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you would approach Exercise 32 above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister. 34. If you were to try to make 100 mL of a 60% acid solution using stock solutions at 20% and 40%, respectively, what would the triangular form of the resulting system look like? Explain. 5... |
�� 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 Solution (1, 3, −2) Consistent independent Solution −3, 1 2, 1 Consis... |
29. Mavis needs 20 pounds of $3 per pound coffee and 30 pounds of $8 per pound coffee. 30. Skippy needs to invest $6000 in the 3% account and $4000 in the 8% account. 31. 22.5 gallons of the 10% solution and 52.5 gallons of pure water. 32. 4 3 − 1 2 t pounds of Type I, 2 3 − 1 2 t pounds of Type II and t pounds of Type ... |
objects with their own algebra in Section 8.3 and introduce them here solely as a bookkeeping device. Consider the system of linear equations from number 2 in Example 8.1.2 2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 We encode this system into a matrix by assigning each equation to a correspondin... |
�rst step of part 1 in Example 8.1.2. 3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→ 3 −1 2 −4 1 −1 1 3 3 16 5 1 Switch R1 and R3 −−−−−−−−−−−→ x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) 1 −1 2 −4 3 −1 5 1 3 16 3 1 Next, we have ... |
�� E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3 3 1 10 4 −9 2 − 1 2 0 −1 2 1 2 2 5 Replace R2 with −10R1 + R2 −−−−−−−−−−−−−−−−−−→ Replace R3 with −4R1 + R3 2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 15 0 −15 ... |
y − z = 4 2x + 3y − 4z = 10 Solution. We first encode the system into an augmented matrix. 3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Encode into the matrix −−−−−−−−−−−−−−→ 3 −1 1 2 1 8 2 −1 4 3 −4 10 Thinking back to Gaussian Elimination at an equations level, our first order of business is to get x i... |
2 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 leading 1 in R2, we get a 0 in the second column of R3. 2 −1 1 1 − 4 0 7 0 −1 −2 4 4 7 2 Replace R3 with R2 + R3 −−−−−−−−−−−−−−−−→ 1 0 0 2 −1 1 − 4 7 0 − 18 7 4 4 7 18 7 F... |
form. 2. The leading 1s are the only nonzero entry in their respective columns. 8.2 Systems of Linear Equations: Augmented Matrices 571 Of what significance is the reduced row echelon form of a matrix? To illustrate, let’s take the row echelon form from Example 8.2.1 and perform the necessary steps to put into reduced ... |
1 + 4x3 = 5 4x2 − x4 = 3 Solution. We first encode the system into a matrix. (Pay attention to the subscripts!) x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Encode into the matrix −−−−−−−−−−−−−−→ −1 3 Next, we get a leading 1 in the first column of R1. −1 3 Replace R1 with − 1 3 R1 −−−−−−−−−−−−−−→ ... |
−−−−−−→ 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 1 we produced and work to get 0’s above it 24 3 19 2 35 24 Replace R2 with −6R3 + R2 −−−−−−−−−−−−−−−−−→ Lastly, we get a 0 above the le... |
(2, 4), (5, −2). Solution. According to Definition 2.5, a quadratic function has the form f (x) = ax2 +bx+c where a = 0. Our goal is to find a, b and c so that the three given points are on the graph of f. If (−1, 3) is on the graph of f, then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to a − b + c = 3, an hone... |
given matrix is in reduced row echelon form, row echelon form only or in neither of those forms. 1 0 3 0 1 3 1. 4. 5. 3 −1 2 −4 1 −1 1 3 3 16. 6 In Exercises 7 - 12, the following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that ... |
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 4x − 4y + 4z = −13 6x − 5y + 7z = −25 22. 24. 26. x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 x − y + z = 8 3... |
◦F; and at 6 AM, the temperature was 70◦F. Use the technique in Example 8.2.3 to fit a quadratic function to these data with the temperature, T, measured in degrees Fahrenheit, as the dependent variable, and the number of hours after 9 PM, t, measured in hours, as the independent variable. What was the coldest temperatu... |
