Given,
A=[10110−3103101],B=[10−i1]
AAT=[10110−3103101][10110310−3101]=[1001]
B2=[10−i1][10−i1]=[10−2i1]
B3=[10−2i1][10−i1]=[10−3i1]
So, we can write
B2023=[10−2023i1]
M=ATBA (given)
M=AATB=[1001][10−i1]=[10−i1]
M2=M⋅M=ATBAATBA=AAT.AAT.B2
=AATB2
M3=AATB3
M2023=AATB2023=B2023 (∵AAT is an identity matrix)
∴M2023=B2023=[10−2023i1]
Inverse of (AM2023AT) is [102023i1]