1}B = I.$$ which is exactly what we want.
It can be that you are missing the following piece of the puzzle. Suppose $A$ and $B$ are matrices such that the product $AB$ is defined and that the two matrices are partitioned conformally, then $$\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} = \begin{pmatrix} A_{11}B_{11} + A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\ A_{21}B_{11} + A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{pmatrix}$$ The term "conformally" merely means that all the product on the right hand side are defined.
The formula itself is the natural extension of how you multiply two matrices square of dimension 2. In reality, it is the just the statement that any inner product $S = x^T y$ can be done gradually, i.e. if $x = (x_1, x_2)^T$ and $y = (y_1,y_2)^T$ are vectors which have been partitioned conformally, i.e. such that $x_1$ and $y_1$ have the same length and similarly for $x_2$ and $y_2$, then $$s = (x_1^Ty_1) + (x_2^T y_2).$$
• Make sure that you get something to eat right away. Afterwards, preserve the factor $(I - BC)^{-1}B$, but eliminate the other pairs of parentheses. – Carl Christian Feb 23 '16 at 10:57
• Here you must strive to create a term of the type $(I - BC)$ which can absorb the "annoying" term $(I-BC)^{-1}$. You do this most efficiently by pulling all terms of the type $(I-BC)^{-1}B$ towards the right of your expression. – Carl Christian Feb 23 '16 at 11:24