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 13 t + 51 for all real numbers t 13 t + 4 23. Inconsistent 25. −2t − 35 2, t for all real numbers t 4, −t − 11 24. (4, −3, 1) 26. (1, 2, 3, 4) 27. This time, 7 diners chose the del... |
there are 2 adults and 1 child in the Zahlenreichs and 1 adult and 3 kids in the Nullsatzs. 5, −t + 4, − 8 15 t + 13 578 Systems of Equations and Matrices 8.3 Matrix Arithmetic In Section 8.2, we used a special class of matrices, the augmented matrices, to assist us in solving systems of linear equations. In this sect... |
, the matrix obtained by adding the corresponding entries of the two matrices is called the sum of the two matrices. More specifically, if A = [aij]m×n and B = [bij]m×n, we define A + B = [aij]m×n + [bij]m×n = [aij + bij]m×n As an example, consider the sum below. 1Recall that means A has m rows and n columns. 2Critics ma... |
argument shows the associative property of matrix addition also holds, inherited in turn from the associative law of real number addition. Specifically, for matrices A, B, and C of the same size, (A + B) + C = A + (B + C). In other words, when adding more than two matrices, it doesn’t matter how they are grouped. This ... |
each of its entries. With the concept of additive inverse well in hand, we may now discuss what is meant by subtracting matrices. You may remember from arithmetic that a − b = a + (−b); that is, subtraction is defined as ‘adding the opposite (inverse).’ We extend this concept to matrices. For two matrices A and B of th... |
real number (a scalar). One may well wonder why the word ‘scalar’ is used for ‘real number.’ It has everything to do with ‘scaling’ factors.5 A point P (x, y) in the plane can be represented by its position matrix, P : (x, y) ↔ P = x y Suppose we take the point (−2, 1) and multiply its position matrix by 3. We have 3P... |
× n matrices A and B, we start by adding A and B, then multiplying by k and seeing how that compares with the sum of kA and kB. k(A + B) = k [aij]m×n + [bij]m×n = k [aij + bij]m×n = [k (aij + bij)]m×n = [kaij + kbij]m×n As for kA + kB, we have kA + kB = k [aij]m×n + k [bij]m×n = [kaij]m×n + [kbij]m×n = [kaij + kbij]m×... |
(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 = 1 5 8 16 1 5 8 16 (1) − 1 (5) 2 (16) (8 1A = A = − 1 2 − 5 2 −4 − 16 2 − 1 2 − 5 2 −4 −8 The reader is encouraged to check our answer in th... |
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 we add to that the product of the last entry in R1 and the last entry in C1. Using entry notation, R1·C1 = a11b11 +a12b21 +a13b31 = (2)(3)+(0)(4)+(−1)(5) = 6+0+(−5)... |
−−−−−→ −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 following definition. Definition 8.9. Product of a Row and a Column: Suppose A = [aij]m×n and B = [bij]n×r. Let Ri denote the ith row of A and let Cj denote the jth ... |
��ned, the number of entries in the rows of A must match the number of entries in the columns of B. This means that the number of columns of A must match7 the number of rows of B. In other words, to multiply A times B, the second dimension of A must match the first dimension of B, which is why in Definition 8.10, Am×n is... |
Matrix Multiplication: (AB)C = A(BC) Associative Property with Scalar Multiplication: k(AB) = (kA)B = A(kB) Identity Property: For a natural number k, the k × k identity matrix, denoted Ik, is defined by Ik = [dij]k×k where dij = 1, if i = j 0, otherwise For all m × n matrices, ImA = AIn = A. Distributive Property of M... |
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 notation becomes quite involved very quickly, so they are left to a course in Linear Algebra. The following example provides some practice with matrix multiplication and its properties. As usual, some valuable ... |
the product AB is the zero matrix. Hence, the the zero product property enjoyed by real numbers and scalar multiplication does not hold for matrix multiplication. Parts 2 and 3 introduce us to polynomials involving matrices. The reader is encouraged to step back and compare our expansion of the matrix product (M − 2I4... |
