Verify that X(t) is a fundamental matrix for the given system and compute X "(1). Then use the result that if X(1) is a fundamental matrix for the system x' = Ax, then x(1) = X(1)X "(0)x, is the solution to the initial value problem x' = Ax, x(0) =XO 0 60 5 12e -3e 21 6e 31 x'= 101 x x(0) = -1 X(t) = -2e -t e -21 3 31 1 10 -10 e -1 e 3 e 31 0 60 (a) if X(1) = [x, (1) x2(t) x3(1)] and A= 1 0 1 , validate the following identities and write the column vector that equals each side of the equation. 1 10 X1 ' = AX1 = x2' = AX2= X3' = AX3 = (b) Next, compute the Wronskian of X(t). Since the Wronskian is never , and each column of X(1) is a solution to x' = Ax, X(1) is a fundamental matrix. (c) Find X ' (1) =] (d) x(1) = x(1)x (0)xo =