Chapter 9: Systems of Equations and Inequalities

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1 Chapter 9: Systems of Equations and Inequalities 9. Systems of Equations Solve the system of equations below. By this we mean, find pair(s) of numbers (x, y) (if possible) that satisfy both equations. We illustrate the techniques of substitution, graphing, and elimination. y= x y= x+ Substitution Solve one of the equations for a single variable. We will solve the first equation to find y = x. Substitute for that variable in the remaining equation(s): the second equation becomes x = x +. Solving, we find: x x = 0 (x 4)(x + 3) = 0. Thus x = 4 or x = 3. Put this back in for x in the substitution y = x to find solutions: (4, 6) and (3, 9) Graphing Graph the equations on the same axes and find their intersections. These values of x and y satisfy both equations simultaneously. (Use menus Math Isect on the TI-86.) Elimination Adjust coefficients of one equation as needed so that when the equations are added, one variable is eliminated. Back substitute to find the remaining variable. x+ y= 5 x 4y= 0 { y= x + 3y = 8 + x + 3y = 8 y= Thus x+ y= 5 x+ 4= 5 x = The solution is (, ). Check in all cases by substituting into all original equations.

2 9. Systems of Linear Equations in Two Variables Many real world problems results in systems of linear equations to be solved simultaneously. There may be one (consistent and independent), none (inconsistent), or infinitely many (consistent but dependent) solutions. Example: This system has one solution. Solve by elimination and graphically. x+ y= 4 x y = Add the two equations to eliminate y: 3x = 6 x = x+ y= 4 + y= 4 y= x,y =, ( ) ( ) Example: This system has no solution. Solve by substitution and graphically. x+ 4y= 8 3x + y = Solve the first equation for x to find x = 8 4y. Substitute into the second equation. 3(8 4y) + y = 4 =. There are no values of x and y that make this true. We say the system has no solution or is inconsistent. (Sometimes we write the solution is the empty set.) Graphically, the lines are parallel (see next page).

3 Example: This system has infinitely many solutions. Solve by elimination and graphically. 6x + 4y = 9x + 6y = 8 Multiply the first equation by / and the second equation by /3. Add the equations. 3x + y = 6 Add: 0 = 0, an identity true for infinitely many (x, y). 3x y = 6 But not all (x, y) are solutions: given any x, only those points with y = 3 (3/)x are 3 solutions. We write t, 3 t for solutions where t is any real number. Graphically, we see the two lines are on top of each other. The system is consistent (there is an answer) but dependent, with one equation being a multiple of the other

4 Example: Word problems can be modeled by systems of equations. The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 7. Find the number. Let u = units digit and t = tens digit. The number is written t u. The value of the number is 0 t + u. With digits reversed, 0 u + t Thus the problem gives two equations: t + u = 7 0 u + t = 0 t + u + 7 These become: t + u = 7 9t + 9u = 7 Multiply the first equation by 9. 9t + 9u = 63 9t + 9u = 7 Add the equations and solve. 8u = 90 u = 5 Back-substitute to find: t + u = 7 t + 5 = 7 t =. Thus the number was 5. Check! 4

5 9.3 Systems of Linear Equations in Several Variables A variation of the method of elimination is used to transform the original system into an equivalent (upper) triangular form. This method is called Gaussian elimination. A triangular system has no x in the second equation, no x and y in the third equation, and so on. We will only allow the following operations:. Add a nonzero multiple of one equation to another: c E + E replaces E. Multiply an equation by a nonzero constant: c E replaces E 3. Interchange the position of two equations: E E swap positions Write the operation as you transform to an equivalent triangular system. Once the system is in triangular form, back-substitute and solve. Again, there may be one solution, infinitely many solutions, or no solution. Example: This system has one solution. First eliminate x in equations and 3. x+ y+ z= 4 x+ y+ z= 4 E+ E x+ 3y+ 3z= 0 y z 6 E+ E + = 3 x y z 3 + = y 3z = 5 Now eliminate y in equation 3. x+ y+ z= 4 x y z 4 x y z + + = + + = 4 E E + E 3 y + z = 6 y + z = 3 y + z = 3 y 3z 5 y 3z 5 = = z= Now equation 3 yields z =. Substitute into equation y + = 3 y =. Substitute x and y into equation : x + + = 4 x =. Thus the solution is: (,, ). Check! Example: This system has no solution. x+ y z= x+ y z= E+ E x + 3y 4z = 3 y z 5 3E+ E = 3 3x 6y 3z 4 + = 0 = The impossible 0 = indicates the system has no solution. Note the planes represented by equations and 3 are parallel. 5

