Let X = begin{bmatrix} 1 & -1 1 & 1 end{bmat | vedclass

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(A) Given $X = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix}$. Let $Y = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ be a $2 	imes 2$ real matrix su

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(A) Given $X = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix}$. Let $Y = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ be a $2 	imes 2$ real matrix such that $XY = YX$. Calculating $XY$: $XY = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} a-c & b-d \ a+c & b+d \end{bmatrix}$. Calculating $YX$: $YX = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} a+b & -a+b \ c+d & -c+d \end{bmatrix}$. Equating $XY = YX$: $a-c = a+b \implies c = -b$. $b-d = -a+b \implies d = a$. Thus,$Y$ must be of the form $\begin{bmatrix} a & b \ -b & a \end{bmatrix}$. The determinant of $Y$ is $\det(Y) = a^2 - (-b^2) = a^2 + b^2$. Since $a, b \in \mathbb{R}$,the smallest possible value of $a^2 + b^2$ is $0$ (when $a=0$ and $b=0$). Therefore,the smallest possible value of $\det(Y)$ is $0$.

Let $X = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$. Let $Y$ be a $2 \times 2$ real matrix satisfying the condition $XY = YX$. Then the smallest possible value of $\det(Y)$ is

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