Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 1 Exercise 1.1 1. Given: 1 0 A 3 5 2 1 and B 4 3 Find: (a) A B 1 1 A B 7 8 (b) 2A 3B 8 3 2A 3B 6 1 (c) AB 2 1 AB 26 18 (d) AB BA 1 4 AB BA 31 33 (e) A TB (where A T denotes the transpose of A ) 10 8 AT B 20 15 2. (f) BT A 10 20 BT A 8 15 (g) A 2B 1 2 A2B 124 87 Given: a1 A a2 a3 and B b1 b2 b3 Find AB and BA OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 2 a1b1 a1b2 a1b3 AB a2b1 a2b2 a2b3 a3b1 a3b2 a3b3 BA b1a1 b2a2 b3a3 3. Given: A a1 a 2 a3 (Note, a scalar) b1 and B b2 b3 Find (if they exist) AB , ATB , BA , BT A , A TB T , and BT A T . In each case, explain why the matrix product does (or does not) exist. AB a1b1 a2b2 a3b3 ATB does not exist because number of elements in a row of AT (=1) does not equal number of elements in a column of B (=3). (Or, equivalently, because number of columns in AT (=1) does not equal number of rows in B (=3) b1a1 b1a2 BA b2a1 b2a2 b3a1 b3a2 b1a3 b2a3 b3a3 BT A does not exist because number of elements in a row of BT (=3) does not equal number of elements in a column of A (=1). (Or, equivalently, because number of columns in BT (=3) does not equal number of rows in A (=1) a1b1 a1b2 a1b3 A B a2b1 a2b2 a2b3 a3b1 a3b2 a3b3 (Note, same as BA) BT AT b1a1 b2a2 b3a3 (Note, same as AB) T T 4. Given: a11 a12 b11 b12 A a21 a22 and B b21 b22 a 31 a 32 b31 b32 Find (if they exist) AB , ATB , BA , BT A , A TB T , and BT A T . OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 3 AB does not exist because number of elements in a row of A (=2) does not equal number of elements in a column of B (=3) AT B b11 b12 a11 a21 a31 a b21 b22 a a 22 32 12 b31 b32 a11b12 a21b22 a31b32 a b a b a b 11 11 21 21 31 31 a12 b11 a22 b21 a32 b31 a12 b12 a22 b22 a32 b32 Exercise 1.2 1. Find the inverses (if they exist) of: 1 3 A 1 2 0 4 B 3 2 ; ; 1 4 C 3 12 1 3 ; D 1 4 The inverses are: 1 1 ; B 6 3 ; 41 0 C-1 does not exist, because each element of row 2 equals 3 times the corresponding element of row 1. (Equally, each element of column 2 equals 4 times the corresponding element of column 1.) 0.4 0.6 A 0.2 0.2 -1 1 4 3 D-1 1 1 See section 19.10 of the book. 2. Using your answers from question 1 above, solve where possible the following sets of simultaneous equations. If you find that no solution is possible, explain why. (a) x + 3y = 5 –x + 2y = 2 As explained in section 19.11 of the book, we can write this pair of simultaneous equations in matrix form as 1 3 x 5 Av = b where A ; v ; and b 1 2 2 y OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 4 0.4 0.6 Therefore v = A-1b. From (1) above, we have A-1 , so 0.2 0.2 0.4 0.6 5 0.8 v A-1b ; that is, x = 0.8, y = 1.4. 0.2 0.2 2 1.4 (You can check that these values are correct by substituting them back into the given simultaneous equations.) (b) 4y = 7 3 x 2y 1 1 1 Following the same steps as in (a) above, and using B 6 3 , from 41 0 question (1), we find x 56 , y 1.75 . (Don’t forget to check by substitution). 1 (c) x + 4y = 1 3 x 12y 1 In this case the relevant matrix in question (1) is C, which has no inverse. Therefore the simultaneous equations in this case do not have a solution. (This is because they are not independent of one another; for example, the second equation is 3 times the first.) (d) x + 3y = 1 x 4 y 2 4 3 Following the same steps as in (a) and (b) above, and using D-1 , 1 1 from question (1), we find x 10, y 3 . (Don’t forget to check by substitution). 3. Find the determinant, all minors and cofactors, and the inverse of each of the following matrices: OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises (a) 5 1 1 1 2 2 3 4 0 0 Determinant = 20 Matrix of minors: 0 -12 8 0 4 -4 5 5 0 0 12 8 0 4 4 5 -5 0 Matrix of cofactors (= signed minors): Inverse matrix: (b) 0 0.25 0 0.6 0.2 -0.25 -0.4 0.2 0 2 1 1 2 0 3 4 0 4 Determinant: –4 Matrix of minors: 0 4 0 -4 12 4 -3 8 2 Matrix of cofactors (=signed minors) Inverse matrix: (c) 0 4 0 4 12 -4 -3 -8 2 0 1 0.75 -1 -3 2 0 1 -0.5 5 2 1 2 1 4 2 0 2 OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 6 Determinant: –4 -2 -4 2 4 12 -4 7 22 -9 Minors Cofactors (= signed minors): Inverse matrix: (d) 2 -2 4 -4 12 4 7 -22 -9 0.5 1 1.75 -1 3 5.5 -0.5 -1 2.25 10 