Question

We have `A=((2,-1),(4,-2))`.Number of solutions from `M_2(RR)`of equation `X^25=A`?

Answer 1

See below.

Explanation:

Supposing that `X_0` is a solution for

`X_0^25=A`

According with Cayley-Hamilton theorem

https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem

we have

`X_0^2+alpha X_0+beta I_2=0_2` or

`X_0^2= -alpha X_0-beta I_0`

`X_0^3=-alpha X_0^2-beta X_0 = -alpha( -alpha X_0-beta I_0)-beta X_0 = (alpha^2-beta)X_0+alpha beta I_0`

`cdots`

`X_0^25 = c_1 X_0+c_2 I_2`

then

`X_0^25 = A = c_1 X_0+c_2 I_2` then

`X_0 = 1/c_1 A - c_2/c_1 I_2 = lambda A+mu I_2`

but `A^2=0_2`

then

`X_0*A=lambda A^2+mu A = mu A`

but then

`X_0 = mu I_2` which is not a solution. Concluding the equation

`X^25=A` has not real solutions.