Question

8．設(shè)數(shù)列{a_n}，{b_n}滿足a_n+1=a_n+b_n，b_n+1=2b_n，其中n∈N^*，若$[{\begin{array}{l}{{a_{n+4}}}\\{{b_{n+4}}}\end{array}}]=M[{\begin{array}{l}{a_n}\\{{b_n}}\end{array}}]$，則二階矩陣M=$[\begin{array}{l}{1}&{15}\\{0}&{16}\end{array}]$．

Answer 1

分析 通過數(shù)列相關(guān)各項(xiàng)之間的關(guān)系即可求出矩陣M．

解答 解：∵b_n+1=2b_n，
∴b_n+4=16b_n，b_n+3=8b_n，b_n+2=4b_n，
∵a_n+1-a_n=b_n，
∴a_n+4=a_n+3+b_n+3
=a_n+2+b_n+2+b_n+3
=a_n+1+b_n+1+b_n+2+b_n+3
=a_n+b_n+b_n+1+b_n+2+b_n+3
=a_n+b_n+2b_n+4b_n+8b_n
=a_n+15b_n，
∵$[{\begin{array}{l}{{a_{n+4}}}\\{{b_{n+4}}}\end{array}}]=M[{\begin{array}{l}{a_n}\\{{b_n}}\end{array}}]$，
∴M=$[\begin{array}{l}{1}&{15}\\{0}&{16}\end{array}]$，
故答案為：$[\begin{array}{l}{1}&{15}\\{0}&{16}\end{array}]$．

點(diǎn)評 本題考查矩陣及逆矩陣，注意解題方法的積累，屬于中檔題．