Find a basis for the eigenspace corresponding to the eigenvalue. - 12 J A = 14 ,2=3 3 -39 A basis for the eigenspace corresponding to 2 = 3 is {} (Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)This exercise refers to P, with the inner product given by evaluation at - 1, 0, and 1. Compute | pl| and |q|, for p(t) = 6 - t and q(t) = 4+ 312. IIpl| =(Simplify your answer. Type an exact answer, using radicals as needed.) llall =(Simplify your answer. Type an exact answer, using radicals as needed.)Find the characteristic polynomial and the eigenvalues of the matrix. 3 2 2 - 1 The characteristic polynomial is (Type an expression using ) as the variable. Type an exact answer, using radicals as needed.) Select the correct choice below and, if necessary, fill in the answer box within your choice. O A. The real eigenvalue(s) of the matrix is/are (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.) O B. The matrix has no real eigenvalues.Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix. 6 1 - 3 W 6 0 ; 2 =2,3,5 6 -6 5 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. O A. 20 0 For P = D = 0 30 0 05 (Simplify your answer.) O B. The matrix cannot be diagonalized.Solve the initial value problem x'(t) = Ax(t) for t 2 0, with x(0) = (3,2). Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by x" = Ax. Find the directions of greatest attraction and/or repulsion. 6 A= - 3 4 Solve the initial value problem. x(1) = Classify the nature of the origin as an attractor, repeller, or saddle point. Choose the correct answer below. O Repeller O Saddle Point Attractor Choose the correct graph below that represents the direction(s) of greatest attraction and/or repulsion. O A. O B. O c. OD.0 - 5 5 Find the distance between u = -4 and Z = 1 4 7 The distance between u and z is (Type an exact answer, using radicals as needed.)\fFind the closest point to y in the subspace W spanned by u, and u2. 7 O B. O C. O D. 20 OA. 8 9 N 3 16 AThe set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W. Assume the vectors are in the order x, , X2. 3 4 9 10 6 4 The orthogonal basis produced using the Gram-Schmidt method for W is { (Type a vector or list of vectors. Use a comma to separate vectors as needed. )Find (a) the orthogonal projection of b onto Col A and (b) a least-squares solution of Ax = b. 14 13 A = -1 8 ,b= - 1 1 4 a. The orthogonal projection of b onto Col A is b = (Simplify your answers.) b. A least-squares solution of Ax = b is x = (Simplify your answers.)1 Given A = 2 2 9 eigen value ( A ) = 3 3- 39 We know I = O O , AI = 3 O 0 O AI = 3 0 A - dI = 4 - 1 2 3 O 0 O 3 D 2 1 4 O 30 O O 3-3 9 03 A- dI = 4- 3 - 1- 0 2 - 0 2 - 0 1- 3 4- 0 2 3-0- 3-0 9 - 3 3 1 = [ 1 - 1 2] , R2 = [ 2 -24], R3 =1 3- 35) Change R2 into, R2 - 2R, , 2 R, = [2 - 2 4 ] R2 - 2RI = [ 2 - 2 4] - 52 - 2 4 ] = 0 00 Now R2 is to 0 0] Change R3 into R3 - 3R1, 3R, = [3 - 3 6] R3 - 3 R, = [3 0 ) - 3 6 ] - [ 3 - 3 6 ] =[0 0 0 ] Now A - di = - 1 0 Let v be eigen vector V= Now ( A - I ) V = O 1 x 2 . + 4 x ( 1 ) + 2 x2 0 oxx toxy + 0xz - ) a - 4 +22 = 0 0 x x to xy to * ? Let 4 = 85 z z t *#M-S+ 26 20 a = S- 2t, V = 15 - 2tLet A = For Finding the 2 . character stic Polynomial we need to find determinant of A->I I = [09) , AI = [! ! ] = [do ) A- dI = [- 3 3 ] - [8 8 ] = [- 3-2 1-0 - 1 -2 dot ( $A- dI ) = ( 3-2) ( - 1-2) - (2) ( 1) = 3 + 3 1+ 2+ 2 2- 2 =2792+1 It is characterstic Polynomial, for finding I equate equation to zero d2 + ud +1 = 0 , add 3 on both sides d' tud+ 4 = 43, ( X ) 2 + 2. d . 2 + ( 2 ) = 3 ( d+ 2 ) = 3, = ) d+ 2 = +53, 1= 2+53 1 = 2+53, 2 - 53 3. A = 6 - 3 2 = 2, 3, 5, 1 = 101 010 6 6 5 Is for d= 2 is 2 O ow o 2 1 1 = 3 WOO NOO 0 O d = 5 is 51 =1 5 0 050 0 5 A - 2I 16 1 - 3 2 4 WN - 1 O 0 O 6 6 -3 0 5 O O 6 - 6 3for finding eigen vector ( A - dI ) V 1 = 0 Let y= ( A - 2I ) VI = 0 QC 2 Plenty , - 32 , 4 1 - 3 O O - 3 O 6 x 1 - 34, to 6 0 6 3 6 21 - 64, +37 , 6 4X1 , + 4 - 321 =0 0 6211 - 34, to = 0 ( 2) =) 6 x, = 34, = ) 4, = 2x, 6 *, - 64, +321= 0 (3) , keep y , in O 421 , + 2 71, - 3 2 , =0 = 7 6 #1, = 321 =) 281 =21 SO 4,4, 21 = 0, age , 271,, 27 VI = = ) Let X 1 = 1 ; Vi = NN (A - 3I) V2 = 0 6 1 - 3 3 0 16 6 0 O 3 6 5 O 2 - 6 3 1 3 let V 2 = :2 2 - 4 6 - 2 CRIS * 2 NOW 7 = ( :) 2 2 3 212 + 4 - 32 2 20 1) 6 01 2 - 4 4 2 to = 0 27 371 2 = 242 - keep in O 6 x 2 - 6 4, + 2 2 2 = 0 (3 2 4 , + 4 , - 3 2 , = 0 = ) 4, = 22212 1 42, 2 2 = X 2 1 3 2 2 2 V2 = 3/2 #12 , Let 72 = 1 1 3 / 2 3 /2 X 2 312 A - 51 - 6 5 O O 1 +1 -3 6 O O 50 6 - 6 O O O 5 - 6 6 -6 5 ( A - SI ) Vg = 0 2 23 + 1 72+ 4 , - 32, = 0 (1) 6 6 6 6*3 - 6 4 2 2 0 68 x3- 642=0 *3 = 42 , keen in O 323 = 2x13 213 + 4 2 - 37 7 = 0 23 = 027(3) ) 3 X31 4 2 1 27 = 23, 263, 0 2 X2 3 = 213 , let 2321 [ V 2 V 2 V - eigen vector Vz - - 2 3 /2 2 3/2 2/3 4 . AEA . A = 9 6 9 A = 6 - 6 - 3 - 6 - 6 9 6 * 6 + 9 x - 3 6 x9 + 9 x -6 - 3 x 9+ ( - 6 ) X (- 6 ) 0 9 L- 3 x 6+ (- 6 ) *(- 3) nowA 0 7 = 9560 7 = 91 [9x6 +0 9x9+ 0 43 ox9 tax Z 9 x - 3 = - 27 - su = 27 [ 2 3 2 ] Given 2' (t ) = A . x (t ) =) 2' ( t ) = A (t ) Integrate on both sides J X " (62 = [ A = ) In (x(=)) = Atb+ c Put t = 0 , In (x (0) ) = A(o )+ C C = en (x 10)) =) In(c(t))= At + In(2101) In ( x( t ) ) - In (x109 ) = At In ( 21 () ) = At [ en a - lub = in =] * ( 0 ) raise Ce Power on both sides X (t ) = eAt = ) x ( t ) = eat 21 ( 0 ) x (0 ) e" expansion = 1+ - - 11 21 0 At = 1+ At + ( At) 2 1! + - . 21 in Matrices expansion is e At = I + At + (At ) + .. .. 21We know AL 9 I At = 9 , At = [ 6 t at ] - 36 -6t ( At ) 2 = AZt2 - 9It2 = 159t 2 0 9/ + 2 2 1 21 2 1 9 6 2 12 At 6t at7. + -3t - 6t 2 9 t2 + - - 2 1+6t+962 9t neplating other - 3t 1 - 6t + 9 + 2 2 teams - 3+ 3 1- 6+ + 9( 2 2 2 *( t ) = e - x (0 ) [3+ 18+ + 27 1 2 + 18t E 27 6+ 36( +3 - at + 2 - 126 + 962 at - 216+2 - To find attractor, repeller or saddle point we need to find eigen value d, for finding that we need to find determinant of ( A-dI) and equate it to zero A - dI = [6 - 3 =[6- 9 - 3 - 6 -d determinant ( A - d I) = ( 6-2) (- 6 -d) - (-3) 9 20 - 36 + 1 2- 6 2 +6 2 +27 = 0 = 