6 Example: This system has infinitely many solutions. 0x + y z = 0 x y + z = x y + z = E E 3 E + E x + 3y + 0z = x + 3y + 0z = y + z = 0 x y z 0x y z 0 + = + = y z= 0 x y+ z= E+ E 3 y + z = 0 0= 0 The identity 0 = 0 is the clue we have infinitely many solutions. Let z be any real number t. Then from equation in our triangular system: y + t = 0 or y = t. Substitute y = t and z = t into equation : x (t) + t = x = 3t +. Write the solution: ( 3t +, t, t) Note: the TI-86 easily solves linear systems using SIMULT. Mathematica can also solve linear systems using matrices (introduced in the next section) and the command LinearSolve. 6

7 9.4 Systems of Linear Equations: Matrices There is no reason to write the variables x, y, z when we solve systems of linear equations as we did in the previous section. They are only placeholders; it is the array or matrix of coefficients that determines the solution. Let us adapt the method of Gaussian elimination from the last section to deal with just the matrix of coefficients. We will illustrate with an example. Rewrite the system as a matrix equation (we will explain matrix multiplication soon). x 3y z = 3 3 x 3 x+ y 5z= 6 5 y = 6 5 z 49 5x y z = 49 Now form the augmented matrix of coefficients augmented by the column of constants: Using a set of operations on rows (or columns), transform this augmented matrix into rowechelon, or reduced row-echelon form. This is similar to upper triangular form of the last section. To be precise, a matrix is in row-echelon form if:. The first nonzero number in each row is (this is the leading entry).. The leading entry in each row is to the right of the leading entry in the row immediately above it. 3. All rows consisting entirely of zeros are at the bottom of the matrix. A matrix in row-echelon form is said to be in reduced row-echelon form if it also satisfies: 4. Every number above and below each leading entry is 0. What operations are used to transform the augmented matrix to row-echelon form? Elementary Row Operations. Add a multiple of one row to another: c R + R replaces R. Multiply a row by a nonzero constant: c R replaces R 3. Interchange two rows: R R swaps two rows Once the augmented matrix has been put in row-echelon from, back-substitution solves the system. This is much simpler if the matrix has been put in reduced row-echelon form. Then the method is called Gauss-Jordan reduction. This is the best method for computer programs. We now complete the example stated above to illustrate the process. 7

8 R R R R+ R 9R+ R R+ R R We see from the last row that z =. Now from the second row, corresponding to y z = 5 we have y ( ) = 5 y = 3. From the top row, x y + 5z = 6 we have: x (3) + 5( ) = 6 x = 0 Thus our solution is: (0, 3, ). Check! If we had continued after the second row above to eliminate the y terms above the second row as well as below, and then to eliminate the z terms above the third row of our matrix, we would have reached reduced row-echelon form. This may be done using the MATRX OPS rref on the TI-86 calculator. The result is: This gives our solution with no back-substitution. What will happen with these matrix methods if the system has no solution or infinitely many? No solution: any row of the transformed matrix in row-echelon form is: 0 = constant Infinitely many solutions: any row of the matrix in row-echelon from is all zeros. In this case, express the solution in terms of arbitrary real numbers as in the last section. 8

