ECON Homework #2 - ANSWERS. Y C I = G 0 b(1 t)y +C = a. ky li = M 0

Transcription

1 ECON Homework #2 - ANSWERS 1. IS-LM model in a closed economy: a Equilibrium in the real goods sector: Y = C + I + G National income = Total planned expenditures. Equilibrium in the monetary sector: M d = M s Money demand = Money supply. b At the equilibrium, the goods market satisfy: Y = C + I + G C = a + b1 ty I = d ei G = G 0 At the equilibrium, the monetary market satisfy: M d = M s M d = ky li M s = M 0 Y C I = G 0 b1 ty +C = a ky li = M 0 M d = M s = M 0 I +ei = d G = G 0 When we combine the 2 sectors at the equilibrium, we have the following system: Y C I = G Y b1 ty +C = a b1 t C I +ei = d e I = ky li = M 0 k 0 0 l i G 0 a d M 0 c To answer, we need to calculate the determinant of the coecient matrix found in b. Since it is a 4,4-matrix, we need to use the expansion in cofactors: we develop 1

2 according to the third row that contains 2 zeros any other expansion acceptable! det b1 t e k 0 0 l = +1 det b1 t 1 0 e det b1 t 1 0 k 0 l = l det ek det b1 t = l[b1 t 1] ek 0 by assumption k 0 0 where the third equality follows from an expansion according to the third column for the rst 3,3-matrix and an expansion according to the third row for the second one. Since the determinant of the coecient matrix of the system is nonzero, Cramer's rule applies and we have: Y = G a det d 0 1 e M l l[b1 t 1] ek The numerator is equal to: G G det a d 0 1 e = a det 0 1 e + det d 1 e 0 0 l M 0 0 l M l 1 e 1 e = a 1 det + G 0 det + det 0 l 0 l = em 0 la + d + G 0 d M 0 e l 2

3 Finally: Y = e ek + l[1 b1 t] M l 0 + ek + [1 b1 t] a + g + G 0 e Note: Coecient is called money-supply multiplier. ek+l[1 b1 t] l Coecient is called government-expenditure coecient. ek+[1 b1 t] 2. We have a linear system with 3 equations and 3 unknowns. So we solve it by gaussian elimination. Note that since there are 2 parameters a and b, we have to make sure transformations of the system are valid and yield to an equivalent system. We will likely have to distinguish dierent cases. x + ay 21z = 2 L2 3x + 7y + az = b L ay 18z = 3 L1 + L2 y + 9 az = 3 b 3L1 L3 y + 9 az = 3 b L ay 18z = 3 L2 Next, I should get rid of y in L3. However, to do so, I have to multiply L2 by 2 + a: this is a valid transformation only when 2 + a 0. We then have 2 cases. - case 1: a = 2. The system writes: y + 11z = 3 b L3 + 18z = 3 L2 x = 67/6 2b y = b 29/6 z = 1/6 There is a unique solution for any real number b. There is 0 degree of freedom. - case 2: a 2. The system writes: y + 9 az = 3 b L2 + [ 9 a2 + a + 18]z = 2 + ab al2 L3 Next, I should be able to deduce z as a function of the parameter from L3. But I can only do that if the coecient of z in L3 is nonzero. First we nd the values of a for 3

4 which this coecient is 0: 9 a2 + a + 18 = 0 a 2 7a = 0 aa 7 = 0 a = 0 or a = 7 Again, we have several cases to distinguish. - case i: a = 0 or a = 7 ie coecient of z in L3 is null. We consider L3 alone for now L3 0 = 2 + ab 3 3 b = 9 2 when a = 0 OR b = 10 3 when a = 7 Here again we have 2 cases: - a = 0 and b 9/2 OR a = 7 and b 10/3 ie L3 does not hold: the system does not have a solution. - a = 0 and b = 9/2 OR a = 7 and b = 10/3 ie L3 holds and can be deleted b/c it states 0=0: the system becomes y + 9 az = 3 b L2 x = 2[9 az + b 3] 3z + 1 y = az + 7 2b x = 2[9 az + b 3] 3z + 1 y = 9 az + b 3 There are innitely many solutions. We have been able to express them as functions of 1 variable only z, so it means that there is 1 degree of freedom. - case ii: a 0 and a 7 ie coecient of z in L3 is nonzero. The system writes: x = 1 2y 3z y = 9 az + b 3 z = 2+ab a2+a the above solution can be simplied more... There is a unique solution, so 0 degree of freedom. x = a2+ab 3 39 a 18 9 a2+a y = 9 a2+ab 3 39 a 18 9 a2+a + b 3 z = 2+ab a2+a 2b ab a2+a To conclude, we can summarize the dierent cases: - when a = 0 and b = 9/2 or when a = 7 and b = 10/3, there are innitely many solutions and one degree of freedom. - when a = 0 and b 9/2 or when a = 7 and b 10/3, there is no solution. - when a 0 and a 7, for any b real number, there is a unique solution and zero degree of freedom. 4

