Matrix algebra. Econ Pre-Session Math Solutions of Problem Sheet 2 30/08/2017

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1 Econ Pre-Session Math Solutions of Problem Sheet /8/7 Matrix algebra (ABC = A(BC In general AB BA!! (A + BC = AC + BC (A T T = A (A + B T = A T + B T (AB T = B T A T rank(a: is the number of pivot elements or number of nonzero rows in echelon form rank(a n c (A and rank(a n r (A. There exists a solution for Ax = b if and only if rank(a = rank(â. If A Mat nn then A if and only if rank(a = n. Use Gauss elimination process to gain the solutions of the following systems! a b x + y + z = x + y + z = 4 x y + z = x + y + z = x y z = 7 x + y z = c d x + 5y = x + y = 5x + 9y = 9 x + y + 5w = y + z + w = x + 5y + z + 9w = 9 a First write the system in matrix form! 4 Then using Gauss elimination process one can nd the following echelon form:. 5 Now we know that rank(a = = rank(â = n c(a (no contradiction so there is a unique solution. To gain the solution there are two possible way to do. One is that we calculate the reduced echelon form or we avalanch back the variables. The reduced echelon form can be the following:. So the solution is x = y = and z =. b The corresponding matrix of this system is the following: Using Gauss elimination one might get the following echelon form 6.

2 Econ Pre-Session Math Solutions of Problem Sheet /8/7 So we conclude that rank(a = = rank(â = n c(a so there is a unique solution. The reduced echelon form can be the following: 9 6. So the unique solution is x = 9 y = 6. c The corresponding matrix of this system is the following: 7. Using Gauss elimination process one might get the following echelon form: Since there is a contradiction (rank(a = < = rank(ã it means that there is no solution of the system. d The corresponding matrix of this system is the following: Using Gauss elimination process one might get the following echelon form: 5. So rank(a = = rank(â which means that the system has solution with freedomness n c (A rank(a = 4 =. (z and w are the free variables. Using Gauss elimination process one might get the following reduced echelon form:. Now a general solution for this system is the following: + z + w z w z w in other words z w can be anything and x = + z + w and y = z w. ( Remark: Observe that if we let B = ( and d = then the basic ( x variables is expressable as follows: y ( x y = ( ( ( z w ( z = d B w

3 Econ Pre-Session Math Solutions of Problem Sheet /8/7. How many solutions can we have for Ax = b if a there are solutions b rank(a = n r (A < n c (A c n r (A > n c (A = rank(a d n r (A = n c (A = rank(a a Since a system can have only or innitely many solution it means there are innitely many solutions. So the number of solutions is innitely many. b Imagine the reduced echelon form of Â. Since rank(a = n r (A it means that we have a pivot in each row in the echelon form of  so there can not be a contradiction. So there is a solution moreover we know there is a free variable so there is innitely many solution. So the number of solutions is innitely many. c Since n c (A = rank(a it means that there is no solution or there is a unique solution. So the number of solutions is or. d Imagine the reduced echelon form of Â. Since ranka = n r (A = n c (A it means that the reduced echelon form of  starts with an identity matrix. So there is no contradiction. But n c (A rank(a = means that there is a unique solution. So the number of solutions is.. Suppose that A Mat nk and has a reduced echelon form E A = (I B where B Mat nk n. What is the rank of A? What can we say about the set of solutions of Ax = b for some b Vect n? Let x be the vector formed by the basic variables and x be the vector formed by the free variables. Using matrix operations determine the set of solutions for Ax = b. The rank of A is n = n r (A. It also means that Ax = b has solution for any b with freedomness k n. A general solution x can be written in the following form: ( ( x Eb Bx x = = where x R k n arbitrary. x 4. What can we state about the number of the solutions of the following system depending on r? rx + y + z = x + ry + z = x + y + rz = There are at least possible way to answer the question one is using only Gauss elimination the other use the concept of the determinants. First we show a solution using Gauss eliminations! Let us consider the following matrix: r r r Then swap the rst and the third row. r r r x

4 Econ Pre-Session Math Solutions of Problem Sheet /8/7 then eliminat with the rst row. r r r r r If r = so r = then we gain. In that case there isn't contradiction and we have = free variable i.e. there is innitely many solution. If r then we can divide the second and the third row with r so r + r then eliminating with the second row we get r. + r If r + = then we gain a contradiction so there isn't any solution otherwise if r ± then there is no contradiction and the number of free variables is therefor there is a unique solution. Summary: If r = then there is innitely many; if r = then there isn't any solution; otherwise there is a unique solution. 5. Find an A and a B in Mat such that AB BA. ( ( For example if A = and B = AB = ( (. Then = BA ( ( a b d b 6. Let A = and let B =. Compute AB. Give a condition for which there c d c a exists A and give a formula for the inverse. AB = ( ad bc ad bc = (ad bci If ad bc then dividing both sides of this equation we get that ( I = (A B = A ad bc ad bc B so A = ad bcb. ( 7. Let A = (AA T if it exists. and let B = (. Calculate A + B AB T A T A and 4

