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oci (1 + 2 √ 2 e = 2 2, 3), (1 − 2 2, 3) 2) √ √ √ √ 2), 530 13. + 2 = 1 4 y − 1 x2 2 9 1 Center 0, 1 2 Major Axis along x = 0 (the y-axis) Minor Axis along y = 1 2 Vertices (0, 2), (0, −1) Endpoints of Minor Axis −1, 1 2 Foci, √ √ 5 5 0, 1− 2, 1, 1 2 √ 0, 1+ 2 5 3 e = 14. Hooked on Conics = 1 (x − 2)2 5 + (y + 2)2 6 √ ... |
��xed two points called foci and looked at points whose distances to the foci always added to a constant distance d. Those prone to syntactical tinkering may wonder what, if any, curve we’d generate if we replaced added with subtracted. The answer is a hyperbola. Definition 7.6. Given two distinct points F1 and F2 in th... |
pictures we have V1 V2 Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse and conjugate axes whose center is the center of the hyperbola and whose diagonals are contained in the asymptotes. This guide rectangle, much akin to the one we saw Sec... |
0)2 = 2a (x − c)2 + y2 = 2a (x − (−c))2 + (y − 0)2 − (x + c)2 + y2 − Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation 7.4, we arrive at the following.1 1It is a good exercise to actually work this out. 534 Hooked on Conics a2 − c2 x2 + a2y2 = a2 a... |
(h, k), we would get the following. Equation 7.6. The Standard Equation of a Horizontala Hyperbola For positive numbers a and b, the equation of a horizontal hyperbola with center (h, k) is aThat is, a hyperbola whose branches open to the left and right (x − h)2 a2 − (y − k)2 b2 = 1 If the roles of x and y were interc... |
hyperbola as well as the corners of the rectangle. This yields the following set up2 −1 1 2 3 4 5 6 x −1 −2 −3 −4 −5 −6 −7 Since the y2 term is being subtracted from the x2 term, we know that the branches of the hyperbola open to the left and right. This means that the transverse axis lies along the x-axis. Hence, the... |
±5, 0), we have a = 5 so a2 = 25. In order to determine b2, we recall that the slopes of the asymptotes are ± b 5 = 2, so b = 10. Hence, b2 = 100 and our final answer is x2 a. Since a = 5 and the slope of the line y = 2x is 2, we have that b 25 − y2 a2 − y2 100 = 1. 7.5 Hyperbolas 537 As with the other conic sections, a... |
the vertical line x = −3 and the conjugate axis lies along the x-axis. Since the vertices of the hyperbola are where the hyperbola intersects and the transverse axis, we get that the vertices are 1 −3, − 1 3 3 of a unit above and below (−3, 0) at −3, 1. To find the foci, we use 3 √ √ c = a2 + b2 = Since the foci lie on... |
distance to Jeff) = 1.9 This is exactly the situation in the definition of a hyperbola, Definition 7.6. In this case, Jeff and Carl are located at the foci,3 and our fixed distance d is 1.9. For simplicity, we assume the hyperbola is centered at (0, 0) with its foci at (−5, 0) and (5, 0). Schematically, we have 2GPS now ru... |
the fixed distance d can be determined as before 760 miles hour × 1 hour 3600 seconds × 18 seconds = 3.8 miles Since Jeff was positioned at (−5, 0), we place Kai at (−5, 6). This puts the center of the new hyperbola at (−5, 3). Plotting Kai’s position and the new center gives us the diagram below on the left. The second... |
approximately at (−0.9629, −0.8113). Each of the conic sections we have studied in this chapter result from graphing equations of the form Ax2 + Cy2 + Dx + Ey + F = 0 for different choices of A, C, D, E, and5 F. While we’ve seen examples6 demonstrate how to convert an equation from this general form to one of the stand... |
+ 2)2 16 = 1 − − − (y + 3)2 9 (y − 4)2 1 (x − 5)2 20 = 1 = 1 = 1 2. 4. 6. 8. − x2 y2 9 16 (y − 3)2 11 (x + 1)2 9 (x − 4)2 8 = 1 − − − (x − 1)2 10 (y − 3)2 4 (y − 2)2 18 = 1 = 1 = 1 In Exercises 9 - 12, put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the ve... |
