Question

please tell thr output of the manufacture and the output of thr energy too

An economy is based on three sectors, agriculture, manufacturing, and energy. Production of a dollar's worth of agriculture requires inputs of $0.30 from agriculture. $0.30 from manufacturing and $0.30 from energy. Production of a dollar's worth of manufacturing requires inputs of $0.30 from agriculture. $0.20 from manufacturing, and $0 20 from energy. Production of a dollar's worth of energy requires inputs of $0.20 from agriculture, $0.30 from manufacturing, and $0.30 from energy Find the output for each sector that is needed to satisty a final demand of $64 billion for agriculture, 567 billion for manufacturing, and 540 billion for energy The output of the agricultural sector is billion dollars. (Round the final answer to three decimal places as needed. Round all intermediate values to six decimal places as needed)

Answer 1

Solution: production of dollar for agriculture, inputs are from agriculture = 80.30 from manufactuning = $ 0:30 from energy = $0.80 dollars production requires for manufacturing inputs are from agri = 10.30 from manu = $0.20 from energy = 80.20 dollars production from energy requiry inputs are from agnia $0.20 from manu = 80.30 from energy = $0.30 • output of sector to satisfy demand of 8 bybillim 867 billion for forogn manu, $40 billion for energy is:

The technology matrix is: m = LA ME 10.30 oigo 0.80 10:30 0.20 0:30 10.30 0.20 0:30 3 IT The demand is: 1 I r by I 67 no - required output for agriculture =X for manufacturing = x2 output required output required for energy = X3 xxx fost calculating (I-m5 : - Toi3o 0.30 L0.30 0.30 0.20 0.20 0.30 0.20 0.30 Tloo o loo, I- M

(~0.30 0-0130 0-0130 0-0.30 0-01209 1-0.20 0-0.30 0-0.20 10.30 = Corto -0.30 -0.30 -0.30 0.80 -0.20 0.20 -0.30 0.70 and | ~~) = 6+9)/C0.fo) (0-70) – 6-0 30/10 vey - 60.30) [C-0.30) (0.4)~6-0-39) €0-20) - 60:20) 60-80) (-0.20) -(-0.30)(0-0) = 6.70) (0.56-0.06) -(-0130) [10:27-001] –60.20) 0.06+0:24] = 0.70) (0:50) - (-0.30) (-0-30) - (-0.20)(030) = .0.35 -0.09 +0.06 0.32)

Now finding a dj (I-M). cofactor of all = (0.80) Co.to) -(0.30) €0.20) = 0056–0.06 = 0.50 Cofactor of 912 = = fo.so] (0.7) - (-0.30) (-0.30) -0.21-0.09 -0.20) cofactor of 013 3 = 60.30) 60.20) – (-0.30).(.80) +0.06+ 0.24 0.30) Cofactor of all = = (-0.30) (0.2o)-(-0.20) (-0.20) -0.24-0.04 0.25 0.70) Co.70) – 60.807 6-0.20) 0.99-0.06 = 10:43] cofactor of 922 = = Cofactor of 923= (0.7o) (-0.20) - (20:30) (-0.30) = -0.14-0.19 = 70.23 (-0.30) (-0.30)-(-0.20) Coedo) Cofactor of a 3 1 = = 0.09 tool6 = 0.25

GRacibr 4032 = (50)(-० •3०) - (-०.१०) (-0.26) = -2 -- 0.06 = ( 0.27 cofacturt-a33 = Co. iv) (0. Po) - (-0 •४०)(-०-36) _ 056-0.०१ - ad (T-m) = 0.50 -0.30 0.30 - • ०.५3 -0 • 23 | Of -00.47 _ 0:50 -0• 30 । 0.30 – 0.५ ०.५3 -0.23 ०.४ -०. 22- 0.47 am) li-mi ag

Р оза - 0.30 (о 30 -о-х р.2 o •43 -o 17- —o.23 отч? Ј в •6 -0.8 o-18 -o •43 1 -3Ч - o •89 о •93 - 0 -- (46 Therefore ae have х= (т-м) p Г. 1. -op o.T8 -0.92 (3 --0-89 о93 -0.7 1.46 С ~ ohy -o + t2 Уt : 0 — һ9 39. S-| оhу bg • o — 62 Х 48 • 1 +h23. -o— гчh 2.9h:) + t9Xtte - h9 2 8Ъ-о- = 878 - 8 bill: y ) 8-3 - 39 billiy (8 то•3s bilm