Question

8．已知矩阵$A=（{\begin{array}{l}1&0\\{-1}&2\end{array}}）$，$B=（{\begin{array}{l}2&4\\ 1&{-3}\end{array}}）$，则A+B=$（\begin{array}{cc}3&4\\ 0&-1\end{array}\right.）$．

Answer 1

分析 直接利用矩阵的和分运算法则求解即可．

解答 解：矩阵$A=（{\begin{array}{l}1&0\\{-1}&2\end{array}}）$，$B=（{\begin{array}{l}2&4\\ 1&{-3}\end{array}}）$，
∴$A+B=（\begin{array}{cc}1&0\\-1&2\end{array}\right.）+（\begin{array}{cc}2&4\\ 1&-3\end{array}\right.）$$+（\begin{array}{cc}2&4\\ 1&-3\end{array}\right.）$=$（\begin{array}{cc}3&4\\ 0&-1\end{array}\right.）$．
故答案为：$（\begin{array}{cc}3&4\\ 0&-1\end{array}\right.）$．

点评 本题考查矩阵的和的求法，是基础题．