Demonstrate solution methods for systems of linear equations. Show that a system of equations can be represented in matrix-vector form.

Transcription

1 Chapter Linear lgebra Objective Demonstrate solution methods for sstems of linear equations. Show that a sstem of equations can be represented in matri-vector form. 4 Flowrates in kmol/hr Figure.: Two distillation columns in series.. Eample Sstem Two distillation columns in series with a additional feed stream miing in with the bottoms stream of the first column. The flow rate of three streams are unknown. s indicated in the Figure., the flow rate of streams,, and are unknown. No reaction is taking place. The steadstate flow rates must be calculated. Basic Mass Balance: accumulation in out + created destroed Mass Balance on first column In this case, assume stead state: accumulation ): 4 9

2 Mass balance on miing point: + Mass balance on second column: Three linear equations: 4 + Note that ou could write too man equations. You could write an overall balance: 4 Ending up with an overspecified sstem of equations, 4 equations, unknowns. Stick with the three equations from above for now. Note that these are linear equations. The unknown variables have constant linear coefficients, nonlinear terms do not appear no, no, no e ). You can rearrange the set of three equations without the overall balance equation) to get all the variable terms on the left side and the constants on the right. fter some The set of equations can be written as: ) s we will see later, this can be more compactl written as: b You ma alread realie that the solution to this problem is 6, 9, and 7. For more comple sstems, this is not quite so eas. To solve the three linear equations simultaneousl in a general manner, ou can perform row reduction using three possible row operations: RULES. dd or subtract) one row to or from) another. Multipl or divide a row b a scalar value an real scalar ). Swap position of rows

3 Tpicall ou would perform these operations until ou have a triangular representation all s above or below the diagonal). The triangular form allows for quick solution. The set of linear equations in Equation. can be compactl written using onl the coefficients as: 6 We need to perform steps - to get the sstem of equations in triangular form with ones on the diagonal and eros below the diagonal, like a b c We can look at the original sstem of equations and realie that we must get eros in position, row, column ) and position, row, column ). You can multipl row b using Rule : d e f 6 Net, swap position of rows and using Rule to get: 6 Then, subtract row from row using Rule to get: 6 9 Then, multipl rows and b using Rule : 6 9 Subtract row from row using Rule again to get: Now, all coefficients below the diagonal are. The solution can be found quickl. From equation row ), 7. Using equation row ), but ou know 6 7

4 that 7 so 9. Equation row ) gives 6, so the overall solution is 6, 9, and 7. CHECK SOLUTIONS: You can plug our solution back into the original three equations and verif that the equations are satisfied. THIS WILL HELP YOU ON EXMS. Note that the general Gaussian elimination or row reduction method specifies that ou start with column and perform operations until all coefficients below the diagonal are, then move to column and perform operations until all coefficients below the diagonal are ero, etc.. Linear Equations - Special Cases In general, there are three possibilities for a square set of linear equations... Case - One solution Consider a simpler sstem: + and. Graphicall, ou can plot the two lines and look for the intersection of two lines which occurs at,. The sstem of equations is: Subtracting row from row gives: This implies or and + or as ou alread realied. In dimensions unknowns) each row represents a plane. Two equations can intersect to give a line, and a line can intersect with a third plane to give a point, the single solution in a single solution case)... Case B - No solution Consider the sstem + and +. Graphicall, this represents two lines that never intersect. Note that column and column are identical. Subtracting row from row gives: You know that + cannot be true. For a square sstem, if Gaussian elimination results in a on the diagonal, this ma be the case.

5 .. Case C - Man solutions Consider the sstem + and +. Graphicall, this represents two lines that are coincident. Subtracting twice the value of row from row gives: These equations are consistent. + and + are consistent. There is no single solution, as man solutions make the equation + + consistent.. Nonsquare Sstems The original eample was for a square sstem with unknowns and equations. You ma often end up with more or fewer) equations than unknowns. Consider the original set of equations: One additional equation can be specified using a mass balance on the entire sstem, ) These four linear equations are not linearl independent. You can test this b using row operations to make two rows identical. Simultaneousl adding row and row to row will make row the same as row ) This set of equations can still be satisfied using the original solution 6, 9, and 7. In other cases, having more equations than unknowns ma complicate the solution process a bit.