9 9.5 The Algebra of Matrices A matrix is a rectangular array of numbers. If it has m rows and n columns, we say it has dimension m x n. The individual elements of matrix A are written a ij. The row entry is always given first; i.e. a is the element in the first row and second column. Matrices A and B are equal if and only if they have the same dimension and corresponding entries are equal. That is, a ij = b ij for i =,,..., m and j =,,..., n. The sum of two matrices A and B is defined if they have the same dimension. A + B = a ij + b ij add corresponding elements. The difference of two matrices A and B is defined if they have the same dimension. A B = a ij b ij subtract corresponding elements. The scalar product c A, is obtained by multiplying each element of the matrix by c. c A = c a ij Matrix multiplication A B is defined if the number of columns in A = number of rows in B. It is only defined for matrixes with dimensions A mn and B nk. The product has dimension m x k. The product C = A B has elements c ij found by summing the products of each element in row i of A multiplied by each element in column j of matrix B. Examples: A= B C = = A + B = = A B = A = B C = = = 3 = = = C B = Note matrix multiplication is not commutative! 9

10 Matrix multiplication is not commutative (CB BC in general). The multiplication may not be defined in one direction, or may give products of different dimensions. Matrix multiplication does obey the Associative and Distributive properties. One may enter matrices in Mathematica with built in templates. On the TI-86, they are entered with the MATRX Edit menu or from the home screen. On the TI-86, enter the matrix A = 3 4 as [ [, ] [3, 4] ] then STO and name the matrix. 0

11 9.6 Inverses of Matrices and Matrix Equations Write a system of linear equations as a matrix equation. 5x + 3y = x 4 = A X= B 3x + y = 0 3 y Above the coefficient matrix A = and the constant matrix B = We will define the inverse of a matrix A such that A A = A A = I, where I is the identity matrix, with all ones on the main diagonal and zeros elsewhere. The identity matrix plays the role of for matrix multiplication: M I = I M = M for any matrix M where the multiplication is defined. If we pre-multiply on the left the matrix equation above we find: A A X = A B I X = A B X = A B Thus if we can find the inverse of the coefficient matrix, the solution of the system of equations reduces to simply matrix multiplication. There is an easy formula for the inverse of a x matrix: If A = d b A = a d b c c a a c b d, then 5 3 Thus for A =, we find A = = = Finally, the solution of the system of equations is X = A B or x y = = and using matrix equality rules, x = 8, y =. Check! The inverse of a matrix and the identity matrix are only defined for square n x n matrices. a b It is not always possible to find the inverse of a matrix. For example, if A =, then c d whenever a d b c = 0, the inverse does not exist. The quantity a d b c is called the determinant of the x matrix. If the determinant of any matrix is zero, the matrix is called singular and is not invertible.

12 How does one find the inverse of an arbitrary n x n matrix? One may use technology: if the matrix has been stored on the TI-86 as matrix A, then from the home screen enter A enter. Mathematica can also take the inverse of a matrix using the function Inverse[A]. x and hit To find the inverse by hand is time-consuming. Augment the coefficient matrix with an identity matrix of the same dimension. Use the elementary row operations to convert the coefficient matrix to an identity matrix, performing the same operations on the augmented identity matrix. When finished, the identity matrix will have been transformed into the desired inverse matrix. We will find the inverse of a x matrix, but the process is the same for any square matrix. Given A = 5 3, form the augmented matrix and transform as: 3 R R R R R 5R We see the identity matrix has been transformed into which we know is A

