Question

help find E^2 and E^8 and E^-1 if E is equal to the matrices
| 1 0 |
| 6 1 |

Answer 1

This may be an application of Hamilton-Cayley:
characteristic polynomial: X²-2X+1
So E satisfies
E²-2E+I=0 [where I is identity matrix and 0 is null matrix]
So I=E(2I-E) from where you get E^-1=2I-E [take into account that the matrices E and 2I-E commute]
E²=I-2E
E^8=E^6-2E^7 [recurrence formula]
:)