Assume that $A$ is a $2 \times 2$ matrix with integer entries. If $B$ is the inverse of $A$ and has integer entries, then choose the correct statement

Question

Assume that $A$ is a $2 \times 2$ matrix with integer entries. If $B$ is the inverse of $A$ and has integer entries, then choose the correct statement:

1. None of the given options

2. $A^2 = I$, where $I$ is the corresponding identity matrix

3. $\det(B) = -2$

4. $A = 2I$ or $A = B$, where $I$ is the corresponding identity matrix

5. $\det(B) = 0$

The options are confusing, I am trying to answer the question by eliminating the options.

Option 5 is incorrect since the inverse of a matrix only exists when the determinant of the original matrix is non-zero. If the determinant of 𝐵 were zero, then 𝐵 would not be invertible and could not be the inverse of 𝐴.

Option 4 is incorrect. If for example
$$
A = \begin{vmatrix}
1 & 2 \\
0 & 1
\end{vmatrix}
$$
Then inverse of this matrix is:
$$
B = \begin{vmatrix}
1 & -2 \\
0 & 1
\end{vmatrix}
$$

Both $A$ and $B$ have integer entries, and $B$ is the inverse of $A$, but $A \neq 2I$ and $A \neq B$.

Option 3 unsure. It could be possible that B is $2 \times 2$ matrix with a determinant of -2 :
$$
B = \begin{vmatrix}
2 & 0 \\
-1 & -1
\end{vmatrix}
$$
which would mean that :
$$
A = \begin{vmatrix}
\frac{1}{2} & 0 \\
\frac{-1}{2} & 1
\end{vmatrix}
$$

Option 2 unsure. It could be possible that A is $2 \times 2$ matrix with a determinant of 0:
$$
A = \begin{vmatrix}
1 & 2 \\
1 & 2
\end{vmatrix}
$$

Answer 1

$\det(A^{-1}) = 1/\det(A)$, so (3) is never true.

Your example for (4) also rules out (2).

What is true is that $\det(A) = \pm 1$.