13 9.7 Determinants and Cramer's Rule A determinant is a function of a square matrix. If A is x, with element a, then det(a) = a. If A is x, with A = a c b, then det(a) or A = a d b c d If A is an n x n matrix, define the minor M ij of the element a ij as the determinant of the matrix obtained by deleting the ith row and jth column of A. The cofactor A ij of the element a ij is A ij = ( ) i + j M ij det(a) is obtained by multiplying each element of a row or column of A by its cofactor, and then summing the results. This is called expanding the determinant along the row or column. 0 Example: Find det(a) where A = 0 4. We will expand along the first row det(a) = ( ) + ( ) + 0 ( ) = = 4 It is best to expand along a row of column with many zero elements. Going down the first column would have been quicker and yielded the same result in the last example. Note the ( ) i + j factor is + if the row + column is even, and if the sum is odd. The signs alternate as one moves along a row or down a column. It is possible to simplify a determinant, for example by adding a multiple of one row to another, or a multiple of one column to another. The determinant is unaffected by these operations. However, interchanging two rows or two columns will multiply the determinant by. Multiplying a row or column by a constant c also multiplies the value of the determinant by c. If any two rows (or any two columns) of a determinant are equal or multiples of each other, then the value of the determinant is zero. There are some geometric interpretations or applications of the determinant. If a triangle has vertices (a, b ), (a, b ), and (a 3, b 3 ), then its area is: a b Area = ± a b where the sign is chosen to make the area positive. a b 3 3 Determinant of A can be found on the TI-86 using MATRX MATH det(a). In Mathematica, use the function Det[A] 3

14 Cramer's Rule: Yet another scheme to solve systems of linear equations. Solve x y = 9 x+ y= 8 Write as a matrix equation and then form three determinants (number of variables + ): x 9 y = D = Dx = Dy = 8 8 D is the system determinant, simply the determinant of the coefficient matrix. D x replaces the x coefficients with the constants, and D y replaces the y coefficients. Dx 0 D y 5 Cramer's Rule tells us: x = = = y= = =5 D 5 D 5 Thus the solution is: (, 5). Check! What happens using Cramer's Rule if there are no solutions? D = 0, but D x and D y 0. If there are infinitely many solutions, all three determinants are zero. 4

15 9.8 Partial Fractions In calculus it is convenient to separate a single fraction into a sum of simpler fractions. (This is just the reverse of combining fractions using common denominators.) By the linear and quadratic factors theorem, every poly with real coefficients can be factored completely into linear and irreducible quadratic factors. Let r(x) be a rational function in lowest terms: r(x) = P(x) / Q(x), where the degree of P is less than the degree of Q. Consider four cases.. Distinct Linear Factors 5 = A + ( x ) ( x+ 4) ( x ) ( x+ 4) B Now multiply both sides by the original denominator and group like powers of x ( ) ( ) 5= A x+ 4 + B x 5= A x+ 4 A+ B x B 0x + 5 = x A + B + 4 A B ( ) ( ) This can only be an identity, true for all allowed values of x, if the coefficients of like powers of x are equal. Thus we solve the system of linear equations: A+ B= 0 5A = 5 A = B = 4A B = 5 5 = + x x+ 4 x x+ 4 The solution is: ( ) ( ) ( ) ( ). Repeated Linear Factors + x 3x 5 A B C + + ( x ) ( x+ 4) ( x ) ( x ) ( x+ 4) ( ) ( ) ( ) ( ) + = x 3x 5 A x x 4 B x 4 C x ( ) ( ) ( ) + = x 3x 5 A x x 8 B x 4 C x 4x 4 + = ( + ) + ( + ) + ( + + ) x 3x 5 x A C x A B 4C 8A 4B 4C Solve the system: A + C =, A + B 4C = 3, 8A + 4B + 4C = 5 We find: A = /, B = /, C = /. x 3x x x+ 4 x x x+ 4 ( ) ( ) ( ) ( ) ( ) 5

16 3. Irreducible Quadratic Factors + = + = x A Bx C ( x 3) ( x + 4) ( x 3) ( x + 4) ( ) ( ) ( ) x A x 4 Bx C x 3 ( ) ( ) ( = = ) x Ax 4A Bx 3Bx Cx 3C x 0x 0 x A B x 3B C 4A 3C = A+ B = 3B + C A = B = C = = 4A 3C 9 4 x x = + ( x 3) ( x + 4) ( x 3) ( x + 4) 4. Repeated Irreducible Quadratic Factors 4 3 x + x + x x+ A Bx+ C Dx+ E = + + ( ) x x + ( ) x x + x + We proceed as above and solve 5 simultaneous linear equations for A, B, C, D, and E. (Actually, we use technology: SIMULT for the TI-86, or Apart[ ] for Mathematica. 4 3 x + x + x x+ The command Apart[ ] yields: x ( x + ) x + + x x + x + ( ) 6

