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In Exercises 1-6, consider the vector sequence {X*), where X* = AXk-1, k = 1, 2, .... For the given starting vector Xo, calculate X1, X2, X3, and x4 by using direct multiplication, as in Example 1.EXAMPLE 1 Let XK = AXk-1, K = 1, 2, .... Calculate X1, X2, X3, X4, and x5, where 8 .2 A = .8 and x. = [2 ]. Solution Some routine but tedious calculations show that 1.2 1.32 1.392 X1 = AXO = X2 = AX1 = X3 = AX2 = 1.8 1.68 1.608 1.4352 1.461 12 X4 = AX3 = and X5 = AX4 = 1.5648 1.53888 In Example 1, the first six terms of a vector sequence {xx} are listed. An inspection of these first few terms suggests that the sequence might have some regular pattern of behavior. For instance, the first components of these vectors are steadily increasing, whereas the second components are steadily decreasing. In fact, as shown in Example 3, this monotonic behavior persists for all terms of the sequence {xx}. Moreover, it can be shown that lim X* = X* , where the limit vector x* is given by 1.5 * * 1.5\f5 .25 128 3. A = XO= .5 .75 64\fIn Exercises 7-14, let X* = AXk-1, k = 1, 2, ..., for the given A and Xo. Find an expression for Xx by using Eq. (6), as in Example 3. With a calculator, compute X4 and X10 from the expression. Comment on limk-co Xk.EXAMPLE 3 Use Eq. (6) to find an expression for Xx, where xx is the kth term of the sequence in Example 1. Use your expression to calculate Xx for k = 10 and k = 20. Determine whether the sequence {xx} converges. Solution The sequence {xx} in Example 1 is generated by XK = AXk-1, k = 1, 2, ..., where .8 .2 A = .2 .8 and xo = _ ]- Now the characteristic polynomial for A is p(t) = 12 - 1.6t + 0.6 = (t - 1)(t - 0.6). Therefore, the eigenvalues of A are 21 = 1 and 12 = 0.6. Corresponding eigenvectors are u1 [:] and we = [- 1] The starting vector xo can be expressed in terms of the eigenvectors as xo = 1.5u1-0.5u2: XO = 2 - 's [ ] ]- os [ _] ]. Therefore, the terms of the sequence {xx} are given by XK = Akxo = Ak(1.5u1 - 0.5u2) = 1.5Aku1 - 0.5Aku2 = 1.5(1)ku1 - 0.5(0.6)ku2 = 1.5u1 - 0.5(0.6)ku2.\f\f\fIn Exercises 15-18, solve the initial-value problem. 15. u'(t) = 5u(t) - 60(t), u(0) = 4 v'(t) = 3u(t) - 4v(t), v(0) = 1