# Invertible Matrices

#### Yankel

##### Active member
I have one more question, I have two matrices A and B, both squared and with the same order. And I have a scalar, a not equal to zero.

why are these statements not correct ?

1. If A and B are invertible, then
$$a\cdot (B^{-1}A^{-1}B)^{^{t}}$$
is not necessarily invertible

2. If A and B are invertible, then
$$a\cdot (A+B)$$
is not necessarily invertible, but if it is, it's inverse is the matrix
$$\frac{1}{a}\cdot (A^{-1}+B^{-1})$$

Thanks...

#### Chris L T521

##### Well-known member
Staff member
I have one more question, I have two matrices A and B, both squared and with the same order. And I have a scalar, a not equal to zero.

why are these statements not correct ?

1. If A and B are invertible, then
$$a\cdot (B^{-1}A^{-1}B)^{^{t}}$$
is not necessarily invertible

2. If A and B are invertible, then
$$a\cdot (A+B)$$
is not necessarily invertible, but if it is, it's inverse is the matrix
$$\frac{1}{a}\cdot (A^{-1}+B^{-1})$$

Thanks...
For the first one, I'm going to assume WLOG that $A$ and $B$ are $n\times n$ matrices. if $A$ and $B$ are invertible, then $\det(A)\neq 0$ and $\det(B)\neq 0$. Furthermore, if $a\neq 0$ is a scalar, then
$\det (a\cdot(B^{-1}A^{-1}B)^t)=a^n\det((B^{-1}A^{-1}B)^t) = a^n \det(B^{-1}A^{-1}B)=\frac{a^n\det(B)}{\det(B)\det(A)}=\frac{a^n}{\det(A)}.$
This tells me that we'll always have $a\cdot(B^{-1}A^{-1}B)^t$ invertible provided that $A$ and $B$ are invertible with a scalar $a\neq 0$.

For the second one, it's true that if $A$ and $B$ are invertible, it may so happen that $a\cdot (A+B)$ isn't invertible. For instance, take $A=I$, $B=-I$ and $a=1$, where $I$ is the identity matrix. Then both matrices are invertible, but $A+B=O$, where $O$ is the zero matrix, which is clearly not invertible.

If $a\cdot(A+B)$ was invertible, then it wouldn't have that form for the inverse. We can see this is the case since
$\left(a\cdot(A+B)\right)\left(\frac{1}{a}\cdot(A^{-1}+B^{-1})\right)=AB^{-1}+BA^{-1}+2I\neq I$

I hope this helps!