Question

11．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^*），a₁≠b₁，且b₁≠0，若$[\begin{array}{l}{{a}_{n+4}}\\{_{n+4}}\end{array}]$=M$[\begin{array}{l}{{a}_{n}}\\{_{n}}\end{array}]$，則二階矩陣M^-1=$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$．

Answer 1

分析 通過(guò)數(shù)列相關(guān)各項(xiàng)之間的關(guān)系求出矩陣M，進(jìn)而計(jì)算可得結(jié)論．

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，
∵$[\begin{array}{l}{{a}_{n+4}}\\{_{n+4}}\end{array}]$=M$[\begin{array}{l}{{a}_{n}}\\{_{n}}\end{array}]$，
∴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=$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$，
故答案為：$[\begin{array}{l}{1}&{-\frac{15}{16}}\\{0}&{\frac{1}{16}}\end{array}]$．

點(diǎn)評(píng) 本題考查矩陣及逆矩陣，求出矩陣M是解決本題的關(guān)鍵，注意解題方法的積累，屬于中檔題．