Question

15．计算：$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$¹⁰．

Answer 1

分析 利用二阶矩阵乘法法则进行求解．

解答 解：$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$¹⁰=$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$⁸=$（\begin{array}{l}{1}&{3}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$⁷=$（\begin{array}{l}{1}&{4}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$⁶
=$（\begin{array}{l}{1}&{5}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$⁵=$（\begin{array}{l}{1}&{6}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$⁴=$（\begin{array}{l}{1}&{7}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$³
=$（\begin{array}{l}{1}&{8}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$²=$（\begin{array}{l}{1}&{9}\\{0}&{1}\end{array}）$$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$=$（\begin{array}{l}{1}&{10}\\{0}&{1}\end{array}）$．
∴$（\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}）$¹⁰=$（\begin{array}{l}{1}&{10}\\{0}&{1}\end{array}）$．

点评 本题考查二阶矩阵的乘法运算，是基础题，解题时要注意二阶矩阵乘法法则的合理运用．