Let C be an n × n consumption matrix whose column sums are less than 1. Let x be the production vector that satisfies a final demand d , and let Δ x be a production vector that satisfies a different final demand Δ d .

a. Show that if the final demand changes from d to d + Δ d , then the new production level must be x + Δ x. Thus Δ x gives the amounts by which production must change in order to accommodate the change Δ d in demand.

b. Let Δd be the vector in Rn with 1 as the first entry and 0’s elsewhere. Explain why the corresponding production Δx is the first column of (I – C)–1. This shows that the first column of (I – C)–1 gives the amounts the various sectors must produce to satisfy an increase of 1 unit in the final demand for output from sector 1.

41*67/2=1373.5

step-by-step explanation:

i just used a calculator

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.