What if matrix AB = BA = -I.

Given that AB = BA = -I, where A, B, and I are $n\times n$ size matrices.
To solve this problem, we can start by taking the determinant of both sides of the equation AB = -I. Using the property that $\det \left(AB\right)=\det \left(A\right)\det \left(B\right)$, we have:

$\det \left(A\right)\det \left(B\right)=\det \left(AB\right)=\det (-I)={(-1)}^n$

where n is the size of the matrices.

Since $\det (-I)={(-1)}^n$, we know that n must be even in order for $\det (-I)$ to be positive. Therefore, we can assume that n is even and that $\det \left(A\right)\text{and}\det \left(B\right)$ are nonzero.

Next, we can take the determinant of both sides of the equation BA = -I:

$\det \left(B\right)\det \left(A\right)=\det \left(BA\right)=\det (-I)={(-1)}^n$

Since $\det \left(A\right)\text{and}\det \left(B\right)$ are both nonzero, we can divide both sides of the equation $AB=BA=-I$ by $\det \left(A\right)\det \left(B\right)$ to obtain:

$AB=BA=-\frac{1}{{\det \left(A\right)}^{2}I}$

Therefore, the matrices A and B commute with each other and are both invertible. Moreover, the inverse of each matrix is given by:

$A^{-1}=-\frac{1}{{\det \left(A\right)}^{2}B}$
$B^{-1}=-\frac{1}{{\det \left(B\right)}^{2}A}$

To see why these formulas hold, we can compute:

$AB{A}^{-1}=-\frac{1}{{\det \left(A\right)}^{2}I{A}^{-1}}=-\frac{1}{{\det \left(A\right)}^2{A}^{-1}}$
$BA{A}^{-1}=-\frac{1}{{\det \left(A\right)}^{2}B{A}^{-1}}=-\frac{1}{{\det \left(A\right)}^2{A}^{-1}}$

Since A and B commute with each other, we can use the same formulas to compute $A^{-1}\text{and}{B}^{-1}$ in terms of each other. Specifically, we have:

$A^{-1}=-\frac{1}{{\det \left(B\right)}^{2}B}$
$B^{-1}=-\frac{1}{{\det \left(A\right)}^{2}A}$

Therefore, we have found explicit formulas for the inverses of A and B in terms of each other, and we have shown that the matrices A and B commute with each other.