+ 16 = 0 24. 4x2 + y2 − 8x + 4 = 0 25. 4x2 + 9y2 − 8x + 54y + 49 = 0 26. x2 + y2 − 6x + 4y + 14 = 0 27. 2x2 + 4y2 + 12x − 8y + 25 = 0 28. 4x2 − 5y2 − 40x − 20y + 160 = 0 29. The graph of a vertical or horizontal hyperbola clearly fails the Vertical Line Test, Theorem 1.1, so the equation of a vertical of horizontal hy... |
bolas in that we can define the eccentricity e of a hyperbola as e = distance from the center to a focus distance from the center to a vertex (a) With the help of your classmates, explain why e > 1 for any hyperbola. (b) Find the equation of the hyperbola with vertices (±3, 0) and eccentricity e = 2. (c) With the help o... |
a circle, parabola, ellipse or hyperbola. 544 7.5.2 Answers Hooked on Conics 1. 2. − = 1 y2 x2 9 16 Center (0, 0) Transverse axis on y = 0 Conjugate axis on x = 0 Vertices (4, 0), (−4, 0) Foci (5, 0), (−5, 0) Asymptotes y = ± 3 4 x − = 1 y2 x2 16 9 Center (0, 0) Transverse axis on x = 0 Conjugate axis on y = 0 Vertice... |
4 + Asymptotes y = ± 1 17, 4), (−4 − √ √ 17, 4) 4 (x + 4) + 4 6. (x + 1)2 9 − (y − 3)2 4 = 1 Center (−1, 3) Transverse axis on y = 3 Conjugate axis on x = −1 Vertices (2, 3), (−4, 3) Foci −1 + Asymptotes y = ± 2 13, 3, −1 − √ 3 (x + 1) + 3 √ 13, 3 7. (y + 2)2 16 − (x − 5)2 20 = 1 Center (5, −2) Transverse axis on x = 5... |
−2 + Foci (3, −2 + Asymptotes y = ± √ √ 5), (3, −2 − √ 23), (3, −2 − 23) √ 10 6 (x − 3) − 2 √ 5) 11. (x − 3)2 25 − (y + 1)2 9 = 1 12. Center (3, −1) Transverse axis on y = −1 Conjugate axis on x = 3 Vertices (8, −1), (−2, −1) Foci 3 + 34, −1, 3 − Asymptotes y = ± 3 √ √ 34, −1 5 (x − 3) − 1 − = 1 (y + 4)2 6 (x + 2)2 5 ... |
y + 4)2 = 4x y −1 1 2 3 4 x 24. −1 −2 −3 −4 −5 −6 −7 −8 548 Hooked on Conics 25. (x − 1)2 9 + (y + 3)2 4 = 1 y −2 −1 1 2 3 4 x 26. (x − 3)2 + (y + 2)2 = −1 There is no graph. −1 −2 −3 −4 −5 27. (x + 3)2 2 + There is no graph. (y − 1)2 1 = − 3 4 28. (y + 2)2 16 − (x − 5)2 20 = 10 11 x 4 3 2 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 ... |
we could check our solutions geometrically by finding where the graphs of y = f (x) and y = g(x) intersect. The x-coordinates of these intersection points correspond to the solutions to the equation f (x) = g(x), and the ycoordinates were largely ignored. If we modify the problem and ask for the intersection points of ... |
of Equations and Matrices in the upcoming sections, we do not consider them linear equations. The key to identifying linear equations is to note that the variables involved are to the first power and that the coefficients of the variables are numbers. Some examples of equations which are non-linear are x2 + y = 1, xy = 5... |
+ 4y = −2 + (−3x − y = 5) 3 3y = This gives us y = 1. We now substitute y = 1 into either of the two equations, say −3x−y = 5, to get −3x − 1 = 5 so that x = −2. Our solution is (−2, 1). Substituting x = −2 and y = 1 into the first equation gives 3(−2) + 4(1) = −2, which is true, and, likewise, when we check (−2, 1) in... |
We leave it to the reader to verify that the solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of 3, − 1 2 5 and 2x 3 − 4y. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coefficients belie something more sinis... |
satisfies both equations. = 6. Simplifying, Substituting x = t and we get 2t − 2t + 6 = 6, which is always true. Similarly, when we make these substitutions in the equation 3x − 6y = 9, we get 3t − 6 1 = 9 which reduces to 3t − 3t + 9 = 9, so it 2 t − 3 2 checks out, too. Geometrically, 2x − 4y = 6 and 3x − 6y = 9 are ... |
the first two equations x − y = 0 + (x + y = 2) 2x = 2 which gives x = 1. Substituting this into the first equation gives 1 − y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the first two equations, when we substitute x = 1 and y = 1 into the third equation, we get −... |
