La calculatrice trouvera la différence de deux matrices (si possible), avec les étapes indiquées. Elle soustrait les matrices de toute taille jusqu'à 10x10 (2x2, 3x3, 4x4, etc.).
Solution [ 1 2 − 3 2 − 3 − 5 1 7 1 ] − [ 2 − 3 0 1 1 5 1 0 − 1 ] = [ ( 1 ) − ( 2 ) ( 2 ) − ( − 3 ) ( − 3 ) − ( 0 ) ( 2 ) − ( 1 ) ( − 3 ) − ( 1 ) ( − 5 ) − ( 5 ) ( 1 ) − ( 1 ) ( 7 ) − ( 0 ) ( 1 ) − ( − 1 ) ] = [ − 1 5 − 3 1 − 4 − 10 0 7 2 ] \left[\begin{array}{ccc}{\color{Purple}1} & {\color{Blue}2} & {\color{Magenta}-3}\\{\color{DarkCyan}2} & {\color{Crimson}-3} & {\color{GoldenRod}-5}\\{\color{Peru}1} & {\color{SaddleBrown}7} & {\color{DeepPink}1}\end{array}\right] - \left[\begin{array}{ccc}{\color{Purple}2} & {\color{Blue}-3} & {\color{Magenta}0}\\{\color{DarkCyan}1} & {\color{Crimson}1} & {\color{GoldenRod}5}\\{\color{Peru}1} & {\color{SaddleBrown}0} & {\color{DeepPink}-1}\end{array}\right] = \left[\begin{array}{ccc}{\color{Purple}\left(1\right)} - {\color{Purple}\left(2\right)} & {\color{Blue}\left(2\right)} - {\color{Blue}\left(-3\right)} & {\color{Magenta}\left(-3\right)} - {\color{Magenta}\left(0\right)}\\{\color{DarkCyan}\left(2\right)} - {\color{DarkCyan}\left(1\right)} & {\color{Crimson}\left(-3\right)} - {\color{Crimson}\left(1\right)} & {\color{GoldenRod}\left(-5\right)} - {\color{GoldenRod}\left(5\right)}\\{\color{Peru}\left(1\right)} - {\color{Peru}\left(1\right)} & {\color{SaddleBrown}\left(7\right)} - {\color{SaddleBrown}\left(0\right)} & {\color{DeepPink}\left(1\right)} - {\color{DeepPink}\left(-1\right)}\end{array}\right] = \left[\begin{array}{ccc}-1 & 5 & -3\\1 & -4 & -10\\0 & 7 & 2\end{array}\right] ⎣ ⎡ 1 2 1 2 − 3 7 − 3 − 5 1 ⎦ ⎤ − ⎣ ⎡ 2 1 1 − 3 1 0 0 5 − 1 ⎦ ⎤ = ⎣ ⎡ ( 1 ) − ( 2 ) ( 2 ) − ( 1 ) ( 1 ) − ( 1 ) ( 2 ) − ( − 3 ) ( − 3 ) − ( 1 ) ( 7 ) − ( 0 ) ( − 3 ) − ( 0 ) ( − 5 ) − ( 5 ) ( 1 ) − ( − 1 ) ⎦ ⎤ = ⎣ ⎡ − 1 1 0 5 − 4 7 − 3 − 10 2 ⎦ ⎤
Réponse [ 1 2 − 3 2 − 3 − 5 1 7 1 ] − [ 2 − 3 0 1 1 5 1 0 − 1 ] = [ − 1 5 − 3 1 − 4 − 10 0 7 2 ] \left[\begin{array}{ccc}1 & 2 & -3\\2 & -3 & -5\\1 & 7 & 1\end{array}\right] - \left[\begin{array}{ccc}2 & -3 & 0\\1 & 1 & 5\\1 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}-1 & 5 & -3\\1 & -4 & -10\\0 & 7 & 2\end{array}\right] ⎣ ⎡ 1 2 1 2 − 3 7 − 3 − 5 1 ⎦ ⎤ − ⎣ ⎡ 2 1 1 − 3 1 0 0 5 − 1 ⎦ ⎤ = ⎣ ⎡ − 1 1 0 5 − 4 7 − 3 − 10 2 ⎦ ⎤ A