17 9.9 Systems of Inequalities Graph the system of inequalities x + y 4 y< x Note we use dashed lines for < or > boundaries, and solid lines for or boundaries. The doubly shaded region is the solution region, where all (x, y) satisfy both inequalities simultaneously. Note points on the dashed line are not part of the solution region, but points on the solid lower circular boundary are part of the solution region. The shaded region is bounded; it can be contained within a suitably large circle. If a region extends in any direction to infinity, it is said to be unbounded. The vertices where the line intersects the circle, are found by solving simultaneous equations. x + y = 4 + = =± =± y = x x x 4 x y The "vertices" are (, ) and (, ). 7

18 Graph the following system of linear inequalities and find the vertices of the shaded region. 3x + 5y 5 3x + y 9 x 0 y 0 The vertices may be discovered (when necessary) using the TI-86 Trace or ISECT commands. Mathematica (which produced theses graphs) can also solve systems of inequalities. Here the vertices of the (bounded) shaded region are: (0. 0), (0, 3), (3, 0), and (5/3, ). 8

19 Supplement: Linear Programming In business, this topic asks one to solve a system of linear inequalities (constraints) and then among the points of the feasible region to find that which maximizes profit or minimizes cost. Example: A furniture factory makes wooden tables and chairs. The process involves two types of labor: carpentry and finishing. A table requires hours of carpentry and hour of finishing, whereas a chair requires 3 hours of carpentry and / hour of finishing. The profit is $35 per table and $0 per chair. The factory workers can supply a maximum of 08 hours of carpentry work and 0 hours of finishing work per day. How many tables and chairs should be made each day to maximize profit? Let x = number of tables made per day and y = number of chairs made per day. The number of tables and chairs made per day cannot be negative. Thus we have constraints x 0, y 0. The hours of carpentry per table times the number of tables plus the number of hours of carpentry per chair times the number of chairs must be less than or equal to the 08 hours of carpentry hours available each day. This constraint is: x + 3y 08. The hours of finishing per table times the number of tables plus the number of hours of finishing per chair times the number of chairs must be less than or equal to the 0 hours of finishing hours available each day. This constraint is: x + (/)y 0. The factory seeks to maximize profit. The profit function may be written as the profit per table times the number of tables plus the profit per chair times the number of chairs. Thus if P = profit, P = 35x+0y. Summarizing, we have a system of constraints that we graph to find a feasible region. We must find the point or points in the region that maximizes profit. The minimax theorem tells us that such a point will always be a vertex of the feasible region for reasonable conditions. Thus we enumerate the vertices, evaluate the profit function, and find our solution. Constraints: (given x = number of tables made per day and y = number of chairs made per day) x + 3y 08 x + y 0 x 0 y 0 Maximize P= 35x+ 0y subject to the above constraints 9

20 The graph of the feasible region is shown below. The vertices of the feasible region and the value of the profit function for each vertex are: Vertex Profit (0, 0) $0 (0,0) $700 (0,36) $70 (3,34) $785 From the table, we see that in order to maximize profit at $785 per day, the factory should produce each day 3 tables and 34 chairs. How are problems handled with more than two products? Graphing cannot be done, but computer programs exist that look for vertices of multidimensional regions. Mathematica uses the command or function below to solve this example problem. NMaximize[{35 x+0 y, x+3 y<=08&&x+(/) y<=0&&x>=0&&y>=0},{x,y}] The solution is given as: {785.,{x->3.,y->34.}} 0

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