129 =0 d = 31 - 3eigen values (1) are 3, -3 one is positive value and other is negative value repelles ) if both I is positive then it is attractor - If both I is negativen then it attractor - As one is positive & one is negative the origin act as saddle point Graph D S 5. u = FO 2 = - 4 7 For finding distance dluiz ) we use Formula d ( M, Z ) = VE ( Mi- zi ) 2 1=1 d( Mit ) = ( Mi - zi ) 2 = ((4,-21 ) +( M2 - Zz ) tellit) M 20 12= -4 13 = 4 212-5 22 = -1 23 = 7 MI- Z 1 = 0 -(-5) =5 12 - 27 = -4 - (-1) = 0-3 49 - 27 - = 4-7 =-3 d ( 14, z ) = 1 5 2 +( -3 ) 2 + ( - 3 )2 = 1 25 + 9+9 = 543 6. 24 wu = 2 10 10For finding the Orthogonal Projection y onto it is given by formula Orthogonal Projection u (4) = you 1141 12 11 4114 - E( 41 ) 2, you = =(y; Mi ) = 4,4, + 4242 ( 1 #1 11 = 542 + 42 = J 27102 = J104 y . u = [ - 7 8 ] [ 3] 2 - 24x 2 + 10 X 10 = -48 +100= 52 Orthogonal Projection (4) = J. 4 52 = 52 104 7 . 8 U 2 = 2 To find closet point to y in Subspace W We use Orthogonal Projection for subspace Projection ( 9) = A. ( ATA ).( ATy ) where A = [4, 12 ) = [ , ? ] A = AT . y = 8 +0 - 2 ) - 18 16 - 2 + 4A . A = [ 1 9 =] [ : 3 ] = 1 X1 + 0 + ( - 1 ) * ( - 1 ) 1 * 2 + 0 + ( - 1 ) * 2 2 O 9 2x1 + 1 x0+ 2X(-1) 2x 2+ 1X 1+ 2x2 ( AT A ) = ( det ( ATA - 1) then K = 1 d - b ( ad-bc ) L-c a (ATA ) = 18 [ 9 2 ] ad- be is determinant ( AT A ) 1 ( ATy ) = = [8 2 ] [ /] = 9 x 6 + 0- 18 0+ 1 8 * 2 = 18 [ 2 X] = [3] A. ( TATA ) ( ATy ) = 1X3+ 2 x27 OX 3 + 1 x 2 -1X3+ 2x2_ The closet 3 + 4 Point 0 + 2 = [ 7 7 - 3+4 18 . 10 4 7 First normalise the vector X, it can be given by X, = XI 11xill 911Xill = V XitXTX/13 X11= 3 212 =-9 9 = the = V 32+1- 9)2762 = 19+ 81+36 = 5126 3 - Now Find Orthogonal Projection X, outoxy Proj ( X 2 ) = X 2. X, X/ 21 = - 4 11 x 1 1 12 1/22 = 10 X23 = -4 Xz. XI = #11: 721 + X12. X22 + 2413. X23 = 3 ( 4) + 1- 9) ( 10) + (6) (-4) = - 12-90-24 =- 126 ( 1 X11/ 2 = (5126 ) 2 = 126 Proj(X 2 ) = - (126 ) XI 126 Subteact the projection from X, Let the result be B B = Now normalise the above vector BB - B 11B11 = 56 B = 1 [ ! ] = - 1/ 56 1/ 56 21 56W is (x B 3 J126 A . 9 6 126 9 . A = b = 13 8 4 Orthogonal Projection of b into col A is given by Proj ( b ) = A ( ATA ) ) ( AT b ) A T =[ # - 1 1 A T A + +1] [ + 1 4 7 4 8 4 - 4 8 4 ) |7 4 Al = AA = 1X 1+ (-1)X (-1) + 1 x1 1X4- 188+ 4X01 4X1 - 1 x8+ 4 X1 4 X 4 + 8 x8 + 4X4 ( A TA ) = 3 0 0 96 , ( ATA ) = 1 96 3 X96 0 3 ( ATA ) = 0 1/96 , ( AT b ) = [ " 3 4 ) [ "; ] ( AT b ) 1 X 13 + ( - 1) * (1 ) + 1 X 1 4 x 13+ 8x (-1) + 4X1 = [ 45 ]A. (AT A ) ( ATb ) = [ 113 8 1/96 48 1/3 x 15 + 0 8 0 + 1X 48 8 4 96 = 1x5+ 4x 1 7 - 1 x5+8 x 1 2 1X5+ 4x1 For least squares Solution of Ax =b * = ( ATA ) 1 ( ATb ) From above we have ( ATA ) ( ATb ) x = [ 50.5) 7 10 . ) P ( t ) = 6 - t . Points are t = -1,0, 1 P ( 1 ) = 6-1 = 5, P(0) = 6-0=6, P ( - 1 ) 2 6 - ( - 1) = 7 ( 1PI) = 5( P(0 ) 1 2 + PU) 2 +(1-1)) = 15 +6 + = 149+ 36+25 = 1 110 2(t ) = 4 + 3 + 2 t = 01 -1, 1 2 1 0 ) = 4 + 3 ( 0 ) 2 = 4, 9 ( 1) = 4+ 3(1) = 7 2 ( -1 ) = 4 + 3 (1)2 = 7\f