题目内容

7.设A=$(\begin{array}{l}{1}&{1}&{1}\\{1}&{1}&{-1}\\{1}&{-1}&{1}\end{array})$,B=$(\begin{array}{l}{1}&{2}&{3}\\{-1}&{-2}&{4}\\{0}&{5}&{1}\end{array})$,求3AB-2A.

分析 直接利用矩阵的运算法则化简求解即可.

解答 解:A=$(\begin{array}{l}{1}&{1}&{1}\\{1}&{1}&{-1}\\{1}&{-1}&{1}\end{array})$,B=$(\begin{array}{l}{1}&{2}&{3}\\{-1}&{-2}&{4}\\{0}&{5}&{1}\end{array})$,
AB=$(\begin{array}{ccc}1&1&1\\ 1&1&-1\\ 1&-1&1\end{array}\right.)(\begin{array}{ccc}1&2&3\\-1&-2&4\\ 0&5&1\end{array}\right.)$$(\begin{array}{ccc}1&2&3\\-1&-2&4\\ 0&5&1\end{array}\right.)$=$(\begin{array}{ccc}0&5&8\\ 0&-5&0\\ 2&9&0\end{array}\right.)$,
3AB-2A=$(\begin{array}{ccc}0&15&24\\ 0&-15&0\\ 6&27&0\end{array}\right.)$-$(\begin{array}{ccc}2&2&2\\ 2&2&-2\\ 2&-2&2\end{array}\right.)$=$(\begin{array}{ccc}-2&13&22\\-2&-17&2\\ 4&29&-2\end{array}\right.)$.

点评 本题考查矩阵的基本运算,考查计算能力.

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