ne what is meant by a linear equation in n variables. Definition 8.2. A linear equation in n variables, x1, x2,..., xn, is an equation of the form a1x1 + a2x2 +... + anxn = c where a1, a2,... an and c are real numbers and at least one of a1, a2,..., an is nonzero. Instead of using more familiar variables like x, y, and ... |
it does explain the choice of terminology ‘overdetermined’. 7We need more than two variables to give an example of the latter. 8Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra. 9That is, a system with the same solution set. 8.1 Systems of Linear Equat... |
ometrically, the system is trying to find the intersection, or common point, of all three planes. If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It tu... |
see that the system is in triangular form.12 An example of a more complicated system in triangular form is x1 − 4x3 + x4 − x6 = 6 x2 + 2x3 = 1 x4 + 3x5 − x6 = 8 x5 + 9x6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. Examp... |
�� x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) To satisfy Definition 8.3, we need to eliminate the x’s from E2 and E3. We accomplish this by replacing each of them with a sum of themselves and a multiple of E1. To eliminate the x from E2, we need to multiply E1 by −2 then add; to eliminate the x fr... |
z = −12 To eliminate the y in E3, we add −2E2 to it. (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 Replace E3 with −2E2 + E3 −−−−−−−−−−−−−−−−−−→ (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Finally, we apply the second move from Theorem 8.1 one last time and multiply E3 by −1 to satisfy t... |
−−−−−−−−−−−−→ E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Now it’s time to take care of the x’s in E2 and E3E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3 2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 3 2 z = 8.1 Systems o... |
an inconsistent system. 3. For our last system, we begin by multiplying E1 by 1 3 to get a coefficient of 1 on x1. 3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→ 3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Next we e... |
in E3. 560 Systems of Equations and Matrices (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Replace E3 −−−−−−−−−−→ with − 1 3 E2 + E3 (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 0 = 0 (E3) Equation E3 reduces to 0 = 0,which is always true. Since we have no... |
substitution in Gaussian Elimination is delayed until all the elimination is done, thus it gets called back-substitution. This may also be inefficient in many cases. Rest assured, the technique of substitution as you may have learned it in Intermediate Algebra will once again take center stage in Section 8.7. Lastly, we... |
have amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution is 0.90y. We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes We first eliminate the x from the secon... |
0 ≤ t ≤ 2500 9 }. Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining and it is due to expire. Can we use all of it to make our required solution? We would have y = 100 so that 1 3. This means the amount of 30% solution required is x = − 3 3 mL. The reader is invited to chec... |
= x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9 10. 12. 14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 4x − y + z = 5 2y + 6z = 30 x + z = 6 x − 2y + 3z = 7 −3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x... |
y + 5z = 39 2x1 + x2 − 12x3 − x4 = 16 −x1 + x2 + 12x3 − 4x4 = −5 3x1 + 2x2 − 16x3 − 3x4 = 25 x1 + 2x2 − 5x4 = 11 x1 − x2 − 5x3 + 3x4 = −1 x1 + x2 + 5x3 − 3x4 = 0 x2 + 5x3 − 3x4 = 1 x1 − 2x2 − 10x3 + 6x4 = −1 27. Find two other forms of the parametric solution to Exercise 11 above by reorganizing the equations so that x... |
30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile? 33. Discuss with your classmates how you would approach Ex... |
