Question

Now suppose that A and B are arbitrary n  n matrices such that A and 2A + BT are invertible.
Using general properties of matrix operations show
that (A-1BT + 2I)-1 = (2A + BT)-1A. Show all the steps.

Answer 1

Let us consider the properties of the inverse matrices. Suppose that $M_1$ and $M_2$ are invertible matrices. Then $(M_1M_2)^{-1} = M_2^{-1}M_1^{-1}.$ (1)

We should show that

$(A^{-1}B^T+2I)^{-1} = (2A+B^T)^{-1}A.$ (2)

The task will be done if we show the analogous equality for the inverted left and right hand-sides of (2):

$(A^{-1}B^T+2I) =\big( (2A+B^T)^{-1}A\big)^{-1}$ .

According to (1) we may rewrite

$\big( (2A+B^T)^{-1}A\big)^{-1} = A^{-1}\big( (2A+B^T)^{-1}\big)^{-1} = A^{-1} (2A+B^T).$

Next we will use the distributive law for matrix multiplication and the commutative law for matrix addition:

$A^{-1} (2A+B^T) = 2A^{-1}A + A^{-1}B^T = 2I + A^{-1}B^T.$

So we obtained the equality

$\big( (2A+B^T)^{-1}A\big)^{-1}=(A^{-1}B^T+2I) .$

We get the analogous equality for the inverse matrices:

$(2A+B^T)^{-1}A=(A^{-1}B^T+2I)^{-1} .$