√ +. 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, y) is on the hyperbola x2 − y2 = 4, we know that this last equation is true. Since all of our steps are reversible, this last equation is equivalent to our original equation, which establi... |
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’ both sides by a. Is it possible to ‘divide’ both sides of the matrix equation AX = B by the matrix A?... |
�ect 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 with more than two rows? 592 Systems of Equations and Matrices In Exercises 23 - 29, consider the following scenario. In the small village of Pedimaxus... |
∞. 27. If you didn’t see the pattern, we’ll help you out. Let Xs = 100 50. Show that QXs = Xs This is called the steady state because the number of people who get each paper didn’t change for the next week. Show that QnX → Xs as n → ∞. 10More specifically, we have a Markov Chain, which is a special type of stochastic p... |
of Equations and Matrices 32. Give an example of a matrix which is neither upper triangular nor lower triangular. 33. Is the product of two n × n upper triangular matrices always upper triangular? 34. Is the product of two n × n lower triangular matrices always lower triangular? 35. Given the matrix A = 1 2 3 4 write ... |
−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 defined BA = 32 −34 7. For A = 2 −3 3 −7 5 1 −2 1 −1 and B = 1 1 2 17 33 19 10 19 11 3A = A2 = 6... |
−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 d e f c a b E1 interchanged R1 and R2 of A. 22. E1A = E2A = E3A = d 5a 5b 5c f a − 2d b − 2e c − 2f f e e d E4 = 1 0 0 −6 E2 multiplied R1 of A by... |
property Multiplicative Identity 3−1 · 3 x = 3−1(5) 1 · x = 3−1(5) x = 3−1(5) If we wish to check our answer, we substitute x = 3−1(5) into the original equation 3x 3 3−1(5) 3 · 3−1 (5 Associative property of multiplication? = 5 = 5 Multiplicative Identity Inverse property Thinking back to Theorem 8.5, we know that ma... |
x3 = 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 system of equations, and in doing so, we have produced two more to solve. Remember, the objective of this discussion is to develop a general method which, when used in... |
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 should by computing AA−1 As an added bonus, AA−1 = 2 −3 4 3 4 17 − 3 17 3 17 2 17 1 0 0 1 = = I2 A−1A = 4 17 − 3 17 3 17 2 17 2 −3 4 3 = 1 0 0 1 = I2 We can now return to the problem at hand. From our discussion at the beginning of th... |
‘super-sized’ augmented matrix and describe the above process as A 1 0 0 1 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 x1 x2 x3 x4 We have the identity matrix I2 appearing as the right hand side of the first super-sized augmented matrix and the left hand side of the second super-sized augmented matrix. To our surpris... |
n matrix. 1. If A is invertible then A−1 is unique. 2. A is invertible if and only if AX = B has a unique solution for every n × r matrix B. The proofs of the properties in Theorem 8.6 rely on a healthy mix of definition and matrix arithmetic. To establish the first property, we assume that A is invertible and suppose t... |
What does all of this mean for a system of linear equations? Theorem 8.6 tells us that if we write the system in the form AX = B, then if the coefficient matrix A is invertible, there is only one solution to the system − that is, if A is invertible, the system is consistent and independent.5 We also know that the proces... |
x + 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. Replace R1 −−−−−−−→ with 1 3 R1 Replace R3 with −−−−−−−−−−→ −2R1 + R3 1 1 3 0 −1 1 2 1 1 3 0 −1 1 0 3 1 0 0 Replace R2 −−−−−−−−→ with (−1)... |
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 13 − 10 − 2 13 13 − 8 13 1 13 13 15 13 3 13 1 3 0 − = I3 AA−1 = 1 3 0 − 13 − 10 − 2 13 13 − 8 13 1 13 2 13 − ... |
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 equations. There are many examples of where this comes in handy ‘in the wild’, and we chose our example for this section from the ... |
