Finding possible values of $\det(X)$ when $AX-XA=A$

determinantlinear algebramatrices

Given matrix $
A=\begin{bmatrix}
1 & 1 \\
-1 & -1 \\
\end{bmatrix}
$ such that
$AX-XA=A \tag{1}\label{1}$
where $X$ is a matrix with integer entries of order 2.
We need to find which of the following options are NOT possible for $\det(X)$.

  1. 10
  2. 6
  3. 2
  4. 0

My attempt: I took
$X=\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
$

and on substituting and solving $\eqref{1}$, I got following:

  • $b+c=1$
  • $a+c=b+d$
  • On solving for $b$ and $c$ ,I got $a$ and $d$ must be of opposite parity (even/odd)

How can I further proceed? Do I have to guess the possible entries of X or there is a brief or better way to solve it quickly. thanks.

Best Answer

The conditions you found can be solved for $c$ and $a$ (for example): \begin{align} c &= 1 - b\\ a &= 2b + d - 1 \end{align} Now given the form of $X$ that you chose, it's clear that $\det(X) = ad - bc$, which can be reduced to: \begin{align} ad-bc&= (2b + d - 1)d + b^2 - b\\ &=d^2 + b^2 - (d + b) + 2bd \\ &=(d + b)^2 - (d + b) \\ &= x^2 - x \end{align} Where $x = d + b$. Hence $x^2 - x - \det(X) = 0$. Now substitute in the potential values for $\det(X)$ and see which result in a polynomial with integer roots.

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