Maybe you meant to say that (i±1,±1) is an eigenvector for the eigenvalue ±i; the + case works, but the − case for your solution does not work, because
* i−1−2=i−3≠−i\*(i−1) and i−1−1=i−2≠−i\*1, while
* i−1+2=i+1=−i\*(i−1) and i−1+1=i=−i\*(−1).
With that said, Symbolab's solution looks like yours, so it looks as if your work following the explicit eigenvectors went as if you had written them out correctly.
An alternative way to find the matrix exponential, the one that I learned, is to first diagonalize the matrix (which you can do in this case because the matrix has a full eigenbasis, because its eigenvalues are distinct), then exponentiate the diagonals, then multiply out; this works because the power series for exponentiation is also valid for matrices, although *why* it's valid is a bit beyond the scope of your course, and then for the diagonalizable case, the P^(−1)P factors cancel out in each term, leaving terms of the form PD^(n)P^(−1) for each whole number n, and then the P and P^(−1) terms can be factored out, each power of a diagonal matrix is the matrix with its diagonal entries raised to that power, and so it reduces to ordinary exponentiation.
Okay, in case you haven't diagonalized a matrix before, the P here is a matrix with each eigenvector as a column, and then the D is a diagonal matrix with the corresponding eigenvalue as the corresponding diagonal entry:
P=⌈i+1 i−1⌉ D=⌈i 0⌉
⌊ 1 −1 ⌋ ⌊0 −i⌋
The idea is that, say, (i+1,1) will be transformed to (1,0), then to (i,0), then to i\*(i+1,1).
Then det(P)=−2i, so
and P^(−1) does work as expected.
Then the diagonal elements of e^(Dt) are e^(it) and e^(−it), and
⌊ e^(it) −e^(−it) ⌋
e^(At)=Pe^(Dt)P^(−1)=⌈−½(i−1)e^(it)+½(i+1)e^(−it) ie^(it)−ie^(−it) ⌉
⌊ −½ie^(it)+½ie^(−it) ½(i+1)e^(it)−½(i−1)e^(−it)⌋
This also agrees with Symbolab.