(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 = 10V and VB2 = 10V 2. Assuming VB1 = 10V and VB2 = 5V, find the possible combinations of resistances which would... |
B 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 (c10 0, 1 (d) B = 10 0 10 0 (a) For VB1 = 10V and VB2 = 5V, the calculator gives i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA, and i4 = 5 mA. We include a cal... |
�ow of electrons in these two regions to cancel out. 2. We now turn the tables and are given VB1 = 10V, VB2 = 5V, i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA and i4 = 5 mA and our unknowns are the resistance values. Rewriting our system of equations, we get 1.25R2 − 4.375R3 + 3.125R4 = 5.625R1 + 4.375R3 = 10 ... |
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 values of s and t. We know R3 = s > 0 and when we combine that with R1 = −0.7s + 1.7 > 0, we get 0 < ... |
�� 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 + 7y = 26 5x + 12y = 39 9. 10. 3x + 7y = 0 5x + 12y = −1 11. 3x + 7y = −7 5 5x + 12y = In Exercises 12 - 14, use the inverse of E from Exercise 5... |
each serving of Sun Berries is 80 calories, contains no protein, but has 15 milligrams of Vitamin X.9 (a) If an adult male Sasquatch requires 3200 calories, 130 grams of protein, and 275 milligrams of Vitamin X daily, use a matrix inverse to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries... |
, 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 = 42 3 12 1 100 −23 38 57 39 36 −12 −42 −152 −46 −133 9Misty Mushrooms and Sun Berries are the only known fictional sources of Vitamin X. 10Vitamin X is needed to sustain Sasquatch longevity only. 8.4 Systems o... |
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. = 39 −13 10. 11. 12 −7 −5 3 0 −1 12 −7 −5 3 −7 5 = = 7 −3 So x = 39 and y = −13. So x = 7 and y = −3. −119 50 So x = −... |
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 57 39 36 −12 −42 −152 −46 −133 = 6 20 2 9 15 7 15 12 0 22 5 9 19 0 0 (c) ‘LOGS RULE’ 614 Systems of Equations and Matrices 8.5 Determinants and Cramer’s Rule 8.5.1 Definition and Properties of the Determinant In this section... |
�ned 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 = a11 det (A11) − a12 det (A12) + −... + (−1)1+na1n det (A1n) ‘det(A)’ and ‘|A|’ There are two commonly used notations for the determinant of a matrix A: We have chosen to use the notation det(... |
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 1 2 2 5 4 we obtain det(A) = det 3 1 0 − det (A11) − 1 det (A12) + 2 det (A13) = 3 det −1 5 1 4 − det 0 5 2 4 + 2 det 0 −1 1 2 = 3((−1)(4) − (5)(1)) − ((0)(4) − (5)(2)) ... |
(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: Let A = [aij]n×n. We may find the determinant by expanding along any row. That is, for any 1 ≤ k ≤ n, det(A) = ak1Ck1 + ak2Ck2 +... +... |
2 2 5 4 We found det(A) = −13 by expanding along the first row. To take advantage of the 0 in the second row, we use Theorem 8.7 to find det(A) = −13 by expanding along that row. det 1 3 0 −1 1 2 2 5 4 = 0C21 + (−1)C22 + 5C23 = (−1)(−1)2+2 det (A22) + 5(−1)2+3 det (A23) = − det 3 2 2 4 − 5 det 3 1 2 ... |
which takes a square matrix to a real number and satisfies some of the properties in Theorem 8.7. From that function, a formula for the determinant is developed. 618 Systems of Equations and Matrices each successive matrix.3 3 1 0 −1 1 2 A 2 5 4 Replace R3 −−−−−−−−−−→ with − 2 3 R1 + R3 3 1 0 − Replace R3 w... |
that the determinant determines whether or not the matrix A is invertible.4 It is worth noting that when we first introduced the notion of a matrix inverse, it was in the context of solving a linear matrix equation. In effect, we were trying to ‘divide’ both sides of the matrix equation AX = B by the matrix A. Just like... |
Aj to that of the determinant of the coefficient matrix. The matrix Aj is found by replacing the column in the coefficient matrix which holds the coefficients of xj with the constants of the system. The following example fleshes out this method. Example 8.5.1. Use Cramer’s Rule to solve for the indicated unknowns. 1. Solve 2... |
encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alternative method for finding the inverse of a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 A = 3 1 0 −1 1 2 2 5 4 We found through a... |
a method in the forthcoming discussion. As with the discussion in Section 8.4 when we developed the first algorithm to find matrix inverses, we ask that you indulge us. 6The reader is encouraged to stop and think this through. 8.5 Determinants and Cramer’s Rule 621 If we expand det (A1) along the first row, we get det (A... |
and surprising formula for A−1, namely A−1 = 1 det(A) C11 C21 C31 C12 C22 C32 C13 C23 C33 To see that this does indeed yield A−1, we find all of the cofactors of A C11 = −9, C21 = −2, C31 = C12 = 10, C22 = C13 = 7 8, C32 = −15 2, C23 = −1, C33 = −3 And, as promised, 7In a solid Linear Algebra course you will le... |
�� This new notation greatly shortens the statement of the formula for the inverse of a matrix. Theorem 8.9. Let A be an invertible n × n matrix. Then A−1 = 1 det(A) adj(A) For 2 × 2 matrices, Theorem 8.9 reduces to a fairly simple formula. Equation 8.2. For an invertible 2 × 2 matrix, a b c d −1 = 1 ad − bc d −b... |
16 0 4 1 −5 1 In Exercises 9 - 14, use Cramer’s Rule to solve the system of linear equations. 3x + 7y = 26 5x + 12y = 39 9. 11. 13. x + y = 8000 0.03x + 0.05y = 250 x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 In Exercises 15 - 16, use Cramer’s Rule to solve for x4. 15. x1 − x3 = −2 2x2 − x4 ... |
asked how many animals of each kind he boards, Daniel answered: ‘We board 49 total animals, and I am responsible for each of their 272 legs and 28 tails.’ How many of each animal does the Rescue board? (Recall: tarantulas have 8 legs and no tails, scorpions have 8 legs and one tail, and snakes have no legs and one tai... |
02×2 does not have a unique solution. So there is a nonzero matrix Y with CY = 02×2. In fact, every matrix of the form Y = − 5 2 t t is a solution to CX = 02×2, so there are infinitely many matrices such that CX = 02×2. But consider the matrix X41 = 3 7 It is NOT a solution to CX = 02×2, but rather, 3 7 6 15 14 35 CX41... |
� = 0 showed up as one of the zeros of the characteristic polynomial just means that C itself had determinant zero which we already knew. Those two numbers are called the eigenvalues of C. The corresponding matrix solutions to CX = λX are called the eigenvectors of C and the ‘vector’ portion of the name will make more ... |
�� 19. Carl owns 78 common cards and 39 rare cards. 20. 3.125 gallons. 21. 20 7 ≈ 2.85 liters. 22. The rescue houses 15 snakes, 21 tarantulas and 13 scorpions. 23. Using Cramer’s Rule, we find we need 53 servings of Ippizuti Fish to satisfy the dietary requirements. The number of servings of Misty Mushrooms required, ... |
are no factors of it that have real coefficients which can contribute to the denominator. The factor x2, however, is not irreducible, since we can think of it as x2 = xx = (x − 0)(x − 0), a so-called ‘repeated’ linear factor.2 This means it’s possible that a term with a denominator of just x contributed to the expressio... |
+ C x2 x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = A x + B x2 + Cx + D x2 + 1 Our next task is to determine the values of our unknowns. Clearing denominators gives x2 − x − 6 = Ax x2 + 1 + B x2 + 1 + (Cx + D)x2 Gathering the like powers of x we have x2 − x − 6 = (A + C)x3 + (B + D)x2 + Ax + B In order for this to ho... |
ratic factors. Once we have this factorization of the denominator of a rational function, the next theorem tells us the form the decomposition takes. The reader is encouraged to review the Factor Theorem (Theorem 3.6) and its connection to the role of multiplicity to fully appreciate the statement of the following theo... |
one but that x = 1 is a zero of multiplicity two due to the two different factors (x − 1) and (1 − x). One way to handle this is to note that (1 − x) = −(x − 1) so 3x − 1 (x + 1)(x − 1)(1 − x)(2 + x) = 3x − 1 −(x − 1)2(x + 1)(x + 2) = 1 − 3x (x − 1)2(x + 1)(x + 2) from which we proceed with the partial fraction decompo... |
