Verify that X(t) is a fundamental matrix for the given system and compute X1(t). Then use the result that if X(t) is a fundamental matrix for the system x' = Ax, then x(t) = x(t)X(0)x is the solution to the initial value problem x' = Ax, x(0) = x0 - 24e-1 -9e-21 060 -5 x'= 10 1x, x(0)= 8e3t x(t)= 1 1 0 -4e-t 3e-21 3t 4e -2 -20e-t 3e-21 3t 4e 060 (a) If x(t) = [×₁ (t) x2(t) ×3(t)] and A = 10 1 validate the following identities and write the column vector that equals each side of the equation. 1 1 0 ×₁' = Ax₁ = x2' = Ax₂ = x3' = Ax3 (b) Next, compute the Wronskian of X(t). W[x()()()]= Since the Wronskian is never and each column of X(t) is a solution to x' = Ax, x(t) is a fundamental matrix. (c) Find x(t) = ☐ (d) x(t) = x(t)x¯1 (0)x₁ =