Verify that X(1) is a fundamental matrix for the given system and compute X "(t). Then use the result that if X(t) is a fundamental matrix for the system x' = Ax, then x(t) = X(1)X "(0)x, is the solution to the initial value problem x' = Ax, x(0) =x- 0 60 6e - 12e -21 6 e 31 X' = 1 01 X, x(0) = 2 X(1) = -t 4e 21 3 e 3t 0 -5e 4e21 3 e 31 0 60 (a) If X(t) = [x, (1) X2(1) x3(t)]and A = 1 0 1 , validate the following identities and write the column vector that equals each side of the equation. 1 0 x1 ' = AX1 = X2' = AX2 = X3' = Ax3 = (b) Next, compute the Wronskian of X(t). W x 1 (t ),* 2 (1).x 3 (1)] =]. Since the Wronskian is never |, and each column of X(t) is a solution to x' = Ax, X(t) is a fundamental matrix