6 .. Reconciliation and Nonsquare Sstems For curve fitting, parameters that appear linearl can be formulated as a nonsquare solution to a linear algebraic sstem of equations. Given that ou have some scalar valued) measured value,, that depends on a process parameter,. ssume the model takes the form: m + b.4) Technicall, ou onl need two data points to find m and b, the model parameters. ssuming that ou have more than two data points, we often desire to determine the best-fit for the line. These parameters minimie the sum of the square of the model error. For an eperiment with four data points: ) m ) + b ) m ) + b ) m ) + b 4) m 4) + b.5) Here, ou know values of and but m and b are our unknown values. This can be written as a set of equations: ) ) ) 4) ) ) ) 4) You can get the best-fit solution to this overspecified set of equations using the psuedo-inverse of the matri: T ) T b m b.4 Vectors group of unknown or known) values can be stacked to form a vector. In the eample problem, the unknowns,, and can be described b the vector : The solution to the problem has a known value and can be written as a vector soln : soln Note that the underbar is used to distinguish between the vector) and the unknown. vector is NOT limited to or unknowns dimension of the vector)

7 .5 The Matri matri is similar to a vector, having dimensions. One ma think of it as a group of vectors augmented together. Matri has a sie, m n representing m rows and n columns. The values for m and n are sometimes written as subscripts for the matri. For eample, the matri with two rows and three columns ma have values: a, a, a, a, a, a, Note that each of the si elements has two indices. The first inde is the row, the second is the column. For the applications in this class, a matri will have constant coefficient values. Some eample matrices: 5. B 6 5 Square Matri - matri with indices equal m n). Note: vector can be seen as a special matri having onl column. Transpose - The transpose operator swaps the indices of a matri or vector). For eample, for as before: Eample. For the matri ) T T a, a, a, a, a, a, 4 Finall, one can take the transpose of a vector. For T Row Vector - The transpose of a vector is also known as a row vector. Dot Product - The dot product of two vectors is the sum of the product of the elements taken individuall. Eamples: 5 T + +

8 Matri Multiplication - Two matrices can be multiplied together. For eample m n can be multiplied b B n j. Matri has m rows and n columns, while B has n rows and j columns. m n... r r r m... Here, each row up to r m is a row vector with n elements.... B n j c c... c j... Here, each column up to column c j is a vector column vector) with n elements. To compute m n B n j or simpl B or just B m n B n j r T c r T c... r T c j r T c r T c... r T c j... rm T c rm T c... rm T c j Method - To compute m n B n j, the result will have j columns. The first column of the result is computed b taking the dot product of B j first column of B) with the transpose of all the rows of. The second column of the result is computed b taking the dot product of B j second column of B) with the transpose of all the rows of. Repeat up to the j th column of B which produces the j th column of the result. Note: In general, B B. Conformable - In order to multipl m n B n j the inner dimensions must be equal. In m n B n j, if the first matri has n columns and the second matri must n rows. Matri Multiplication Eamples:

9 5.6 Column Eample Consider again the equations from the original distillation column eample: Notice that the variables with constant coefficients) are on the left side and constant values are on the right hand side. This set of linear equations can be represented in the compact notation b where b 6 Identit Matri - The identit matri has values of one on the diagonal and eros elsewhere. It is defined as I and for a square matri I and I. I.6. How to solve sets of linear equations We need a solution to the matri equation b. You cannot divide b a matri: b / There is no division operator for a matri. Instead, an inverse is defined for some square matrices such that 7

