Question

4．设x为实数．若矩阵M=$[{\begin{array}{l}1&{-5}\\ 2&x\end{array}}]$为不可逆矩阵，求M²．

Answer 1

分析 由矩阵M为不可逆矩阵，|M|=0，求得x的值，根据矩阵的乘法求得M²．

解答 解：依题意，矩阵M=$[{\begin{array}{l}1&{-5}\\ 2&x\end{array}}]$为不可逆矩阵，
M的行列式|M|=0，解得：x=-10，
∴M²=$[{\begin{array}{l}1&{-5}\\ 2&{-10}\end{array}}][{\begin{array}{l}1&{-5}\\ 2&{-10}\end{array}}]=[{\begin{array}{l}{-9}&{45}\\{-18}&{90}\end{array}}]$，
M²=$[\begin{array}{l}{-9}&{45}\\{-18}&{90}\end{array}]$．

点评 本题考查矩阵可逆的条件，矩阵的乘法，属于基础题．