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Let $X=\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]$, Let $Y$ be a $2 \times 2$ real matrix satisfying the condition $X Y=Y X$. Then the smallest possible value of $\operatorname{det}(Y)$ is
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$X=\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right], Y_{2 \times 2}=?, X Y=Y X$
Let $Y=\left[\begin{array}{ll}x & y \\ z & t\end{array}\right]$ such that
$X Y=Y Z$
$\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}x & y \\ z & t\end{array}\right]=\left[\begin{array}{cc}x & y \\ z & t\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}x-z & y-t \\ x+z & y+t\end{array}\right]=\left[\begin{array}{ll}x+y & -x+y \\ z+t & -z+t\end{array}\right]$
$\Rightarrow \quad x+z=z+t,$
$y-t=-x+y, x-z=x+y$
$\Rightarrow \quad x=t \Rightarrow y=-z$
$Y=\left[\begin{array}{cc}t & -z \\ z & t\end{array}\right]$
$\therefore|Y|=t^2+z^2$ which is always non-negative for
$t, Z \in R$.
$\therefore$ Smallest value of $|Y|=0$
Let $Y=\left[\begin{array}{ll}x & y \\ z & t\end{array}\right]$ such that
$X Y=Y Z$
$\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}x & y \\ z & t\end{array}\right]=\left[\begin{array}{cc}x & y \\ z & t\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}x-z & y-t \\ x+z & y+t\end{array}\right]=\left[\begin{array}{ll}x+y & -x+y \\ z+t & -z+t\end{array}\right]$
$\Rightarrow \quad x+z=z+t,$
$y-t=-x+y, x-z=x+y$
$\Rightarrow \quad x=t \Rightarrow y=-z$
$Y=\left[\begin{array}{cc}t & -z \\ z & t\end{array}\right]$
$\therefore|Y|=t^2+z^2$ which is always non-negative for
$t, Z \in R$.
$\therefore$ Smallest value of $|Y|=0$
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