10 lso, ) I ) I Now, to solve b for First, multipl on the left b ) ) ) b Realiing that ) I replace ) with I. Now, realiing I is, the solution is I ) b ) b Note that multipling on the right will not lead to a solution. ) b ).6. How determine a matri inverse To solve b, ou need to know ). We are going to use row reduction to calculate ). Start with I. use row reduction techniques until is I. ) if it eists will be on the right where I was originall. Inverse Eample Solve the following for using ) : 5 6 For this procedure, one must first calculate ). Set up I as: Use row reduction to get???? Then verif that ) I. Use ) to calculate using ) b. Verif solution again to be safe. 8

11 STRT Start b using row reduction on Multipl row b / to get : 4 Then subtract row from row to get: Now, multipl row b -/ to get: To get the left side looking like the identit matri, subtract times row from row. Note that this is a compound use of row reduction rules. You now have ) Now verif that ) I ) + ) ) + ) ) + 4 ) ) + 4 ) You ma also verif that ) I Now, compute the solution, ) b gain, verif the solution is the solution to the original equations: Just as epected

12 .6. Stead State Control Eample Two pumps are used to fill two tanks. The pumps usuall operate at 5%, keeping the tanks at levels of 75 inches and 8 inches respectivel.it is known that a % increase in pump increases the height of tank b 5 inches and the height of tank b inches. For a % change in pump, the height of tank increases b 4 inches. It is desired to change the operating levels of the tanks to inches and 89 inches. P H H P What do ou know: Figure.: Pump / Tank eample 5 P %) H inches) P %) + 4 P %) H inches) You know the target reference, setpoint) for H and H as and 89. This translates into H 75 5 and H You need to increase tank b 5 inches and increase tank b 9 inches. You do not know the final values of the pump speeds. You do know the original steadstate values, 5% and 5%, realiing that: P final P ss + P You can now set up linear equations to solve for P and P, then calculate the final values for the pump speeds..7 Visualiation 5 P P H H Each row in b is a single linear equation. For a D problem with elements / unknowns) the equation defines a line in the, ) plane. Two equations define two

13 lines, and the unique solution to b is the point where the lines intersect. In some cases, there ma be man solutions to b and in some cases there ma be no solutions to b. Figure.: Three D eamples with two equations. Each equation row) represents a line. The first case has one solution, the second case has no solution, and the third case has man solutions. For a D problem, each row defines the equation for a plane in space. The intersection of non-parallel planes is a line in space, and the intersection of a line and a plane in space is a point. gain, in some cases there ma be a single solution, man solutions, or no solutions. For higher dimensions, each equation defines a hperplane in a n dimensional space, R n..7. Linear Transform vector in R n means has n elements. Matri multiplication of a matri of sie m n times a vector of sie n maps the vector from R n to R m. _ m R n R Figure.4: Matri multiplication as a mapping from R n to R m..7. Range The range of a matri is the space of all possible points that ma be mapped to in a matri multiplication of that matri times an unknown vector.

14 Range Eample For eample, the matri can onl map to points on the line + in D as follows. + + The columns of the matri define possible directions for the matri to transform a vector. In this eample, columns and are the same, and column is the ero vector. where takes an real value will alwas be on the line defined b the direction. Range Eample In another eample, the matri can onl map to a variet of points in D as follows. + + gain, the columns of the matri define possible directions for the matri to transform a vector. In this eample, onl points in the directions of and can be reached when multipling. These two directions form a plane in dimensional space. m R Range of _ n R Figure.5: Range of as space in R m of all possible mappings from R n using matri multiplication.

15 Range Eample In another eample, the matri can onl map to a variet of points in D as follows. + + Here, column is linearl dependent upon columns and. This means that ou can find some combination of columns and that give column. Column lies in the plane defined b columns and column. Underling point: For b to have a solution, the b must be in the range of. For the last eamples, if b?? if b has element in the position) there will not be a solution to b. In such a case, the possible range of does not include b. Range Eample 4 In another eample, the matri can map to all of the points in D as follows. + + Here, column is NOT linearl dependent upon columns and. This means that ou can find some combination of columns,, and that give an point in dimensions. Rank - The rank of a matri is the number of linearl independent columns. For a square matri of sie n n, there is a unique solution if there are n independent columns. The matri would have rank n.