. 19. 20. 21. 22. 23. 24. 25. 0 x + y + 2z = 4 y − 7z = 17 z = −2 x − 2y + 2z = − − 3y − 4z = 3 13 z = 4 13 0 = 0 y + 11 − 2z = − = − 11 2 0 0 = Consistent independent Solution (1, 3, −2) Inconsistent no solution Consistent independent So... |
= 0 0 = Inconsistent No solution 27. If x is the free variable then the solution is (t, 3t, −t + 5) and if y is the free variable then the solution is 1 3 t, t, − 1 3 t + 5. 28. 13 chose the basic buffet and 14 chose the deluxe buffet. 29. Mavis needs 20 pounds of $3 per pound coffee and 30 pounds of $8 per pound coffee. ... |
A, B, C, etc.) while their entries are usually denoted by the corresponding letter. So, for instance, if we have A = 3 0 −1 10 2 −5 then a11 = 3, a12 = 0, a13 = −1, a21 = 2, a22 = −5, and a23 = 10. We shall explore matrices as mathematical objects with their own algebra in Section 8.3 and introduce them here solely as ... |
in solving the systems of linear equations in Example 8.1.2 of Section 8.1. The reader is encouraged to perform the indicated operations on the rows of the augmented matrix to see that the machinations are identical to what is done to the coefficients of the variables in the equations. We first see a demonstration of swi... |
−2 0 Finally, we have an example of replacing a row with itself plus a multiple of another row using the second step from part 2 in Example 8.1.2. 1We shall study the coefficient and constant matrices separately in Section 8.3. 8.2 Systems of Linear Equations: Augmented Matrices 569 E1) 2 10x − z = 2 (E2) (E3)... |
to the rows to get the matrix into row-echelon form. We then decode the matrix and back substitute. The next example illustrates this nicely. Example 8.2.1. Use an augmented matrix to transform the following system of linear equations into triangular form. Solve the system. 3x − y + z = 8 x + 2y − z = 4 2x + 3y ... |
−1 1 0 −7 0 −1 −2 4 4 −4 2 Now we repeat the above process for the variable y which means we need to get the leading entry in R2 to be 1. 2 −1 1 0 −7 0 −1 −2 4 4 −4 2 Replace R2 with − 1 7 R2 −−−−−−−−−−−−−−→ 2 −1 1 1 − 4 0 7 0 −1 −2 4 4 7 2 To guarantee the leading 1 in R3 is to the right of the le... |
lon form. If we also require that 0’s are the only numbers above a leading 1, we have what is known as the reduced row echelon form of the matrix. Definition 8.5. A matrix is said to be in reduced row echelon form provided both of the following conditions hold: 1. The matrix is in row echelon form. 2. The leading 1s are... |
have put a matrix into reduced row echelon form is called Gauss-Jordan Elimination. Example 8.2.2. Solve the following system using an augmented matrix. Use Gauss-Jordan Elimination to put the augmented matrix into reduced row echelon form. x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Solution. We first encode th... |
−−−−−−−−→ Next, it’s time for a leading 1 in R3. 1 − 1 3 3 19 1 0 2 0 −24 −5 −35 0 0 − 1 3 − 2 1 6 Replace R3 with − 1 24 R3 −−−−−−−−−−−−−−−→ 1 − 1 3 3 19 1 0 2 0 −24 −5 −35 24 3 19 2 35 24 The matrix is now in row echelon form. To get the reduced row echelon form, we start with the last leading... |
ations: Augmented Matrices 573 Like all good algorithms, putting a matrix in row echelon or reduced row echelon form can easily be programmed into a calculator, and, doubtless, your graphing calculator has such a feature. We use this in our next example. Example 8.2.3. Find the quadratic function passing through the po... |
(x) = − 7 18 x + 37 with the points (−1, 3), (2, 4) and (5, −2) 18 x2 + 13 9 4We’ve tortured you enough already with fractions in this exposition! 574 Systems of Equations and Matrices 8.2.1 Exercises In Exercises 1 - 6, state whether the given matrix is in reduced row echelon form, row echelon form only or in neither... |
3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = 8.2 Systems of Linear Equations: Augmented Matrices 575 21. 23. 25. 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9 x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1... |
