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

La calculadora multiplicará la matriz $$$3$$$ x $$$3$$$ $$$\left[\begin{array}{ccc}4 & 5 & 7\\2 & 1 & 0\\1 & 2 & 3\end{array}\right]$$$ por la matriz $$$3$$$ x $$$3$$$ $$$\left[\begin{array}{ccc}- \frac{1}{2} & 0 & \frac{1}{2}\\5 & -1 & -1\\- \frac{7}{2} & 1 & \frac{1}{2}\end{array}\right]$$$, y se muestran los pasos.

Calculadora relacionada: Calculadora de matrices

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Si la calculadora no calculó algo o ha identificado un error, o tiene una sugerencia/comentario, escríbalo en los comentarios a continuación.

Tu aportación

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

Solución

$$$\left[\begin{array}{ccc}{\color{DarkBlue}4} & {\color{BlueViolet}5} & {\color{Chartreuse}7}\\{\color{SaddleBrown}2} & {\color{GoldenRod}1} & {\color{Brown}0}\\{\color{Crimson}1} & {\color{Violet}2} & {\color{Purple}3}\end{array}\right]\cdot \left[\begin{array}{ccc}{\color{DeepPink}- \frac{1}{2}} & {\color{Blue}0} & {\color{Peru}\frac{1}{2}}\\{\color{BlueViolet}5} & {\color{Purple}-1} & {\color{Chartreuse}-1}\\{\color{Magenta}- \frac{7}{2}} & {\color{Crimson}1} & {\color{GoldenRod}\frac{1}{2}}\end{array}\right] = \left[\begin{array}{ccc}{\color{DarkBlue}\left(4\right)}\cdot {\color{DeepPink}\left(- \frac{1}{2}\right)} + {\color{BlueViolet}\left(5\right)}\cdot {\color{BlueViolet}\left(5\right)} + {\color{Chartreuse}\left(7\right)}\cdot {\color{Magenta}\left(- \frac{7}{2}\right)} & {\color{DarkBlue}\left(4\right)}\cdot {\color{Blue}\left(0\right)} + {\color{BlueViolet}\left(5\right)}\cdot {\color{Purple}\left(-1\right)} + {\color{Chartreuse}\left(7\right)}\cdot {\color{Crimson}\left(1\right)} & {\color{DarkBlue}\left(4\right)}\cdot {\color{Peru}\left(\frac{1}{2}\right)} + {\color{BlueViolet}\left(5\right)}\cdot {\color{Chartreuse}\left(-1\right)} + {\color{Chartreuse}\left(7\right)}\cdot {\color{GoldenRod}\left(\frac{1}{2}\right)}\\{\color{SaddleBrown}\left(2\right)}\cdot {\color{DeepPink}\left(- \frac{1}{2}\right)} + {\color{GoldenRod}\left(1\right)}\cdot {\color{BlueViolet}\left(5\right)} + {\color{Brown}\left(0\right)}\cdot {\color{Magenta}\left(- \frac{7}{2}\right)} & {\color{SaddleBrown}\left(2\right)}\cdot {\color{Blue}\left(0\right)} + {\color{GoldenRod}\left(1\right)}\cdot {\color{Purple}\left(-1\right)} + {\color{Brown}\left(0\right)}\cdot {\color{Crimson}\left(1\right)} & {\color{SaddleBrown}\left(2\right)}\cdot {\color{Peru}\left(\frac{1}{2}\right)} + {\color{GoldenRod}\left(1\right)}\cdot {\color{Chartreuse}\left(-1\right)} + {\color{Brown}\left(0\right)}\cdot {\color{GoldenRod}\left(\frac{1}{2}\right)}\\{\color{Crimson}\left(1\right)}\cdot {\color{DeepPink}\left(- \frac{1}{2}\right)} + {\color{Violet}\left(2\right)}\cdot {\color{BlueViolet}\left(5\right)} + {\color{Purple}\left(3\right)}\cdot {\color{Magenta}\left(- \frac{7}{2}\right)} & {\color{Crimson}\left(1\right)}\cdot {\color{Blue}\left(0\right)} + {\color{Violet}\left(2\right)}\cdot {\color{Purple}\left(-1\right)} + {\color{Purple}\left(3\right)}\cdot {\color{Crimson}\left(1\right)} & {\color{Crimson}\left(1\right)}\cdot {\color{Peru}\left(\frac{1}{2}\right)} + {\color{Violet}\left(2\right)}\cdot {\color{Chartreuse}\left(-1\right)} + {\color{Purple}\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]$$$

Respuesta

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