Let $A$, $B$, and $C$ be $n \times n$ matrices such that all of the following conditions hold:

Question

Let $A$, $B$, and $C$ be $n \times n$ matrices such that all of the following conditions hold: a) $\det(A) \leq 0$

b) $A^{2}=I$ (identity matrix)

c)$ AB^{3}=I$

d) $AB=B^{3}A$

e) $ACA=BC$

Questions:

1) $\det(A)= ?? $

2) Find $(A+B)^{-1}$ in terms of the given matrices, or prove it does not exist.

3) Find $C^{-1}$ in terms of the given matrices, or prove it does not exist.

Answer 1

For question 1, since $\det(A)^2 = \det(A^2) = \det(I) = 1$, $\det(A) = -1$ by condition (a).

Multiply equation (b) by $A$ to get $A = A^2B^3 = B^3$. Substituting into equation (c) gives $AB = B^3A = A^2 = I$, multiplying by $A$ gives $A = A^2B = B$. Now we can solve question 2: $(A+B)^{-1} = (2A)^{-1} = \frac{1}{2}A^{-1}$ ($A$ is invertible since $\det(A) \neq 0$. Multiplying equation (b) by $A^{-1}$ gives $A = A^{-1}$ so the answer can be further simplified to $\frac{1}{2}A$.

Equation (e) can be written as $ACA = AC$, so we have $CA = C$. Suppose $C$ were invertible. Then multiplying by $C^{-1}$ gives $A = I$, which is a contradiction as $\det{A} = -1$. Hence $C$ is not invertible, answering question 3.

Answer 2

Hints:

($1$)

$$\det(A^2)=\det(A)^2=1$$

You are told that $\det(A) \leq 0$, can you conclude what is $\det(A)$?

($2$) One possible way is to prove that $A=B$. (Credit: @JeffreyDawson )

($3$) Take determinant of the last equation and cancel out similar terms. You should be able to conclude that $\det(C)=0$

Edit: Proving $A=B$.

Comparing (b) and (c), we conclude that $A=B^3$ since they are both inverse of $A$ and inverse is unique. Substitute this into (d), we have $AB=A^2$ which leads to $A=B$ since $A$ is invertible.