Math 307 Spring, 2005 Hentzel Wednesday, April 20, 2005 Test IV Actual test is Wednesday, April 20, 2005 1. Find an orthonormal basis of the Null Space of the matrix A where | 1 0 1 1 1 | A = | 0 1 1 1 0 | -------------------------------------------- x y z=a w=b u=c | 1 0 1 1 1 | | 0 1 1 1 0 | x -1 -1 -1 y -1 -1 0 z = a 1 b 0 c 0 w 0 1 0 u 0 0 1 V1 V2 V3 -1 -1 W1 = 1 0 0 -1 -1 -1 1 V2.W1 -1 2 -1 -1 1 W2 = V2 - ----- W1 = 0 - --- 1 = (1/3) -2 USE 2 W1.W1 1 3 0 3 -3 0 0 0 0 -1 -1 1 V3.W1 V3.W2 0 1 -1 -1 1 W3 = V3 - ------ W1 - ------ W2 = 0 - --- 1 - ---- 2 W1.W1 W2.W2 0 3 0 15 -3 1 0 0 -15 -5 1 -9 | -3 | 1 0 1 -5 -1 1 1 6 | 2 | ---- 0 - --- 5 - ---- 2 = ---- -3 USE | -1 | 15 0 15 0 15 -3 15 -3 | -1 | 15 0 0 15 | 5 | -1 1 -3 -1 1 2 1 2 -1 0 -3 -1 0 0 5 ------- ----- ------- Sqrt[3] Sqrt[15] Sqrt[40] <=======Solution 2. Find the best parabola fit to the points ( -2,0), (-1, 3), (0,4), ( 1, 2), ( 2,-1). Solution: 1 x x^2 | 1 -2 4 | | co | | 0 | | 1 -1 1 | | c1 | | 3 | | 1 0 0 | | c2 | = | 4 | | 1 1 1 | | 2 | | 1 2 4 | | -1 | | 1 1 1 1 1 | | 1 -2 4 | | co | | 1 1 1 1 1 | | 0 | | -2 -1 0 1 2 | | 1 -1 1 | | c1 | | -2 -1 0 1 2 | | 3 | | 4 1 0 1 4 | | 1 0 0 | | c2 | =| 4 1 0 1 4 | | 4 | | 1 1 1 | | 2 | | 1 2 4 | |-1 | A = {{1,-2,4},{1,-1,1},{1,0,0},{1,1,1},{1,2,4}}; B = {0,3,4,2,-1}; Inverse[Transpose[A].A].Transpose[A].B 131 3 15 --- + -(--)x -(--) x^2 35 10 14 p1 = Plot[131/35 - 3/10 x -15/14 x^2,{x,-2,2}]; p2 = ListPlot[{{-2,0}, {-1, 3}, {0,4}, { 1, 2}, { 2,-1}}]; Show[p1,p2]; 3. Write down the solution to the differential Equation | y1 | | 3 0 0 0 0 0 | | y1 | | y2 | | 0 3 1 0 0 0 | | y2 | | y3 | = | 0 0 3 1 0 0 | | y3 | | y4 | | 0 0 0 3 1 0 | | y4 | | y5 | | 0 0 0 0 3 0 | | y5 | | y6 | | 0 0 0 0 0 4 | | y6 | --------------------------------------------------------- Solution: 1 0 0 0 0 C1 e^(3t) 0 1 1+t 1+t+t^2/2 0 C2 e^(3t) 0 0 1 1+t 0 C3 e^(3t) 0 0 0 1 0 C4 e^(3t) 0 0 0 0 1 C5 e^(4t) ------------------------------------------------------------------ -1 4. Find a matrix P such that P A P is in Jordan canonical form where | -3 -5 -4 -4 | A = | 1 3 0 0 | | 5 6 7 5 | | -3 -4 -3 -1 | Hint: A has 0 and 2 as eigen values. Solution: eigenvalue 0 A = {{-3,-5,-4,-4}, { 1, 3, 0, 0}, { 5, 6, 7, 5}, {-3,-4,-3,-1}}; | 1 0 0 -3 | | 3 | | 0 1 0 1 | Use | -1 | | 0 0 1 2 | | -2 | | 0 0 0 0 | | 1 | A-2I = {{-5,-5,-4,-4,r}, { 1, 1, 0, 0,s}, { 5, 6, 5, 5,t}, {-3,-4,-3,-3,u}}; | 1, 0, 0, 0, (-5 r - s - 4 t)/4 | | 0, 1, 0, 0, (5 r + 5 s + 4 t)/4 | | 0, 0, 1, 1, (-r - 5 s)/4 | | 0, 0, 0, 0, (r + s + 2 t + 2 u)/2 | | r | | (-5r -s-4t)/4 | | 0 | | s | has solution | ( 5r+5s+4t)/4 | + a | 0 | | t | | (-r-5s)/4 | | -1 | |(-r-s-2t/2| | 0 | | 1 | r=0 s= 0 t=-1 r=1 s=-1 t=0 | 0 | | 1 | | -1 | | 0 | | -1 | | 0 | 0 <=== |-1 | <===== | 0 | <======= | 1 | | 1 | | 0 | | 0 | P = {{ 0, 1,-1, 3 }, { 0,-1, 0,-1 }, {-1, 0, 1,-2 }, { 1, 0, 0, 1 }}; A = {{-3,-5,-4,-4}, { 1, 3, 0, 0}, { 5, 6, 7, 5}, {-3,-4,-3,-1}}; MatrixForm[ Inverse[P].A.P] | 2 1 0 0 | | 0 2 1 0 | | 0 0 2 0 | | 0 0 0 0 | 5. Tidbits. 1. Write down the definition of the determinant. Sum Sgn(pi) a1p1 a2p2 ... anpn All p | x 5 1 | 2. Solve Det|x+1 2 0 | = 2 | 2x 1 0 | |1 5 1 | |0 5 1 | x |1 2 0 | + |1 2 0 | = 2 |2 1 0 | |0 1 0 | -3x + 1 = 2 x = -1/3. Det[ {{x,5,1},{x+1,2,0},{2x,1,0}} /. {x->-1/3}] | 1 | 3. If A has an eigen value of 3 with eigenvector | 2 | | 3 | 2 Find an eigen value and vector of A + 5 A. | 1 | 24, | 2 | | 3 | T T 4. If A and B are orthogonal matrices, what is A B B A The identity matrix. 5. Find a unit vector in the direction of | 5 | | 3 | | 1 | | 5 | 1/Sqrt[35] | 3 | | 1 | 6. Express {1,1,1} as a linear combination of the orthonormal vectors | 2/3 | |-1/3 | | 2/3 | | 2/3 | | 2/3 | |-1/3 | | 1/3 | |-2/3 | |-2/3 | (5/3, -1/3, -1/3} 7. Write the solution for x using Cramer's rule. Do not expand the determinants. x + y - z = 61 x - y = 92 x + y + z = 88 8. If A is a 3x3 matrix with determinant 3 -1 What is the determinant of 2 A 8/3 9. Find x so that (x,1,1) and (3,1,1) are orthogonal. x = -2/3 T | 3 2 1 | 10. Find | 1 0 1 | | 0 1 2 | | 3 1 0 | | 2 0 1 | | 1 1 2 |