$$$\left[\begin{array}{ccc}4 & 5 & 7\\2 & 1 & 0\\1 & 2 & 3\end{array}\right]\cdot \left[\begin{array}{ccc}- \frac{1}{2} & 0 & \frac{1}{2}\\5 & -1 & -1\\- \frac{7}{2} & 1 & \frac{1}{2}\end{array}\right]$$$

$$$\left[\begin{array}{ccc}4 & 5 & 7\\2 & 1 & 0\\1 & 2 & 3\end{array}\right]\cdot \left[\begin{array}{ccc}- \frac{1}{2} & 0 & \frac{1}{2}\\5 & -1 & -1\\- \frac{7}{2} & 1 & \frac{1}{2}\end{array}\right]$$$을(를) 계산하세요.

$$$\left[\begin{array}{ccc}{\color{Red}4} & {\color{DarkCyan}5} & {\color{OrangeRed}7}\\{\color{DarkMagenta}2} & {\color{DarkBlue}1} & {\color{SaddleBrown}0}\\{\color{Green}1} & {\color{Magenta}2} & {\color{Violet}3}\end{array}\right]\cdot \left[\begin{array}{ccc}{\color{Violet}- \frac{1}{2}} & {\color{Peru}0} & {\color{Magenta}\frac{1}{2}}\\{\color{DarkBlue}5} & {\color{Fuchsia}-1} & {\color{Chocolate}-1}\\{\color{DarkMagenta}- \frac{7}{2}} & {\color{Blue}1} & {\color{GoldenRod}\frac{1}{2}}\end{array}\right] = \left[\begin{array}{ccc}{\color{Red}\left(4\right)}\cdot {\color{Violet}\left(- \frac{1}{2}\right)} + {\color{DarkCyan}\left(5\right)}\cdot {\color{DarkBlue}\left(5\right)} + {\color{OrangeRed}\left(7\right)}\cdot {\color{DarkMagenta}\left(- \frac{7}{2}\right)} & {\color{Red}\left(4\right)}\cdot {\color{Peru}\left(0\right)} + {\color{DarkCyan}\left(5\right)}\cdot {\color{Fuchsia}\left(-1\right)} + {\color{OrangeRed}\left(7\right)}\cdot {\color{Blue}\left(1\right)} & {\color{Red}\left(4\right)}\cdot {\color{Magenta}\left(\frac{1}{2}\right)} + {\color{DarkCyan}\left(5\right)}\cdot {\color{Chocolate}\left(-1\right)} + {\color{OrangeRed}\left(7\right)}\cdot {\color{GoldenRod}\left(\frac{1}{2}\right)}\\{\color{DarkMagenta}\left(2\right)}\cdot {\color{Violet}\left(- \frac{1}{2}\right)} + {\color{DarkBlue}\left(1\right)}\cdot {\color{DarkBlue}\left(5\right)} + {\color{SaddleBrown}\left(0\right)}\cdot {\color{DarkMagenta}\left(- \frac{7}{2}\right)} & {\color{DarkMagenta}\left(2\right)}\cdot {\color{Peru}\left(0\right)} + {\color{DarkBlue}\left(1\right)}\cdot {\color{Fuchsia}\left(-1\right)} + {\color{SaddleBrown}\left(0\right)}\cdot {\color{Blue}\left(1\right)} & {\color{DarkMagenta}\left(2\right)}\cdot {\color{Magenta}\left(\frac{1}{2}\right)} + {\color{DarkBlue}\left(1\right)}\cdot {\color{Chocolate}\left(-1\right)} + {\color{SaddleBrown}\left(0\right)}\cdot {\color{GoldenRod}\left(\frac{1}{2}\right)}\\{\color{Green}\left(1\right)}\cdot {\color{Violet}\left(- \frac{1}{2}\right)} + {\color{Magenta}\left(2\right)}\cdot {\color{DarkBlue}\left(5\right)} + {\color{Violet}\left(3\right)}\cdot {\color{DarkMagenta}\left(- \frac{7}{2}\right)} & {\color{Green}\left(1\right)}\cdot {\color{Peru}\left(0\right)} + {\color{Magenta}\left(2\right)}\cdot {\color{Fuchsia}\left(-1\right)} + {\color{Violet}\left(3\right)}\cdot {\color{Blue}\left(1\right)} & {\color{Green}\left(1\right)}\cdot {\color{Magenta}\left(\frac{1}{2}\right)} + {\color{Magenta}\left(2\right)}\cdot {\color{Chocolate}\left(-1\right)} + {\color{Violet}\left(3\right)}\cdot {\color{GoldenRod}\left(\frac{1}{2}\right)}\end{array}\right] = \left[\begin{array}{ccc}- \frac{3}{2} & 2 & \frac{1}{2}\\4 & -1 & 0\\-1 & 1 & 0\end{array}\right]$$$

$$$\left[\begin{array}{ccc}4 & 5 & 7\\2 & 1 & 0\\1 & 2 & 3\end{array}\right]\cdot \left[\begin{array}{ccc}- \frac{1}{2} & 0 & \frac{1}{2}\\5 & -1 & -1\\- \frac{7}{2} & 1 & \frac{1}{2}\end{array}\right] = \left[\begin{array}{ccc}- \frac{3}{2} & 2 & \frac{1}{2}\\4 & -1 & 0\\-1 & 1 & 0\end{array}\right] = \left[\begin{array}{ccc}-1.5 & 2 & 0.5\\4 & -1 & 0\\-1 & 1 & 0\end{array}\right]$$$A