题目内容

8.设数列{an},{bn}满足an+1=an+bn,bn+1=2bn,其中n∈N*,若$[{\begin{array}{l}{{a_{n+4}}}\\{{b_{n+4}}}\end{array}}]=M[{\begin{array}{l}{a_n}\\{{b_n}}\end{array}}]$,则二阶矩阵M=$[\begin{array}{l}{1}&{15}\\{0}&{16}\end{array}]$.

分析 通过数列相关各项之间的关系即可求出矩阵M.

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

点评 本题考查矩阵及逆矩阵,注意解题方法的积累,属于中档题.

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