This is a way:

DSolve[{Subscript[x, 1]'[t] == -Subscript[x, 1][t] + 
    Subscript[x, 2][t], 
  Subscript[x, 2]'[t] == 
   Subscript[x, 1][t] - Subscript[x, 2][t]}, {Subscript[x, 1][t], 
  Subscript[x, 2][t]}, t]

General solution:

DSolve[{x1'[t] == -x1[t] + x2[t], x2'[t] == x1[t] - x2[t]}, {x1[t], x2[t]}, t]
  (*{{x1[t] -> 
     1/2 E^(-2 t) (1 + E^(2 t)) C[1] + 1/2 E^(-2 t) (-1 + E^(2 t)) C[2],
     x2[t] -> 
     1/2 E^(-2 t) (-1 + E^(2 t)) C[1] + 
      1/2 E^(-2 t) (1 + E^(2 t)) C[2]}}*)

Or:

A = {{-1, 1}, {1, -1}};
DSolve[x'[t] == A . x[t], x[t] \[Element] Vectors[2], t]
(*{{x[t] -> {1/2 E^(-2 t) (1 + E^(2 t)) C[1] + 
     1/2 E^(-2 t) (-1 + E^(2 t)) C[2], 
    1/2 E^(-2 t) (-1 + E^(2 t)) C[1] + 
     1/2 E^(-2 t) (1 + E^(2 t)) C[2]}}}*)

Solution by series:

AsymptoticDSolveValue[{x1'[t] == -x1[t] + x2[t], 
  x2'[t] == x1[t] - x2[t]}, {x1[t], x2[t]}, {t, 0, 5}]
(*{(1 - t + t^2 - (2 t^3)/3 + t^4/3 - (2 t^5)/15) C[
   1] + (t - t^2 + (2 t^3)/3 - t^4/3 + (2 t^5)/15) C[
   2], (t - t^2 + (2 t^3)/3 - t^4/3 + (2 t^5)/15) C[
   1] + (1 - t + t^2 - (2 t^3)/3 + t^4/3 - (2 t^5)/15) C[2]}*)