Kalkulator znajdzie iloczyn dwóch macierzy (jeśli to możliwe), z pokazanymi krokami. Mnoży macierze o dowolnym rozmiarze do 10x10 (2x2, 3x3, 4x4 itd.).
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Rozwiązanie [ 4 5 7 2 1 0 ] ⋅ [ 2 3 8 9 1 1 ] = [ ( 4 ) ⋅ ( 2 ) + ( 5 ) ⋅ ( 8 ) + ( 7 ) ⋅ ( 1 ) ( 4 ) ⋅ ( 3 ) + ( 5 ) ⋅ ( 9 ) + ( 7 ) ⋅ ( 1 ) ( 2 ) ⋅ ( 2 ) + ( 1 ) ⋅ ( 8 ) + ( 0 ) ⋅ ( 1 ) ( 2 ) ⋅ ( 3 ) + ( 1 ) ⋅ ( 9 ) + ( 0 ) ⋅ ( 1 ) ] = [ 55 64 12 15 ] \left[\begin{array}{ccc}{\color{DarkBlue}4} & {\color{Purple}5} & {\color{DarkMagenta}7}\\{\color{SaddleBrown}2} & {\color{Brown}1} & {\color{BlueViolet}0}\end{array}\right]\cdot \left[\begin{array}{cc}{\color{GoldenRod}2} & {\color{Brown}3}\\{\color{Magenta}8} & {\color{SaddleBrown}9}\\{\color{Purple}1} & {\color{DarkCyan}1}\end{array}\right] = \left[\begin{array}{cc}{\color{DarkBlue}\left(4\right)}\cdot {\color{GoldenRod}\left(2\right)} + {\color{Purple}\left(5\right)}\cdot {\color{Magenta}\left(8\right)} + {\color{DarkMagenta}\left(7\right)}\cdot {\color{Purple}\left(1\right)} & {\color{DarkBlue}\left(4\right)}\cdot {\color{Brown}\left(3\right)} + {\color{Purple}\left(5\right)}\cdot {\color{SaddleBrown}\left(9\right)} + {\color{DarkMagenta}\left(7\right)}\cdot {\color{DarkCyan}\left(1\right)}\\{\color{SaddleBrown}\left(2\right)}\cdot {\color{GoldenRod}\left(2\right)} + {\color{Brown}\left(1\right)}\cdot {\color{Magenta}\left(8\right)} + {\color{BlueViolet}\left(0\right)}\cdot {\color{Purple}\left(1\right)} & {\color{SaddleBrown}\left(2\right)}\cdot {\color{Brown}\left(3\right)} + {\color{Brown}\left(1\right)}\cdot {\color{SaddleBrown}\left(9\right)} + {\color{BlueViolet}\left(0\right)}\cdot {\color{DarkCyan}\left(1\right)}\end{array}\right] = \left[\begin{array}{cc}55 & 64\\12 & 15\end{array}\right] [ 4 2 5 1 7 0 ] ⋅ ⎣ ⎡ 2 8 1 3 9 1 ⎦ ⎤ = [ ( 4 ) ⋅ ( 2 ) + ( 5 ) ⋅ ( 8 ) + ( 7 ) ⋅ ( 1 ) ( 2 ) ⋅ ( 2 ) + ( 1 ) ⋅ ( 8 ) + ( 0 ) ⋅ ( 1 ) ( 4 ) ⋅ ( 3 ) + ( 5 ) ⋅ ( 9 ) + ( 7 ) ⋅ ( 1 ) ( 2 ) ⋅ ( 3 ) + ( 1 ) ⋅ ( 9 ) + ( 0 ) ⋅ ( 1 ) ] = [ 55 12 64 15 ]
Odpowiedź [ 4 5 7 2 1 0 ] ⋅ [ 2 3 8 9 1 1 ] = [ 55 64 12 15 ] \left[\begin{array}{ccc}4 & 5 & 7\\2 & 1 & 0\end{array}\right]\cdot \left[\begin{array}{cc}2 & 3\\8 & 9\\1 & 1\end{array}\right] = \left[\begin{array}{cc}55 & 64\\12 & 15\end{array}\right] [ 4 2 5 1 7 0 ] ⋅ ⎣ ⎡ 2 8 1 3 9 1 ⎦ ⎤ = [ 55 12 64 15 ] A