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Applied Finite Mathematics HW Solutions 35

# Applied Finite Mathematics HW Solutions 35 - EXERCISES 3.2...

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Unformatted text preview: EXERCISES 3.2 . 33 - EXERCISES 3.2 _ . (1(1) 1(—3) 1(0)) (1 —3 0) 8. AD: 3(1) 3(—3) 3(0) : 3 —9 0 . 2(1) 2(_3) 2(0) 2 _3 0 7. DA = (1(1) — 3(3) + 0(2)) = (—8) - . (1(1)+0(3)+0(2)) (1) ' 8. EA: 0(1)+2(3)+0(2) = 6 _1(1) + 0(3) + 3(2) 5 9. Since the number of columns of A is not equal to the number of rows of 'E, this product is not deﬁned. ( 1(3) + 0(0) + 0(0) 1(2) + 0(2) + 0(0) 1(—1) + 0(0) + 0(0) ) 10. ER = 0(3) + 2(0) + 0(0) 0(2) + 2(2) + 0(0) 0(-1) + 2(0) + 0(0) _ —1(3) + 0(0) + 3(0) —1(2) + 0(2) + 3(0) —1(.—1) + 0(0) + 3(0) 3 2 —1 = 0 4 0 ' —3 —2 ' 1 11. Since FE and 0 do not have the same dimensions, they cannot be added. 12. Since B is an identity matrix 23K = 2K = (3 g). 13. Since B is an identity matrix 2K B 2 2K = (i g). 1 0 0 ‘- 14. ATE=(1,3,2) ( 0 2 0) =(1(1)+3(0)+2(—1), 1(0)+3(2)+2(0), 1(0)+3(0)+2(3)) —1 0 3 = (—1: 616) - 15.199: (1(1)43(0)+0(0), 1(0) —3(0) +0(0), 1(0) —3(0) 40(1)) = (1,0,0) ‘ 16. Using EF from Exercise 10. ' — 3 2—1 100 (EF)G.= 0 4 0 000 _ _—3 —21 001 ( 3(1)+2(0)—1(0) 3(0)+2(0)—1(0) 3(0)+2(0) —1(1) ) ( 3 0 —1) = 0(1) + 4(0) + 0(0) 0(0) + 4(0) + 0(0) 0(0) + 4(0) + 0(1) = 0 0 0 43(1) — 2(0) + 1(0) —3(0) 4 2(0) + 1(0) —3(0) —2(0) + 1(1) —3 0 1 17. Since matrix multiplication is associative, this matrix is equal to that in Exercise 16. . (_1(—1) + 2(—4) + 3(—7) —1(2) + 2(5) + 3(8) _1(3) + 2(6) + 3(9)) 18. J2 —4(41) + 5(—4) + 0(—7) —4(2) + 5(5) + 0(8) "4(3) + 5(6) + 6(9) _7(~1) + 8(—4) + 9(4) 4(2) + 8(5) + 9(8) —7(3) + 8(6) + 9(9) —28 32 36 = —58 65 72 #88 98 108 (3(1) + 2(0) — 1(0) 3(3) + 2(1) — 1(0) 3(2) + 2(4) — 1(0) 3(5) + 2(3)-— 1(2)) - 19. F0 = 0(1) +2(0) + 0(0) 0(3) + 2(1) + 0(0) 0(2) + 2(4) _+ 0(0) 0(5) + 2(3) + 0(2) . 0(1) + 0(0) + 0(0) 0(3) + 0(1) + 0(0) 0(2) + 0(4) + 0(0) 0(5) + 0(3) + 0(2) 3111419 =0286 0000 _ __: _1_. u L. . .L- a I'_:-..-_.~:h. .91 Innank. nudism» qu‘lcrl'lur mmuhrdim in Mr! or in ﬁll]. is 5111‘le mohibited. ...
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