ECON 33 Homework #2 - Solution. (Leontief model) (a) (i) The matrix of input-output A and the vector of level of production X are, respectively:.2.3.2 x A =.5.2.3 and X = y.3.5.5 z In a closed model the vector of external demand is zero, so the matrix equation writes: X = AX (I 3 A)X = O 3.8.3.2.5.8.3.3.5.5 x y z = (ii) We show that it is not possible to solve the equation by using Cramer's rule because the determinant of the matrix (I 3 A) is zero. I do the expansion by co-factors according to the rst row (there are no zero entry in the matrix (I 3 A) so all expansions are equally cumbersome!):.8.3 det(i 3 A) =.8.5.5 +.3.5.3.3.5.2.5.8.3.5 =.8.25 +.3 (.34).2.49 = We then have to use Gaussian elimination to solve the above equation. First, we rewrite the matrix equation as a system and we work our way towards an upper triangular matrix with ones on the main diagonal as suggested by Gaussian elimination (Note that since the above matrix is not invertible, the solution cannot be unique, which
means that we will not be able to exactly get this upper triangular matrix): +.8x.3y.2z =.5x +.8y.3z =.3x.5y +.5z = x +3/8y 2/8z = L L /8 +4.9y 3.4z = L 2 5L + 8L 2 4.9y +3.4z = L 3 3L + 8L 3 { { x = 3/8y + /4z y = 34/49z x = 25/49z y = 34/49z And this is as far as we can go. The system has one degree of freedom and an innite number of solutions that can be expressed as a function of one of the unknown. In the above resolution, I chose z and the solutions are: for any z real number, x = 25/49z and y = 34/49z. Note: if you decide to choose another unknown instead of z, you would nd an equivalent set of solutions: i) choosing x, the solutions are for any real number x, y = 34/25x and z = 49/25x; ii) choosing y, the solutions are for any real number y, x = 25/34y and z = 49/34y. (iii) First, we have to normalize the matrix of input-output A as follows. We calculate the sum of each row and we divide each element of the row by it to get Ã:.2.3.2.5.2.3.3.5.5 Ã = =.7.3 2/7 3/7 2/7.5.2.3 3/3 5/3 5/3 Consider now the row vector of relative prices, P = (p x p y p z ). The matrix equation is 2
the following: 5/7 3/7 2/7 P Ã = P P (I 3 Ã) = ( ) (p x p y p z ).5.8.3 = ( ) 3/3 5/3 8/3 Note that if you choose to work with the column vector of prices, say P = then the matrix equation writes: (I 3 Ã) P = p x p y p z,. And this leads to the same system of equations. Similarly to what we have done in the previous question, the determinant of (I 3 Ã) is zero and we have to solve with Gaussian elimination. Note that this time the system is simpler since we already know that p x =. The system writes: 5/7 /2p y 3/3p z = 3/7 +.8p y 5/3p z = 2/7.3p y +8/3p z = { p y = 34/343 99.3 p z = 325/343 94.75 For a given price of energy equal to $, we nd that the prices of manufacturing and sector are approximately equal to $99.3 and $94.75 respectively. (b) Each unit of petroleum produced requires an input of. units of petroleum, so of the 9 units produced, (.*9) or 9 units of petroleum are required; each unit of textiles produced requires an input of.4 units of petroleum, so of the 3 units produced, (.4*3) or 2 units of petroleum are required; each unit of transportation produced requires an input of.6 units of petroleum, so of the 85 units produced, (.6*85) or 5 units of petroleum are required; and, each unit of chemicals produced requires an input of.2 units of petroleum, so of the 8 units produced, (.2*8) or 6 units of petroleum are required. A total of 88 (9+2+5+6) dollars of 3
petroleum are needed as input (internally consumed) to produce 9, 3, 85, and 8 million dollars of petroleum, textiles, transportation, and chemicals, respectively. Note that the totals needed for input of petroleum, textiles, transportation, and chemicals can be found by multiplying the technology matrix (or input-output matrix) by the production vector:..4.6.2 9.. A =.2.5..3 3 85 =.4.3.25.2 2. (Linear regression model) 8 88 55 822.5 (a) Consider rst the matrix P = X(X X) X. X is a (n,2)-matrix and X is then (2,n)-matrix. It makes sense to calculate (X X) which is a square matrix of size 2. You can assume it is invertible, since this is used in the exercise (Note that it would be an issue if (X X) was not a square matrix!). And (X X) is still a (2,2)-matrix and it can be pre-multiply by X and post-multiply by X ; in the end we get a (n,n)-matrix. Consider now the matrix M = I n P. We subtract 2 (n,n)-matrices. We can always do that, and we also get a (n,n)-matrix. (b) M is symmetric if and only if M = M. We calculate M and check that it is equal to M: M = (I n P ) = I n P = I n (X(X X) X ) (b/c I n is symmetric) = I n (X ) ((X X) ) X (b/c (ABC) = C B A ) = I n X((X X) ) X (b/c (A ) = (A ) ) = I n X(X (X ) ) X (b/c (AB) = B A ) = I n X(X X) X = M 4
(c) We calculate M 2 and check that it is equal to M: M 2 = (I n P ) 2 = (I n P ) (I n P ) = I 2 n P I n I n P + P 2 = I n P P + P 2 (use the properties of I n ) = I n 2P + P 2 ( ) We calculate separately P 2 (and if we can show it is equal to P then we are done!): P 2 = (X(X X) X ) 2 = (X(X X) X ) (X(X X) X ) = X(X X) X X(X X) X = XI n (X X) X (b/c (X X) X X = I n ) = X(X X) X (use the properties of I n ) = P We plug this back into (*) to get: M 2 = I n 2P + P = I n P = M. (d) We calculate MX and check that it is equal to the zero matrix: MX = (I n P )X = I n X P X = X P X ( ) We calculate separately P X (and if we can show it is equal to X then we are done!): P X = (X(X X) X ) X = X(X X) X X = XI n (b/c (X X) X X = I n ) = X (use the properties of I n ) We plug this back into (**) to get: MX = X X = O. (e) Consider b = (X X) X y. We plug this b into the denition of RSS and check that it is equal to y M: RSS = y y 2b X y + b X Xb 5
As a preliminary step, we calculate the transpose of b (so we can directly plug b ): b = ((X X) X y) = y (X ) ((X X) ) = y X(X X) (check the calculations of P 2 is needed) We now replace b and b in RSS by their expressions to get: RSS = y y 2(y X(X X) )X y + (y X(X X) )X X((X X) X y) = y y 2y X(X X) X y + y X(X X) X X(X X) X y = y y 2y X(X X) X y + y XI n (X X) X y = y y 2y X(X X) X y + y X(X X) X y = y y y X(X X) X y = y (y X(X X) X y) (factorize y on the left) = y (I n X(X X) X )y (factorize y on the right) = y (I n P )y = y My 6