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 were zero, then would not be invertible and could not be the inverse of .

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 $\det(A) = \pm 1$.