, cashews, and peanuts? (You may find it helpful to review Example 8.1.3 in Section 8.1.) 29. Find the quadratic function passing through the points (−2, 1), (1, 4), (3, −2) 30. At 9 PM, the temperature was 60◦F; at midnight, the temperature was 50◦F; and at 6 AM, the temperature was 70◦F. Use the technique in Example 8... |
. Inconsistent 11. (8s − t + 7, −4s + 3t + 2, s, t) for all real numbers s and t 12. (−9t − 3, 4t + 20, t) for all real numbers t 13. (−2, 7) 15. (−t + 5, −3t + 15, t) for all real numbers t 17. (1, 3, −2) 19. (1, 3, −2) 14. (1, 2, 0) 16. (2, −1, 1) 18. Inconsistent 20. −3, 1 2, 1 21. 1 3, 2 3, 1 22. 19 13, t 13, − 11 ... |
. This system is 5, t. Our variables repreconsistent dependent and its solution is 8 15 t + 2 sent numbers of adults and children so they must be whole numbers. Running through the values t = 0, 1, 2, 3, 4 yields only one solution where all four variables are whole numbers; t = 3 gives us (2, 1, 1, 3). Thus there are 2... |
52/3 32 · 13 log(0.001) Now that we have an agreed upon understanding of what it means for two matrices to equal each other, we may begin defining arithmetic operations on matrices. Our first operation is addition. Definition 8.7. Matrix Addition: Given two matrices of the same size, the matrix obtained by adding the corr... |
+ [aij]m×n = B + A where the second equality is the definition of A + B, the third equality holds by the commutative law of real number addition, and the fourth equality is the definition of B + A. In other words, matrix addition is commutative because real number addition is. A similar argument shows the associative pr... |
, its additive inverse, which we call −A, does exist and is unique and, moreover, is given by the formula: −A = [−aij]m×n. The long and short of this is: to get the additive inverse of a matrix, 3A technical detail which is sadly lost on most readers. 580 Systems of Equations and Matrices take additive inverses of each... |
�cally, if k is a real number and A = [aij]m×n, we define kA = k [aij]m×n = [kaij]m×n aThe word ‘scalar’ here refers to real numbers. ‘Scalar multiplication’ in this context means we are multiplying a matrix by a real number (a scalar). One may well wonder why the word ‘scalar’ is used for ‘real number.’ It has everythi... |
then kA = 0m×n if and only if k = 0 or A = 0m×n As with the other results in this section, Theorem 8.4 can be proved using the definitions of scalar multiplication and matrix addition. For example, to prove that k(A + B) = kA + kB for a scalar k and m × n matrices A and B, we start by adding A and B, then multiplying b... |
3 −5 + − −2 1 −3 −5 = = (−2)A + 02×2 = (−2)A = (−2)A = −4 2 6 −2 −4 2 6 −2 −4 2 6 −2 1 3 + + + 9 12 −3 39 1 (9) 1 3 3 (−3) 1 1 3 3 4 3 −1 13 (12) (39) −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 + − −2 1 −3 −5 −1 6 5 11 −1 − (−2) − −2 1 −3 −5 6 − 1 5 − (−3) 11 − (−5) − 1 2 ((−2)A) = − 1 2 − 1 2 (−2) A =... |
−12 Let R1 denote the first row of A and C1 denote the first column of B. To find the ‘product’ of R1 with C1, denoted R1 · C1, we first find the product of the first entry in R1 and the first entry in C1. Next, we add to that the product of the second entry in R1 and the second entry in C1. Finally, we take that sum and... |
−12 6See this article on the Hadamard Product. 584 Systems of Equations and Matrices −−−−−−−−−→ −10 3 5 2 −5 −2 a21b13 = (−10)(2) = −20 + −−−−−−−−−→ −10 5 3 2 −5 −2 −−−−−−−−−→ −10 3 5 2 −5 −2 a22b23 = (3)(−5) = −15 a23b33 = (5)(−2) = −10 + Generalizing this process, we have the follow... |
There are a number of subtleties in Definition 8.10 which warrant closer inspection. First and foremost, Definition 8.10 tells us that the ij-entry of a matrix product AB is the ith row of A times the jth column of B. In order for this to be defined, the number of entries in the rows of A must match the number of e... |
