Application of linear systems Linear Algebra with Computer Science Application

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1 Linear Algebra with Computer Science Application February 5, 208 Review. Review: Vectors We introduced vectors as ordered list of numbers [ ] [ ] 3 E.g., u = and v = are vectors in R 4 2 One can identify vectors with points in R n for geometric interpretation Vector arithmetic Vector addition Scalar multiplication.2 Review: Linear combinations Linear combination of vectors v, v 2,..., v p in R n with scalars weights c, c 2,..., c p is w = c v + + c p v p When the weights are unknown for a given w, we have a vector equation Is w a linear combination of vectors v,..., v p? Span of a set of vectors is the set of all linear combinations of vectors v,..., v p, denoted by span{v,..., v p } Span of one vector? Two vectors? We can rephrase the linear combination question as a span question, i.e., is b in the span{v,..., v p }?.3 Review: Linear combinations Linear combination (or span ) question is answered by forming and solving the augmented matrix: has the same solution as x v + + x p v p = b [ v... v p b ]. /6

2 .4 Review: Matrix-vector product Matrix-vector product Ax is Linear combination of columns of A with weights from x x Ax = [ ] x 2 a a 2... a n. = x a + x 2 a x n a n. x n This is only defined if the number of columns of A is equal to the number of entries in x. This implies that a matrix equation Ax = b has a solution if and only if b is in the span of the columns of A..5 Review: Matrix-vector product We showed that A(u + v) = Au + Av and A(cu) = c(au). Matrix-vector product can be computed easier using the row-vector rule. For y = Ax we have y i = row i (A)x row i is a n matrix (row vector) and x is an n matrix (column vector)..6 Review: Homogeneous system If the RHS is 0, the linear system of equations is homogeneous There is always the trivial solution It has a non-trivial solution if and only if it has at least one free variable We learned how to express the solution of homogeneous systems as span of some vectors (parametric vector form). 0 is always in the span..7 Review: Non-homogeneous system We know how to solve (homogeneous and non-homogeneous) linear systems using Gaussian elimination. When a non-homogeneous system, Ax = b has many solutions, the solutions can be obtained by Finding a particular solution p to the system, i.e., Ap = b. Find the solution set of the homogeneous system Ax = 0 as span{v,..., v k }. All vectors of the form w = p + v h where v h span{v,..., v k } are solutions to Ax = b.8 Review: Example Find all solutions to the system Ax = b for A = 4 8 2, b = /6

3 Solution. Reducing the augmented matrix we have /2. A particular solution to the non-homogeneous system is p = [ 0 ] T..9 Review: Example Solutions to the homogeneous system (zero RHS) are in the form Therefore.0 Review: Example (difficult) 0 v h = s.5, s R. 0 x = p + v h = + s.5. 0 Construct a non-homogeneous system of three equations and four unknowns that has as its general solution.. Review: Example (difficult) x x Solution. By inspection, we see that the homogeneous system should have x 2 and x 4 as free variables, therefore, the system should have pivots in columns and 3. In pivot rows, the values in the non-pivot columns (right of the pivot) are the negative of values in the arrays above: Since the matrix is in reduced form, we pick the free variables to be zero and read off a particular solution. 2 Applications 2. A homogeneous system in economics Suppose a nation s economy is divided into many sectors. E.g., various manufacturing, communication, entertainment, and service industries. Suppose that for each sector we know its total output for one year. 3/6

4 exactly how this output is divided or exchanged among the other sectors of the economy Let the total dollar value of a sector s output be called the price of that output. Leontief proved the following result: There exist equilibrium prices that can be assigned to the total outputs of the various sectors in such a way that the income of each sector exactly balances its expenses. Let s consider a simple example of this model. 2.2 Example: an economy with three sectors Suppose an economy consists of the coal, electric (power), and steel sectors The output of each sector is distributed among the various sectors as shown in the table The entries are the fractional parts of a sector s total output. Coal Electric Steel Purchased by Coal.6..2 Electric Steel Denote the prices (i.e., dollar values) of the total annual outputs of the coal, electric, and steel sectors by p C, p E, and p S respectively. If possible, find equilibrium prices that make each sector s income match its expenditures. For a a sector we look down a column to see where its output goes, and look across a row to see what it needs as inputs. Coal Electric Steel Purchased by Coal.6..2 Electric Steel The first row of the table says that coal receives (and pays for) 40% of the electric output and 60% of the steel output. To make coal s income, p C, equal to its expenses, we want p C =.4p E +.6p S The other rows of the exchange table shows that p E =.6p C +.p E +.2p S Solving this system of equations, we have p S =.4p C +.5p E +.2p S p C.4p E.6p S = 0.6p C +.9p E.2p S = 0.4p C.5p E.8p S = 0 4/6

5 Row reduction is next. For simplicity here, decimals are rounded to two places The general solution is p C =.94p S, p E =.85p S, and p S is free. The equilibrium price vector for the economy has the form p C.94p S.94 p = p E =.85p S = p S.85. p S p S Any (nonnegative) choice for p S results in a choice of equilibrium prices. For instance, if we take p S to be $00, then $p_c = 94 and $p_e = Network flow As a second example, consider the network flow Road network Water pipes 2.4 Network flow A network consists of a set of points called junctions, or nodes, with lines or arcs called branches connecting some or all of the junctions. The direction of flow in each branch is indicated, and the flow amount (or rate) is either shown or is denoted by a variable. The basic assumption of network flow is that the total flow into the network equals the total flow out of the network and that the total flow into a junction equals the total flow out of the junction. The figure shows that 30 units flowing into a junction through one branch, with x and x 2 denoting the flows out of the junction through other branches. Since the flow is conserved at each junction, we must have x + x 2 = 30. In a similar fashion, the flow at each junction is described by a linear equation. The problem of network analysis is to determine the flow in each branch when partial information (such as the flow into and out of the network) is known. 5/6

6 2.5 Example: traffic flow The network in the figure below shows the traffic flow (in vehicles per hour) over several one-way streets in downtown Baltimore during a typical early afternoon. Determine the general flow pattern for the network. Write equations that describe the flow, and then find the general solution of the system. At each intersection, set the inflow equal to the outflow. Intersection Inflow Outflow A = x + x 2 B x 2 + x 4 = x 3 C = x 4 + x 5 D x + x 5 = 600 Also, the total flow into the network ( ) equals the total flow out of the network (300 + x ), which simplifies to x3 = 400. Combine this equation with a rearrangement of the first four equations to obtain the following system of equations: x +x 2 = 800 x 2 x 3 +x 4 = 300 x 4 +x 5 = 800 x +x 5 = 600 x 3 = 400 The general flow pattern for the network is described by { x = 600 x 5, x 2 = x 5, x 3 = 400, x 4 = 500 x 5, x 5 is free. A negative flow in a network branch corresponds to flow in the direction opposite to that shown on the model. Since the streets in this problem are one-way, none of the variables here can be negative. This fact leads to certain limitations on the possible values of the variables. For instance, x because x 4 cannot be negative. If the maximum capacity of Calvert St. is 400, what can we say about other variables? 6/6