Question

Dadas as matrizes A= | 3 0 | e B= | 2 1 |

| 1 -4 | | -1 0|

Então A.B-B.A é igual a:


A) | 0 0| B) | 2 -3| C) | -1 7 |

| 0 0 | | 5 0| | 9 1 | 7 |




D) | -3 1 |

| 2 7 |

Answer 1

$A~.~B~=~\left[\begin{array}{ccc}3&0\\1&-4\end{array}\right]~.~\left[\begin{array}{ccc}2&1\\-1&0\end{array}\right]\\\\\\\\A~.~B~=~\left[\begin{array}{ccc}3~.~2~+~0~.~(-1)&&3~.~1~+~0~.~0\\1~.~2~+~(-4)~.~(-1)&&1~.~1~+~(-4)~.~0\end{array}\right]\\\\\\\\A~.~B~=~\left[\begin{array}{ccc}6+0&3+0\\2+4&1+0\end{array}\right]\\\\\\\\A~.~B~=~\left[\begin{array}{ccc}6&3\\6&1\end{array}\right]$

$B~.~A~=~\left[\begin{array}{ccc}2&1\\-1&0\end{array}\right]~.~\left[\begin{array}{ccc}3&0\\1&-4\end{array}\right]\\\\\\\\B~.~A~=~\left[\begin{array}{ccc}2~.~3~+~1~.~1&&2~.~0~+~1~.~(-4)\\(-1)~.~3~+~0~.~1&&(-1)~.~0~+~0~.~(-4)\end{array}\right]\\\\\\\\B~.~A~=~\left[\begin{array}{ccc}6+1&0+(-4)\\-3+0&0+0\end{array}\right]\\\\\\\\B~.~A~=~\left[\begin{array}{ccc}7&-4\\-3&0\end{array}\right]$

$A.B~-~B.A~=~\left[\begin{array}{ccc}6&3\\6&1\end{array}\right]~-~\left[\begin{array}{ccc}7&-4\\-3&0\end{array}\right]\\\\\\\\A.B~-~B.A~=~\left[\begin{array}{ccc}6-7&3-(-4)\\6-(-3)&1-0\end{array}\right]\\\\\\\\A.B~-~B.A~=~\left[\begin{array}{ccc}-1&7\\9&1\end{array}\right]$