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$$$\left[\begin{array}{cc}3 & 2\\2 & 3\\2 & -2\end{array}\right]\cdot \left[\begin{array}{cc}\frac{17}{225} & - \frac{8}{225}\\- \frac{8}{225} & \frac{17}{225}\end{array}\right]$$$

The calculator will multiply the $$$3$$$x$$$2$$$ matrix $$$\left[\begin{array}{cc}3 & 2\\2 & 3\\2 & -2\end{array}\right]$$$ by the $$$2$$$x$$$2$$$ matrix $$$\left[\begin{array}{cc}\frac{17}{225} & - \frac{8}{225}\\- \frac{8}{225} & \frac{17}{225}\end{array}\right]$$$, with steps shown.

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Calculate $$$\left[\begin{array}{cc}3 & 2\\2 & 3\\2 & -2\end{array}\right]\cdot \left[\begin{array}{cc}\frac{17}{225} & - \frac{8}{225}\\- \frac{8}{225} & \frac{17}{225}\end{array}\right].$$$

Solution

$$$\left[\begin{array}{cc}{\color{Violet}3} & {\color{SaddleBrown}2}\\{\color{Red}2} & {\color{OrangeRed}3}\\{\color{Purple}2} & {\color{DarkMagenta}-2}\end{array}\right]\cdot \left[\begin{array}{cc}{\color{Fuchsia}\frac{17}{225}} & {\color{Violet}- \frac{8}{225}}\\{\color{Purple}- \frac{8}{225}} & {\color{DarkBlue}\frac{17}{225}}\end{array}\right] = \left[\begin{array}{cc}{\color{Violet}\left(3\right)}\cdot {\color{Fuchsia}\left(\frac{17}{225}\right)} + {\color{SaddleBrown}\left(2\right)}\cdot {\color{Purple}\left(- \frac{8}{225}\right)} & {\color{Violet}\left(3\right)}\cdot {\color{Violet}\left(- \frac{8}{225}\right)} + {\color{SaddleBrown}\left(2\right)}\cdot {\color{DarkBlue}\left(\frac{17}{225}\right)}\\{\color{Red}\left(2\right)}\cdot {\color{Fuchsia}\left(\frac{17}{225}\right)} + {\color{OrangeRed}\left(3\right)}\cdot {\color{Purple}\left(- \frac{8}{225}\right)} & {\color{Red}\left(2\right)}\cdot {\color{Violet}\left(- \frac{8}{225}\right)} + {\color{OrangeRed}\left(3\right)}\cdot {\color{DarkBlue}\left(\frac{17}{225}\right)}\\{\color{Purple}\left(2\right)}\cdot {\color{Fuchsia}\left(\frac{17}{225}\right)} + {\color{DarkMagenta}\left(-2\right)}\cdot {\color{Purple}\left(- \frac{8}{225}\right)} & {\color{Purple}\left(2\right)}\cdot {\color{Violet}\left(- \frac{8}{225}\right)} + {\color{DarkMagenta}\left(-2\right)}\cdot {\color{DarkBlue}\left(\frac{17}{225}\right)}\end{array}\right] = \left[\begin{array}{cc}\frac{7}{45} & \frac{2}{45}\\\frac{2}{45} & \frac{7}{45}\\\frac{2}{9} & - \frac{2}{9}\end{array}\right]$$$

Answer

$$$\left[\begin{array}{cc}3 & 2\\2 & 3\\2 & -2\end{array}\right]\cdot \left[\begin{array}{cc}\frac{17}{225} & - \frac{8}{225}\\- \frac{8}{225} & \frac{17}{225}\end{array}\right] = \left[\begin{array}{cc}\frac{7}{45} & \frac{2}{45}\\\frac{2}{45} & \frac{7}{45}\\\frac{2}{9} & - \frac{2}{9}\end{array}\right]\approx \left[\begin{array}{cc}0.155555555555556 & 0.044444444444444\\0.044444444444444 & 0.155555555555556\\0.222222222222222 & -0.222222222222222\end{array}\right]$$$A