2. An economy has three markets with supply and demand...

2. An economy has three markets with supply and demand functions for the three goods given by q1^S = -20 + 3p1 - p2 q1^D = 255 - 2p1 - p3 q2^S = -125 + 2p2 - p1 - p3 q2^D = 200 - p2 q3^S = 4p3 - 2p1 q3^D = 225 - 2p3 - p1 (a) (2 pts) Find a matrix equation of the form A [p1 p2 p3] = b that will allow us to solve for the equilibrium prices. (b) (4 pts) Solve for the equilibrium prices by using Gaussian Elimination.

Transcript

00:01 In this problem we are provided with the supply and demand equations.

00:07 We have q1s to be equal to negative 20 plus 3 times p 1 minus p 2.

00:17 Q1 d equals to 2505 minus 2 times p 1 minus p 3.

00:25 Q2 s equals to negative 125 plus 2 times p 2 times p 2.

00:32 Minus p1 minus p3 q2d equals to 200 minus p2 2 q3s equals to 4 times p3 minus 2 and q3 d equals to 22 25 minus 2 times p3 minus p 1 so here in subpart a we are asked just form the matrix equation a x equals to v so that we saw can solve for the equilibrium prices pe so here in order to get the equilibrium prices the demand must equal to the supply at this price so let us equate each of these and let us form the equation in the standard form so equating each of the qis with the respective qid for i equals to 1, 2 and 3, we have the first equation to be 5 times p 1 minus p 2 plus p 3 equals to 275.

01:49 The next equation is p 1 minus p 2 sorry p 1 minus 3 p 2 plus p 3 p 3 plus p 3 equals to negative 325 and the third equation is p1 minus 6 times p3 equals to negative 225 so now the matrix a equals to the coefficients so we have the coefficients to be 5 negative 1 1 1 negative 3 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 and the matrix 6 and the matrix matrix x is the matrix of unknowns.

02:34 So the unknowns here are p1, p2 and p3 and this equals to b, which is the column vector of the constants on the right hand side.

02:46 So that is 275 minus 325 and negative 225.

02:54 So therefore this is the required answer for subpart a.

03:06 In subpart b we are asked to solve the matrix equation.

03:12 So first we form the augmented matrix.

03:15 We have 5 negative 1 1, 275.

03:22 1, negative 3 1, negative 3 1, negative 325, 1 0 negative 6, negative 225.

03:34 So this is the augmented matrix.

03:36 When we augment a and b.

03:40 Now let us perform the row operation.

03:42 R1 implies r1 divided by 5.

03:45 So we get the equivalent augmented matrix 1, negative 1 over 5, 1 over 5, 55 and the other two rows remain the same since we have not performed any row operation for the rows 2 and 3...

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