BA. Although there is no commutative property of matrix multiplication in general, several other real number properties are inherited by matrix multiplication, as illustrated in our next theorem. Theorem 8.5. Properties of Matrix Multiplication Let A, B and C be matrices such that all of the matrix products below are ... |
5 3 In order for the product IkA to be defined, k = 2; similarly, for AIk to be defined, k = 3. We leave it to the reader to show I2A = A and AI3 = A. In other words, and 1 0 0 1 2 −10 0 −1 5 3 = 2 −10 0 −1 5 3 2 −10 0 −10 0 −1 5 3 While the proofs of the properties in Theorem 8.5 are computational in nature, the notati... |
I4M ) + 3 (M I4) − 2 (I4 (3I4)) = M 2 − 2M + 3M − 2 (3 (I4I4)) = M 2 + M − 6I4 Example 8.3.2 illustrates some interesting features of matrix multiplication. First note that in part 1, neither A nor B is the zero matrix, yet the product AB is the zero matrix. Hence, the the zero product property enjoyed by real numbers ... |
−1 −1 1 2 3 Q x T −2 −3 −4 P RT For a generic point P (x, y) on the hyperbola x2 − y2 = 4, we have RP = = √ which means R takes (x, y) to y = 2 x, we replace x with √, √ 2 2 y and y with √ √ +. To show that this point is on the curve √ 2 2 y and simplify. 8.3 Matrix Arithmetic 589 √ √ = 2 2 x − y2 x2 x2 − y2 Since (x,... |
B 3 −1 3 −4 x y z 3x − y + z x + 2y − z 2x + 3y − 4z = = 8 4 10 8 4 10 We see that finding a solution (x, y, z) to the original system corresponds to finding a solution X for the matrix equation AX = B. If we think about solving the real number equation ax = b, we would simply ‘divide’ bo... |
− 2I2 18. E2 + 5E − 36I3 19. EDC 20. CDE 22. Let A = a b c d e f E1 = 21. ABCEDI2 0 1 1 0 E2 = 5 0 0 1 E3 = 1 −2 1 0 Compute E1A, E2A and E3A. What effect did each of the Ei matrices have on the rows of A? Create E4 so that its effect on A is to multiply the bottom row by −6. How would you extend this idea to matrices w... |
week to week, then Q remains the same and we have what’s known as a Stochastic Process10 because Week n’s numbers are found by computing QnX. Choose a few values of n and, with the help of your classmates and calculator, find out how many people get each paper for that week. You should start to see a pattern as n → ∞. ... |
below the main diagonal are zero and it is said to be a lower triangular matrix if all of its entries above the main diagonal are zero. For example9 0 −5 from Exercises 8 - 21 above is an upper triangular matrix whereas F = 1 0 3 0 is a lower triangular matrix. questions with your classmates. (Zeros are allowed on the... |
3A = 7 8 9 and B = 1 2 3 21 24 27 A2 is not defined AB = 7 14 21 8 16 24 9 18 27 6. For A = −3 1 −2 4 5 −6 and B = −5 1 8 3A = 3 −6 −9 12 15 −18 A2 is not defined AB is not defined Systems of Equations and Matrices −B = −1 −2 −3 A − 2B is not defined BA = [50] −B = 5 −1 −8 A − 2B is not... |
+ D is undefined 13. CD + 2I2A = 238 3 −126 361 863 5 15 15. A2 − B2 = 17. A2 − 5A − 2I2 = −8 16 3 25 0 0 0 0 18. E2 + 5E − 36I3 = −30 20 −15 0 −36 0 −36 0 0 19. EDC = 3449 15 − 407 15 − 101 99 6 − 9548 3 −648 −324 −35 −360 20. CDE is undefined 21. ABCEDI2 = − 90749 15 − 156601 15 − 28867 5 − 47033 5... |
attention to the properties of real numbers being used at each step. Recall that dividing by 3 is the same as multiplying by 1 3 = 3−1, the so-called multiplicative inverse 1 of 3. 3x = 5 3−1(3x) = 3−1(5) Multiply by the (multiplicative) inverse of 3 Associative property of multiplication Inverse property Multiplicati... |
−1 = I2. We have AA−1 = I2 x1 x2 2 −3 x3 x4 4 3 2x1 − 3x3 2x2 − 3x4 3x1 + 4x3 3x2 + 4x4 = = 1 0 0 1 1 0 0 1 This gives rise to two more systems of equations 2x1 − 3x3 = 1 3x1 + 4x3 = 0 2x2 − 3x4 = 0 3x2 + 4x4 = 1 At this point, it may seem absurd to continue with this venture. After all, the intent was to solve one sys... |
