Question

11．设数列{a_n}，{b_n}满足a_n+1-a_n=b_n，b_n+1=2b_n（其中n∈N^*），a₁≠b₁，且b₁≠0，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^-1=$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$．

Answer 1

分析 通过数列相关各项之间的关系求出矩阵M，进而计算可得结论．

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=$[\begin{array}{l}{1}&{15}\\{0}&{16}\end{array}]$，
∵$[\begin{array}{l}{1}&{15}&{1}&{0}\\{0}&{16}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{1}&{-\frac{15}{16}}\\{0}&{16}&{0}&{1}\end{array}]$→$[\begin{array}{l}{1}&{0}&{1}&{-\frac{15}{16}}\\{0}&{1}&{0}&{\frac{1}{16}}\end{array}]$，
∴M=$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$，
故答案为：$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$．

点评 本题考查矩阵及逆矩阵，求出矩阵M是解决本题的关键，注意解题方法的积累，属于中档题．