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Introduction to Linear Algebra - Instructor's Manual (3rd Edition)

By Gilbert Strang

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B) Av 1 =  2 eight four/ 17 2 17 2/ five  √   √   √  1/ 17 2/ five 1/ five u1 =  √  for the column area, v 1 =  √  for the row house, u2 =  √  for two/ five four/ 17 −1/ five  √  four/ 17 the nullspace, v 2 =  √  for the left nullspace. −1/ 17   √ √ 2 1 T T  has eigenvalues σ12 = three + five and σ22 = three − five . A A = AA =  2 2 1 1  1 2 three four T five 20   σ12 seeing that A = on the eigenvectors of AT A are almost like for A. on account that λ2 = √ 1− five 2 is damaging, σ1 = λ1 yet σ2 = −λ2 . The eigenvectors are almost like in part 6.

Fifty four, . fifty five, . forty three, zero. The MATLAB code is E = diag(ones(6, 1), 1); okay = sixty four ∗ (2∗ eye(7) − E − E ); D = eighty ∗ (E− eye(7)); (K + D)\ones(7, 1), (K − D )\ones(7, 1). challenge Set eight. 2, web page 366  −1 1   1 A = −1 zero  zero −1   c       1 ; nullspace comprises  c ;    1 c zero    1      zero  isn't orthogonal to that nullspace.   zero 2 AT y = zero for y = (1, −1, 1); present = 1 alongside area 1, side three, again on facet 2 (full loop). sixty seven  −1 1   three U =  zero −1  zero zero zero    1 ; tree from edges 1 and a couple of.

1 1 1 1 1                 17   −1 1  = −1  = E;  1 1  = L = E −1  1 1 1         −1 1 −1 1 1 zero −1 1 1 1 1 after reversing the order and altering −1 to +1. 18 A2 B = i will be written as A(AB) = I. accordingly A−1 is AB. 19 The (1, 1) access calls for 4a − 3b = 1; the (1, 2) access calls for 2b − a = zero. Then b = a= 2 . five For the five by means of five case 5a − 4b = 1 and 2b − a = zero provide b = 1 6 and a = 1 five and 2 . 6 20 A ∗ ones(4, 1) is the 0 vector so A can't be invertible.

Their elements nonetheless upload to at least one.        . eight . three . 6 . 6 . eight . three    =   = regular kingdom s. No switch while accelerated by means of  . 32  . 2 . 7 . four . four . 2 . 7     eight three four 5+u 5−u+v 5−v         34 M =  1 five nine  =  five − u − v five five + u + v ; M3 (1, 1, 1) = (15, 15, 15);     6 7 2 5+v 5+u−v 5−u M4 (1, 1, 1, 1) = (34, 34, 34, 34) as the numbers 1 to sixteen upload to 136 that's 4(34). challenge Set 2. 2, web page forty 1 Multiply by way of l = 10 2 = five and subtract to discover 2x + 3y = 14 and −6y = 6. 2 y = −1 after which x = 2.

17 those bases will not be designated! (a) (1, 1, 1, 1) (c) (1, −1, −1, 0), (1, −1, zero, −1) (b) (1, −1, zero, 0), (1, zero, −1, 0), (1, zero, zero, −1) (d) (1, 0)(0, 1); (−1, zero, 1, zero, 0), (0, −1, zero, 1, 0), (−1, zero, zero, zero, 1). 18 Any bases for R2 ; (row 1 and row 2) or (row 1 and row 1 + row 2). 19 (a) The 6 vectors will possibly not span R4 (b) The 6 vectors are usually not self reliant (c) Any 4 will be a foundation. 20 self sustaining columns ⇒ rank n. Columns span Rm ⇒ rank m. Columns are foundation for Rm ⇒ rank = m = n. 21 One foundation is (2, 1, 0), (−3, zero, 1).

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