its augmented matrix into reduced row echelon form (Think about why that works.) and we obtain 17 and x3 = − 3 2 −3 0 4 1 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 1 0 0 1 3 17 2 17 which means x2 = 3 17 and x4 = 2 17. Hence, A−1 = x1 x2 x3 x4 = 4 17 − 3 17 3 17 2 17 We can check to see that A−1 behaves as it shoul... |
x2 x4 Since the row operations for both processes are the same, all of the arithmetic on the left hand side of the vertical bar is identical in both problems. The only difference between the two processes is what happens to the constants to the right of the vertical bar. As long as we keep these separated into columns,... |
not all matrices are square, not all matrices are invertible. However, just because a matrix is square doesn’t guarantee it is invertible. (See the exercises.) Our first result summarizes some of the important characteristics of invertible matrices and their inverses. Theorem 8.6. Suppose A is an n × n matrix. 1. If A ... |
justifies our quest to find A−1 using our super-sized augmented matrix approach A In Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ In A−1 We are, in essence, trying to find the unique solution to the equation AX = In using row operations. What does all of this mean for a system of linear equations? Theorem 8.6 tells us that... |
Use A−1 to solve the following systems of equations (a) 3x + y + 2z = 26 −y + 5z = 39 2x + y + 4z = 117 (b) 3x + y + 2z = 4 −y + 5z = 2 2x + y + 4z = 5 (c) 3x + y + 2z = 1 −y + 5z = 0 2x + y + 4z = 0 Solution. 1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination. ... |
0 3 1 0 − 10 0 1 − 2 13 13 1 13 Replace R1 with −−−−−−−−−−→ 3 R2 + R1 − 1 2 13 − 7 9 1 0 0 13 0 1 0 − 10 0 0 1 − 2 13 13 − 8 13 1 13 13 15 13 3 13 13 15 13 3 13 We find A−1 = 2 13 − 7 13 15 13 3 13 9 13 − 10 − 2 13 13 − 8 13 1 13 A−1A = and . To check our answer, we compute 2 13 − 7 9... |
3 13 13 15 13 3 13 13 15 13 3 13 = −39 91 . Our solution is (−39, 91, 26). 26 . We get 5 13, 19 13, 9 13. 5 13 19 13 9 13 . We find 9 13, − 10 13, − 2 13.6 9 13 − 10 13 − 2 13 In Example 8.4.1, we see that finding one inverse matrix can enable us to solve an entire family of systems of linear ... |
R1i4 = VB1 −R3i1 + (R2 + R3 + R4) i2 − R4i3 − R2i4 = 0 −R4i2 + (R4 + R6) i3 − R6i4 = −VB2 −R1i1 − R2i2 − R6i3 + (R1 + R2 + R5 + R6) i4 = 0 1. Assuming the resistances are all 1kΩ, find the mesh currents if the battery voltages are (a) VB1 = 10V and VB2 = 5V (b) VB1 = 10V and VB2 = 0V (c) VB1 = 0V and VB2 = 10V (d) VB1 ... |
matrix A into the calculator, we find from which we have A−1 = 1.625 1.25 1.125 1 1.25 1.125 1.5 1.25 1.25 1.625 To solve the four systems given to us, we find X = A−1B where the value of B is determined by the given values of VB1 and VB2 1 (a) B = 10 0 −5 0, 1 (b) B = 10 0 0 0, 1... |
B1 = 10V and VB2 = 10V, we get i1 = 5 mA, i2 = 0 mA, i3 = −5 mA, and i4 = 0 mA. The mesh currents i2 and i4 being zero is a consequence of both batteries ‘pushing’ in equal but opposite directions, causing the net flow of electrons in these two regions to cancel out. 2. We now turn the tables and are given VB1 = 10V, VB... |
� R1 + 0.7R3 = 1.7 R2 − 3.5R3 − 1.5R6 = −4 R4 + 0.6R6 = 1.6 1 R5 = We can solve for R1, R2, R4 and R5 leaving R3 and R6 as free variables. Labeling R3 = s and R6 = t, we have R1 = −0.7s + 1.7, R2 = 3.5s + 1.5t − 4, R4 = −0.6t + 1.6 and R5 = 1. Since resistance values are always positive, we need to restrict our... |
that the matrix is not invertible. 1. A = 3. C = 1 2 3 4 6 15 14 35 2. B = 4. D = 12 −7 −5 3 2 −1 16 −9 5. E = 7. G = 0 3 2 −1 4 3 2 −5 3 2 3 11 4 19 −3 1 2 3 6. F = 83 4 −3 6 2 1 2 −2 0 0 0 −3 8 16 4 − In Exercises 9 - 11, use one matrix inverse to solve the following systems of linear equations. 3x +... |