5 1 2 1 4 4 0 4 Determinant: 4 Minors: -4 -8 4 20 36 -20 21 38 -20 Cofactors: 8 4 -4 -20 36 20 21 -38 -20 inverse matrix: -1 -5 5.25 2 9 -9.5 1 5 -5 Exercise 1.3 1. Use Cramer's rule to find x in the equation systems below. (a) y z 2 3 x 2y z 4 5 x 4y 1 OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 7 As explained in section 19.11 of the book, we can write this set of simultaneous equations in matrix form as 0 1 1 x 2 Av = b where A 3 2 1 ; v y ; and b 4 5 4 0 z 1 Cramer’s rule tells us that we can solve for each variable as follows. (i) To solve for x, we form a new matrix, A1, obtained by replacing the first column of A with the vector b. Thus 2 1 1 A1 4 2 1 1 4 0 The solution value of the variable x is then given by A1 ; that is, the ratio of the A determinants of the two matrices. In this example, the two determinants are A 3 and A1 21, so x A1 A 21 3 7 (ii) To solve for y, we form a new matrix, A2, obtained by replacing the second column of A with the vector b. Thus 0 2 1 A2 3 4 1 5 1 0 The solution value of the variable y is then given by A2 ; that is, the ratio of the A determinants of the two matrices. In this example, the two determinants are A 3 and A1 27 , so y A1 A 27 3 9 (iii) To solve for z, we form a new matrix, A3, obtained by replacing the third column of A with the vector b. Thus 0 1 2 A2 3 2 4 5 4 1 OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises The solution value of the variable y is then given by 8 A3 ; that is, the ratio of the A determinants of the two matrices. In this example, the two determinants are A 3 and A3 21, so z A1 A 21 3 7. So our solutions are: x = –7, y = 9, z = 7 (You can check that these values are correct by substituting them back into the given simultaneous equations.) OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 9 x z 3 (b) 3x 2y z 4 21 5x 4y 1 Using the method of (a) above: (i) Solution for x. We have 3 0 1 1 0 1 A 3 2 1 ; A1 4.5 2 1 ; A 2 and A1 4 1 4 0 5 4 0 So x A1 A 4 2 2 (ii) Solution for y. We have 1 3 1 1 0 1 A 3 2 1 ; A2 3 4.5 1 ; A 2 and A2 5.5 5 5 4 0 1 0 So y A2 A 5.5 2 2.75 (iii) Solution for z. We have 1 0 3 1 0 1 A 3 2 1 ; A3 3 2 4.5 ; A 2 and A3 10 5 4 1 5 4 0 So z A3 A 10 5 2 So our solutions are: x = –2, y = 2.75, z = 5 (You can check that these values are correct by substituting them back into the given simultaneous equations.) OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 2. 10 Solve the following system of linear equations using (a) matrix inversion, and (b) Cramer's rule. 3 0 x 4 1 0 0.5 1 y 1 0.5 0 1 z 3 (a) Solution by matrix inversion. We can write this set of simultaneous equations in matrix form as 3 0 1 x 4 Av = b where A 0 0.5 1 ; v y ; and b 1 0.5 0 1 z 3 0.25 -1.5 1.5 The solution is v = A-1b. We find A-1 as A 1 0.25 0.5 -0.5 . So -0.125 0.75 0.25 0.25 -1.5 1.5 4 4 v A 1b 0.25 0.5 -0.5 1 0 -0.125 0.75 0.25 3 1 So our solutions are: x = 4, y = 0, z = 1 (You can check that these values are correct by substituting them back into the given simultaneous equations.) OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 11 (b) Solution by Cramer’s rule. Using the method of question 1 above: (i) Solution for x. We have 3 0 1 4 3 0 A 0 0.5 1 ; A1 1 0.5 1 ; A 2 and A1 8 0.5 0 1 3 0 1 So x A1 A 8 2 4 (ii) Solution for y. We have 3 0 1 1 4 0 A 0 0.5 1 ; A2 0 1 1 ; A 2 and A2 0 0.5 0 1 0.5 3 1 So y A1 A 0 2 0 (iii) Solution for z. We have 3 0 3 4 1 1 A 0 0.5 1 ; A3 0 0.5 1 ; A 2 and A3 2 0.5 0 1 0.5 0 3 So z A3 A 2 2 1 So our solutions are: x = 4, y = 0, z = 1, as before. 3. In the economy of the Kingdom of Monomania, households spend their incomes on domestically produced goods, Cd, and imported goods, M. This spending is observed to follow the relationship Cd M aY (1) where a is a parameter and Y = households’ income. Imports are observed to be related to household consumption by the relationship M mCd (where m is a parameter) OXFORD © Oxford University Press, 2005. All rights reserved. (2) Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises 12 The equilibrium condition for this economy is that aggregate output, Y, must equal the demand for output for domestic consumption, Cd, and domestic investment, I, plus demand for exports