Finding a 5×5 Matrix such that the sum of it and its inverse is a 5×5 matrix of ones

[escobarro]

Finding a 5×5 Matrix such that the sum of it and its inverse is a 5×5 matrix of ones

[dont worry, done]

 

[stapel]

How can I do this question:

Is it possible to find a 5×5 invertible matrix B over ℤ 2 such that:
B+B-1 = 5×5 matrix where every element is 1?

I assume that the question requires a proof if such a matrix B does not exist and working/solution(maybe?) if B exists

Is it correct to assume that, by "B - 1", you mean "B^-1"? And, by "ℤ 2", you mean "\(\displaystyle \, \mathbb{Z}_2\)? Also, how far have you gotten based on what was posted here?

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No, this isn't possible. Denote the all-one matrix by E. If B + B-¹ = E, then BE = B² + I = EB. E. Therefore B² + I has identical columns and identical rows. Hence either B² + I = 0 or else B² + I = E.

In the first case, what is BE? Why is this case impossible?

In the second case, what are BE, B^-1E, and E(B + B^-1)E? How do they violate the condition that B + B^-1?

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Please be complete. Thank you!

 

[escobarro]

yep, both of those assumptions are correct
I responded to that guy's answer but I still don't understand why BE cant be 0 and why E^3 cant equal 2E^2 (is this because the scalar 2 does not exist in Z2?)

thanks