consequence of Theorem 3.14. Define p(x) to be the difference of the left hand side of the equation in Theorem 8.11 and the right hand side. Then p(x) = 0 for all x in the open interval I. If p(x) were a nonzero polynomial of degree k, then, by Theorem 3.14, p could have at most k zeros in I, and k is a finite number. Si... |
Our answer is easily checked by getting a common denominator and adding the fractions. x + 5 2x2 − 2x + 1 632 Systems of Equations and Matrices 2. Factoring the denominator gives x3 − 2x2 + x = x x2 − 2x + 1 = x(x − 1)2 which gives x = 0 as a zero of multiplicity one and x = 1 as a zero of multiplicity two. We have 3 ... |
B = −3. From −A + C = 0, we get C = A = 3. We get 3 x3 − x2 + x = 3 x + 3 − 3x x2 − x + 1 4. Since 4x3 x2−2 isn’t proper, we use long division and we get a quotient of 4x with a remainder of 8x. That is, 4x3 x2−2 so we now work on resolving 8x x2−2 into partial fractions. The quadratic x2 −2, though it doesn’t factor ... |
x − 1 = (Ax + B) x2 + 3 + Cx + D which yields x3 + 5x − 1 = Ax3 + Bx2 + (3A + C)x + 3B + D. Our system is Cx + D (x2 + 3)2 Ax + B x2 + 3 + = A = 1 B = 0 5 3A + C = 3B + D = −1 We have A = 1 and B = 0 from which we get C = 2 and D = −1. Our final answer is x3 + 5x − 1 x4 + 6x2 + 9 = x x2 + 3 + 2x − 1 (x2 + 3)2 ... |
− 2C √ 2 + D)x2 + (4A + 2B √ 2 + 4C − 2D √ 2)x + 4B + 4D which gives the system √ 2A 4A + 2B 2 + B − 2C √ 2 + 4C − 2D √ 4B + 4D = 0 √ We choose substitution as the weapon of choice to solve this system. From A + C = 0, we get A = −C; from 4B + 4D = 0, we get B = −D. Substituting these into the remaining two ... |
2 + 15 4x4 + 40x2 + 36 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 8. 10. 12. 14. 16. 18. −7x + 43 3x2 + 19x − 14 −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 −21x2 + x − 16 3x3 + 4x2 − 3x + 2 x6 + 5x5 + 16x4 + 80x3 − 2x2 + 6x − 43 x3 + 5x2 + 16x + ... |
2 + Ex + F 3x2 + 7x + 9 −7x + 43 3x2 + 19x − 14 11x2 − 5x − 10 5x3 − 5x2 = = 5 3x − x2 − 4 5(x − 1) −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 = − 9 x + 4 + 7x − 8 x2 + 4 −x2 + 15 4x4 + 40x2 + 36 = 1 2(x2 + 1) − 3 4(x2 + 9) = − −21x2 + x − 16 6 3x3 + 4x2 − 3x + 2 x + 2 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 − 3x + 5 3x2 − 2x + ... |
all of the usual hazards of non-linear equations like extraneous solutions and unusual function domains are once again present. Along with the tried and true techniques of substitution and elimination, we shall often need equal parts tenacity and ingenuity to see a problem through to the end. You may find it necessary ... |
6 and a minor axis along the y-axis of length 4. We see from the graph that the two curves intersect at their y-intercepts only, (0, ±2). 9 + y2 2. We proceed as before to eliminate one of the variables (E1) x2 + y2 = 4 (E2) 4x2 − 9y2 = 36 Replace E2 with −−−−−−−−−−→ −4E1 + E2 (E1) x2 + y2 = 4 (E2) −13y2 = 20 638 Syst... |
a line through the origin with slope 2. Though we cannot verify the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required. While it may be tempting to solve y − x2 = 0 as y = x2 and substitute, we note that this syste... |
� or ‘inconsistent,’ we generally don’t use the labels ‘dependent’ or ‘independent’. Secondly, as we saw with number 4, sometimes making a quick sketch of the problem situation can save a lot of time and effort. While in general the curves in a system of non-linear equations may not be easily visualized, it sometimes pa... |
x yields the solution. We leave it to the reader to show that both points satisfy both equations and now turn to verifying our solution graphically. We begin by solving x2+2xy−16 = 0 for y to obtain y = 16−x2 2x. This function is easily graphed using the techniques of Section 4.2. Solving the second equation, y2 + 2xy... |
2. We is a zero and use synthetic division to factor the left hand side as u − 1 2 √ use the quadratic formula to solve 8u2 + 4u − 2 = 0 and find u = −1± 5. Since u = ex, we 4 = − ln(2). As now must solve ex = 1 for ex = −1± has no real solutions. We are 4 2 and ex = −1± 4, we first note that −1−. From ex = 1 4 < 0, so ... |