5 3. a M I = α α α 1 The determinant is obtained after an expansion in cofactors according to the rst row other expansions possible!: 1 1 M I = α α α α = α 1 b When α = 1, we have: x +y +z = x y +z = 0 Mv = v x y z = y x 2y z = 0 x +y +z = z x +y = 0 x +2y +z = 0 L2 y +z = 0 L1 x +y = 0 L3 x +2y +z = 0 y +z = 0 y +z = 0 L1 L3 x = 2y z y = z x = z y = z For any real number z, the vectors z z satisfy the system Mv = v the system z has 1 degree of freedom!. Here we are looking for a vector of length 1, so we need 1 z = ± 1 3, so v = ±

6 c M 2 v = MMv = Mv = v,, so M n v = v for all n. We can easily demonstrate this formula by recurrence. 4. a We are looking for symmetric matrix A of order 2, solution of the equation A 2 a b = O 2. Therefore, let A =. Then, we have: b d A 2 a 2 + b 2 ba + d 0 0 = O 2 = ba + d b 2 + d a 2 + b 2 = 0 i ba + d = 0 ii b 2 + d 2 = 0 iii We study these equations one by one: i a 2 + b 2 = 0 a = 0 and b = 0 b/c the square of a real number is always nonnegative. ii Similarly, b 2 + d 2 = 0 b = d = 0. iii Finally: ba + d = 0 b = 0 or a = d. Hence, for the three above equations to hold simultaneously, there is only one solution: a = b = d = 0, that is the only symmetric matrix solution of the equation is A = O 2. b We are now looking for matrix A of order 2 non necessarily symmetric, solution of the equation A 2 a b = O 2. Therefore, let A =. Then, we have: c d a 2 + bc = 0 A 2 a 2 + bc ba + d 0 0 ba + d = 0 = O 2 = ca + d bc + d ca + d = 0 bc + d 2 = 0 a 2 = bc i b = 0 or a = d ii c = 0 or a = d iii d 2 = bc iv 6

7 There are several cases: - case 1: b = 0 then from equations i and iv, we deduce that a = d = 0; c is a free parameter ie it can be any real number. So the family of matrices solutions of the equation can be written as: A c = 0 0 c 0 where c is any real number Note1: the unique symmetric solution we found in a belongs to this family when c = 0. - case 2: c = 0 and b 0 then from equations i and iv, we deduce that a = d = 0; b is a free non-zero parameter. So the family of matrices solutions of the equation can be written as: A b = 0 b 0 0 where b is any non-zero real number Note2: we exclude the case b = 0 because it already belongs to the previous family. - case 3: b 0 and c 0 then from equation ii d = a; from equation iv c = d 2 /b = a 2 /b; a and b are 2 free parameters. So the family of matrices solutions of the equation can be written as: A a,b = a a 2 /b b a where a, b are any non-zero real numbers Note3: I chose here as free parameters a and b ie all the other variables are expressed as functions of a and b. This is not the only possibility: for instance, you can choose a and c and you get family of matrices: A = a a 2 /c c a real numbers. And the 2 families are equivalent in the sense that they give the exact same family of matrices when their parameters take all the possible values. Note4: all the solutions we have found are non-symmetric matrices except family 1 when c = 0. So there is no contradiction with question a. Note5: this system cannot be solved by Gaussian elimination since it is nonlinear. 7 where a, c are any non-zero

8 c We have to check that all solutions can be written in general as A = pq with p, q column-vectors of size 2 s.t. p q = 0. The three families can be reexpressed as follows: c A c = = c 0 and 0 1 = 0 c b 1 0 A b = = 0 b and 1 0 = b a b 1 a A a,b = = a b and 1 a/b = 0 a 2 /b a a/b b All three cases are of the form A = pq with p q = 0. Conversely, if A = pq then A 2 = pq pq = pq pq = pp q q = O 2 whenever pq = 0. 8