by foreigners, X. Therefore the equilibrium condition is Y Cd I X (3) (where I and X are assumed to be exogenous). Note that household income is necessarily equal to output because households earn their incomes (wages, salaries and profits) by producing output. (a) Show that, after suitable rearrangement, the set of simultaneous equations 1 to 3 above can be written as Ax = b, where A is a matrix containing the parameters of the model (a and m); x is a vector containing the endogenous variables (Y, Cd and M) and b is a vector containing the exogenous variables (I and X). We can present the equations of this model in table form as: Y -Cd -mCd a Y 1 m -Cd +M = I + X (from equation 3) = 0 (from equation 2) = 0 (from equations 1 & 2) This helps us to see that the equations can be expressed in matrix form. For example, the coefficients of Y in column 1 are 1, 0 and a 1 m . This gives us the first column of the coefficient matrix. Similarly the coefficients of Cd in the second column are –1, –m, and –1. This gives us the second column of the coefficient matrix. After some fiddling around, we find that the equations can be expressed in matrix form as follows: 1 0 a 1 m 1 0 Y I X m 1 Cd 0 1 0 M 0 or in matrix notation, Ax = b. The key point here is that we have now the endogenous variables Y, Cd and M on the left hand side (the vector x). These endogenous variables are determined by the values of the exogenous variables I and X (the vector b), together with the values of the parameters (the matrix A.) OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises (b) 13 By calculating A-1, find the equilibrium values of the endogenous variables in terms of the values of the parameters and exogenous variables. Given Ax b , we want to find x A 1b . To find A-1, we follow the procedure explained in section 19.13 of the book. The matrix of transposed cofactors, D' , is 1 D' 1am ma 1 m 1 1 m a (1 m ) 1 m 0 0 The determinant of A is A 1( 1) 1( 1am ) 0 = 1 1 1 m a D' So A 1 A a m 1 1 m ma 1 m 1 1 m a - m -1 1 m a (1 m ) 1 m 0 0 Since x A 1b , we find the solution values for the vector of endogenous variables, x, as x A 1b , that is: Y 1 C 1 m a d a m 1 1 m M ma 1 m 1 1 m I X 0 0 a (1 m ) 1 m 0 0 Multiplying out, we get Y (c) 1 m (I 1 m a X ) ; Cd a (I 1 m a X) ; M am (I 1 m a X) Using your answer to (b), find the equilibrium values of the endogenous variables when a 0.75, m 0.25, and I X 1000. Check your answer by solving the model by simultaneous equation methods. From (b) above, Y Cd a (I 1 m a X) = 1 m (I 1 m a X) = 0.75 (1000) 0.5 1.25 (1000) 0.5 2500 1500 OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education Renshaw: Maths for Economics Chapter 19: Matrix algebra Answers to additional exercises M (d) am (I 1 m a 14 3 8 X ) = (1000) 375 Use your answer to (b) to calculate the effect on Y, C, and M of a small increase in each of the exogenous variables, I or X. Our matrix solution in (b) gives us the levels of Y, Cd, and M, as functions of the levels of I + X. However, we can also calculate the effects of small changes, simply by replacing I +X with dI + dX, and so on. Then we get dY 1 dC 1 m a d a m 1 1 m dM ma 1 m dY 1 m (dI 1 m a 1 1 m dX ) ; dCd dI dX 0 0 a (1 m ) 1 m 0 0 a (dI 1 m a dX ) ; dM am (dI 1 m a dX ) (If you distrust the validity of this, try calculating Y, Cd, and M when I = I0, X = X0. Then calculate Y, Cd, and M when I = I0 + dI, X = X0 + dX (where of course dI and dX denote small changes). Then subtract the first solution values from the second solution values, to arrive at the changes in Y, Cd, and M. You will find that you have the same result as above.) (e) What is the relationship between exports and imports in this model? Exports are an exogenous variable; that is, they are determined by factors lying outside of this model. There is a weak causal link running from exports to imports, through the reduced form equation M am (I 1 m a X), but it is only the sum of investment plus exports that appears on the right hand side, so a change in exports need have no effect on imports if there is a compensating change in investment at the same time. OXFORD © Oxford University Press, 2005. All rights reserved. Higher Education