Final10Prob1_5_11_15

# Final10Prob1_5_11_15 - O with LinearAlgebra Problem 3 4 1 0...

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Unformatted text preview: O with LinearAlgebra : Problem 3 4, 1, 0 , K 1, 1 , 1, 1, 2 2, O A d Matrix A := ; 4 1 0 (1) 1 1 2 K 1 1 2 Im Maple I denotes the imaginary number i. Hence we denote the identity matrix by II. O II d DiagonalMatrix 1, 1, 1 1 0 0 II := O B d A K3\$II 1 B := O C d MatrixMatrixMultiply B, B K K 1 1 C := 3 K K 2 2 1 (4) 2 3 3 K 1 1 0 1 (3) K K 2 2 0 1 0 0 0 1 (2) 1 K 1 We solve the system BX=0. The row reduction leaves only the first row, producing -x-y+z=0. Hence x=s, y=t, z=s+t yielding X=sX1+tX2 where O X1 d Matrix 1 , 0 , 1 ; X2 d Matrix 0 , 1 , 1 ; 1 X1 := 0 1 0 X2 := 1 1 Since B2X1 = 0, et\$AX1 = e3\$t ICt\$B X1. Hence O x1 d exp 3\$t \$ X1 Ct\$MatrixMatrixMultiply B, X1 ; x2 d exp 3\$t \$ X2 Ct \$MatrixMatrixMultiply B, X2 ; e3 t 1 Ct x1 := K3 t t e e3 t (5) e t x2 := e3 t 1 Kt e O Y1 d Matrix 1 , K , 2 3 3t 3t (6) ; y1 d exp 3\$t \$Y1; 1 Y1 := K 3 2 e3 t y1 := K e 3 3t (7) 2 e3 t The general solution is O X d c1\$x1 Cc2\$x2 Cc3\$y1 c1 e X := 3t 3t 1 Ct Cc2 e3 t t Cc3 e3 t 3t K e t Cc2 e c1 1 Kt K3 c3 e 3t (8) c1 e3 t Cc2 e3 t C2 c3 e3 t Problem 4 O A d Matrix 4, 1, 1 , 0, 4, 2 , 0, 0, 4 A := O B d A K4\$II 0 1 1 B := 0 0 2 0 0 0 O EA d exp 4\$t \$ IICt\$B Ct2\$ MatrixMatrixMultiply B, B 2 e4 t e4 t t e4 t t Ct2 EA := 0 0 We find the general solution: O X d Matrix c1 , c2 , c3 e4 t 0 2 e4 t t e4 t (11) (10) ; 4 1 1 0 4 2 0 0 4 (9) ; x d MatrixMatrixMultiply EA, X c1 X := c2 c3 e c1 Ce t c2 Ce x := 4t 4t 4t 4t t Ct 2 c3 (12) e c2 C2 e t c3 e4 t c3 4t ...
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Final10Prob1_5_11_15 - O with LinearAlgebra Problem 3 4 1 0...

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