5 Econ Pre-Session Math Solutions of Problem Sheet /8/7 AB T = A T A = AA T = ( 5 A + B = 5 = ( ( ( = = ( ( 4 Since AA T is a diagonal it is easy to see what will be the inverse matrix but we will determine it with Gauss elimination process. So take the following matrix and try to get an echelon form. ( 4 Since it is already an echelon form we can conrm that n r (AA T = = rank(aa T so it is invertible. Then use Gauss elimination to obtain the identity matrix on the left side. Then we gain that ( 4 so (AA T = ( 4 8. Let M be the transition matrix of a Markov process which is discribed on the gure. Suppose that x is distribution on the states i.e. x x x and x + x + x = and let y = Mx be the generated state. a Find a stable state of this process! b If x = ( T then verify that the generated state is also a distribution on the states! c Is the previous statement holds for any transition matrix and for any distribution?.7. employed unemployed.5 subsidized..7. 5

6 Econ Pre-Session Math Solutions of Problem Sheet /8/7 Let x denote the unemployed x the employed and x the subsidized The transition matrix has the following form:.7.. M = a To nd the stable distribution we have to nd a nonzero solution for the Mx = x equation and then normalize to i.e. x is a nonzero solution for (M Ix = equation. We use Gauss elimination on the following matrix: b One might have the following echelon form: From that one might have the following reduced echelon form: Then choosing x = (x is a free variable the others are basic variables we get a nonzero solution of this system so 4.5 x = To gain a distribution we have to normalize the sum to. Since the sum of the coordinates is ( x =.5 then.4 x = = y = Mx = = We have to check that each coordinate is nonnegative (it is clear and the sum is zero in other words ( y =. But so y is a distribution =. + =. +.8 = 9. Suppose that there are two stocks and three possibility of change of the related economy and politics. Suppose that the payo matrix is the following (i.e. R sj = y sj v j where v j is the current value of a unit of asset j and y sj is the value of one unit of asset j at the end of the investment period: R = R =

7 Econ Pre-Session Math Solutions of Problem Sheet /8/7 a ( What are the payos depending on the states if we invest $ 6 with portfolio vector x =.5?.75 b Is the previous portfolio duplicable? c Is there a riskless portfolio? d Which states are insurable? Case R : a R x = (.5.75 = So our return at the end of the investment if we invested $ 6 then if the rst state occurs then we will have $.5 6 if the second state occurs then we will have $.5 6 and if the third state occurs than we will have $.5 6. b It is enough to check that rank( R < A = so do Gauss elimination on the following matrix: R = One might get the following echelon form: so the rank( R = = A so x is not duplicable. c To answer this question it is enough to conrm that R y = is solvable or not. So let's take the corresponding matrix: and do Gauss elimination. One might get the following echelon form: 5. 5 We can see there is a contradiction so there isn't any riskless portfolio. d To answer this question look the following matrix: and do Gauss elimination on the left side. One might get the following echelon form: 5..4 If we keep any column on the right side we gain contradiction so non of the state is insurable. 7

8 Econ Pre-Session Math Solutions of Problem Sheet /8/7. Suppose that a payo matrix is the following: ( R = 5.5 a Is the portfolio vector x = ( T duplicable? b Is there a riskless portfolio? c Which states are insurable? a We have to check that rank( R <. So take the following matrix R = 5.5 and do Gauss elimination. One might get the following echelon form: From that we can derive that rank( R = = A so non of the portfolios can be dupblicated. b We have to check the solvability of Ry =. ( 5.5 For this matrix one might get the following echelon form: ( 9.95 and there isn't any contradiction so it means there exists a riskless portfolio. c Let ( 5.5 then do Gauss elimination. Then one might get the following echelon form: ( 9.95 Deleting any column on the right we gain a solvable system so it means any state is insurable.. Let R Mat SA be a payo matrix. a Suppose that rank(r = S. What can we say about riskless portfolios and insurable states? b If any state is insurable then what can we state about the rank of R? c Suppose that we have a duplicable portfolio. What can we state about the rank of R? 8

9 Econ Pre-Session Math Solutions of Problem Sheet /8/7 a Since rank(r = S it means that S A. But in that case from I//bd we know that Ry = b always have solution for any b. Since to verify that there exists a riskless portfolio or to say that state s is insurable we have to check that Ry = b is solvable for some special b's. b It means that rank(r = S. c It means that rank(r rank( R < A since rank(r rank( R < A 9