Let  A be an invertible $$5 \times 5$$​ matrix over a field  F. Suppose that the characteristic polynomials of A and  $$A^{-1}$$​ are the same.
Which of the following is necessarily true?

Correct option is A

Let A be any  $$5 \times 5$$​  matrix with eigenvalues  $$d_1, d_2, d_3, d_4, d_5$$​​
The eigenvalues of  $$A^{-1}$$​ will be:

​$$\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3}, \frac{1}{d_4}, \frac{1}{d_5}.$$​​

Now, since the characteristic polynomial of  A  and  $$A^{-1}$$​  is the same:

​$$A \text{ and } A^{-1} \text{ have the same eigenvalues.}$$​​

Thus:

​$$d_1 = \frac{1}{d_1}, \quad d_2 = \frac{1}{d_2}, \quad \dots, \quad d_5 = \frac{1}{d_5}.$$​​

This implies:

​$$d_1^2 =d_2^2 = \quad \dots, \quad d_5^2 = 1.$$​​

​$$\implies d_1=d_2=d_3=d_4=d_5=\pm1$$

​$$\textbf{Verifying Options:}\\[10pt]\textbf{Option A:}\\[10pt]\det(A)^2 = (\text{product of eigenvalues of } A)^2 = [d_1 d_2 d_3 d_4 d_5]^2.\\[10pt]\text{Since} \ d_i^2 = 1 \ \text{for all } i , det(A)^2 = 1.\\\implies \text{Option A is correct.}$$

$$\textbf{Option B:}\\[10pt]\det(A)^5 = (\det(A))^5 = 1^5 = 1.\\\text{ However, eigenvalues can be +1 or -1, and the product may yield -1}.\\\text{ For example, if det(A) = -1 }$$

then , Option B becomes incorrect.
​$$\textbf{Options C and D :} \\\text{ Trace(A)}=d_1+d_2+\cdots+d_5 \neq \pm1\\\text{So, both options are incorrect.}\\[10pt]$$​​