n(2), 0) due to its issues in graphing square root functions. If we mentally connect the two branches of the thicker curve, we see the intersection. 642 Systems of Equations and Matrices z(x − 2) = x E1 E2 yz = y E3 (x − 2)2 + y2 = 1 The easiest equation to start with appears to be E2. While it may be tempting to... |
��nd the most efficient method to solve it. Sometimes you just have to try something. 8.7 Systems of Non-Linear Equations and Inequalities 643 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we first introduced in Section 2.4. Before we embark on some example... |
.3. Sketch the solution to the following nonlinear inequalities in the plane. 1. y2 − 4 ≤ x < y + 2 Solution. 2. x2 + y2 ≥ 4 x2 − 2x + y2 − 2y ≤ 0 1. The inequality y2 − 4 ≤ x < y + 2 is a compound inequality. It translates as y2 − 4 ≤ x and x < y + 2. As usual, we solve each inequality and take the set theoretic inter... |
curves. That is, we need to solve the system of nonlinear equations (E1) y2 = x + 4 y = x − 2 (E2) Solving E1 for x, we get x = y2 − 4. Substituting this into E2 gives y = y2 − 4 − 2, or y2 − y − 6 = 0. We find y = −2 and y = 3 and since x = y2 − 4, we get that the graphs intersect at (0, −2) and (5, 3). Putting all of... |
y2 = 4 which lie on or √ 8.7 Systems of Non-Linear Equations and Inequalities 645 inside the circle (x − 1)2 + (y − 1)2 = 2. To produce the most accurate graph, we need to find where these circles intersect. To that end, we solve the system (E1) x2 + y2 = 4 (E2) x2 − 2x + y2 − 2y = 0 We can eliminate both the x2 and y2... |
= 4 (x − 2)2 + y2 = 1 x2 + 4y2 = 4 √ x + 1 − y = 0 x2 + 4y2 = 4 x2 + y2 = 25 y − x = 1 8. 11. 13. y = x3 + 8 y = 10x − x2 14. x2 − xy = 8 y2 − xy = 8 x + 2y2 = 2 x2 + 4y2 = 4 9. 12. 15. x2 + y2 = 25 x2 + (y − 3)2 = 10 x2 + y2 = 25 4x2 − 9y = 0 3y2 − 16x = 0 16. A certain bacteria culture follows the Law of Uninb... |
log2(z) = 6 y + 2 log2(z) = 5 y + 4 log2(z) = 13 In Exercises 21 - 26, sketch the solution to each system of nonlinear inequalities in the plane. 21. 23. 25. x2 − y2 ≤ 1 x2 + 4y2 ≥ 4 (x − 2)2 + y2 < 1 x2 + 4y2 < 4 x + 2y2 > 2 x2 + 4y2 ≤ 4 22. 24. 26. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y < y > 10x − x2 x3 + 8 x2 + y2 ≥ 25... |
+ 6. 5If using λ bothers you, change it to w when you solve the system. 648 Systems of Equations and Matrices 8.7.2 Answers √ 1. (±2, 0), ± y 3, −1 2 1 −2 −1 1 2 x −1 −2 −3 −4 3. (0, ±4) y 4 3 2 1 2. No solution y 2 1 −2 −1 1 2 x −1 −2 −3 −4 4. (±4, 0) y 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −... |
2 + 4y2 ≤ 4 y 1 −1 18. No solution 19. e−5, ± √ 7 22. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 24. y > 10x − x2 x3 + 8 y < y 16 9 −4−3−2−1 1 2 x −56 26. x2 + y2 ≥ 25 5 −3 −1 1 3 5 x −3 −5 650 Systems of Equations and Matrices 27. x = 10, y = 10, z = 10, λ = 2 5 28. (c) x4 + 4 = (x... |
− 3 4, 9 8, − 27 16,... (1) As usual, the periods of ellipsis,..., indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call 1 8 the ‘third term’ and so forth. In numbering them this way, we are setting up a function, which we’ll call a per tradition, between t... |
6. f0 = 1, fn = n · fn−1, n ≥ 1 Solution. 32 = 5 1. Since we are given n ≥ 1, the first four terms of the sequence are a1, a2, a3 and a4. Since the notation a1 means the same thing as a(1), we obtain our first term by replacing every occurrence of n in the formula for an with n = 1 to get a1 = 51−1 3. Proceeding similar... |
successive terms. When n = 1, this equation becomes a1 + 1 = 2 − a1 which simplifies to a2 = 2−a1 = 2−7 = −5. When n = 2, the equation becomes a2 + 1 = 2−a2 so we get a3 = 2 − a2 = 2 − (−5) = 7. Finally, when n = 3, we get a3 + 1 = 2 − a3 so a4 = 2 − a3 = 2 − 7 = −5. 6. As with the problem above, we are given a place t... |
��nition 9.1, we need to shift the variable k so it starts at k = 1 instead of k = 0. To see how we can do this, it helps to think of the problem graphically. What we want is to shift the graph of y = b(k) to the right one unit, and thinking back to Section 1.7, we can accomplish this by replacing k with k − 1 in the d... |
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