’s diet consists of three primary foods: Ippizuti Fish, Misty Mushrooms, and Sun Berries. Each serving of Ippizuti Fish is 500 calories, contains 40 grams of protein, and has no Vitamin X. Each serving of Misty Mushrooms is 50 calories, contains 1 gram of protein, and 5 milligrams of Vitamin X. Finally, each serving of... |
and make them our first column, the next three numbers, 6 15 15, and make them our second column, and so on. We put 0’s to round out the matrix. M = 6 20 2 9 15 7 15 12 9 19 0 0 0 22 5 To encode the message, we find the product AM AM = 2 −3 3 −7 5 1 −2 1 −1 6 20 2 9 15 7 15 12 9 19 0 0 0 22 5 =... |
3. C is not invertible 5. E−1 = −1 8 4 1 −3 −1 1 −6 −3 2. B−1 = 3 7 5 12 4. D−1 = 9 2 − 1 2 8 −1 6. F − 16 3 0 2 − 35 −90 − 1 2 0 1 5 0 −7 −36 0 7 2 0 1 7. G is not invertible 8. H −1 = The coefficient matrix is B−1 from Exercise 2 above so the inverse we need is (B−1)−1 = B. 12 −7 −5 3 26 39 9. ... |
of Ippizuti Fish and 20 servings of Sun Berries daily. (No Misty Mushrooms are needed!) (c) The adolescent Sasquatch requires 10 servings of Ippizuti Fish daily. (No Misty Mush- rooms or Sun Berries are needed!) 17. (a) A−1 = 1 2 1 17 33 19 10 19 11 (b) 1 1 2 17 33 19 10 19 11 42 3 12 1 100 −23 38 ... |
Delete R2 and C3 −−−−−−−−−−−→ A23 = 3 1 2 1 We are now in the position to define the determinant of a matrix. Definition 8.13. Given an n × n matrix A the determinant of A, denoted det(A), is defined as follows If n = 1, then A = [a11] and det(A) = det ([a11]) = a11. If n > 1, then A = [aij]n×n and det(A) = det [aij]n×n ... |
1]) + 3 det([2]) = 4(1) + 3(2) = 10 For a generic 2 × 2 matrix A = a b c d we get det(A) = det a b c d = a det (A11) − b det (A12) = a det ([d]) − b det ([c]) = ad − bc This formula is worth remembering Equation 8.1. For a 2 × 2 matrix, det a b c d = ad − bc Applying Definition 8.13 to the 3 × 3 matrix A = 3 1 0 −1 ... |
− a12 det (A12) + −... + (−1)1+na1n det (A1n) can be rewritten as a11(−1)1+1 det (A11) + a12(−1)1+2 det (A12) +... + a1n(−1)1+n det (A1n) which, in the language of cofactors is a11C11 + a12C12 +... + a1nC1n We are now ready to state our main theorem concerning determinants. Theorem 8.7. Properties of the Determinant: ... |
what few properties we can after we have developed some more tools such as the Principle of Mathematical Induction in Section 9.3.2 For the moment, let us demonstrate some of the properties listed in Theorem 8.7 on the matrix A below. (Others will be discussed in the Exercises.) A = 3 1 0 −1 1 2 2 5 4 We found... |
transform our 3 × 3 matrix A into an upper triangular matrix, keeping track of the row operations, and labeling 2For a very elegant treatment, take a course in Linear Algebra. There, you will most likely see the treatment of determinants logically reversed than what is presented here. Specifically, the determinant is d... |
reduced row echelon form must be 1. That is, A’s reduced row echelon form is In, and A is invertible. Conversely, if A is invertible, then A can be transformed into In using row operations. Since det (In) = 1 = 0, our same logic implies det(A) = 0. Basically, we have established that the determinant determines whether... |
2,... xn is given by: xj = det (Aj) det(A), where Aj is the matrix A whose jth column has been replaced by the constants in B. In words, Cramer’s Rule tells us we can solve for each unknown, one at a time, by finding the ratio of the determinant of Aj to that of the determinant of the coefficient matrix. The matrix Aj is ... |
A3 = Az = 2 −3 −1 1 −1 1 0 0 3 Expanding both det(A) and det (Az) along the third rows (to take advantage of the 0’s) gives z = det (Az) det(A) = −12 −10 = 6 5 The reader is encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alterna... |
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