This document contains a mathematics chapter about matrices, determinants, and linear systems. It includes examples of matrix operations and solving systems of linear equations. It also contains practice problems related to these topics with solutions.
The document defines:
1) Powers and exponentiation, including rules for multiplying, dividing, and raising powers of numbers.
2) Examples are provided to illustrate the rules.
3) It notes that care must be taken with negative bases and even/odd exponents.
The summary provides the high level definition of powers/exponentiation and notes some key rules and examples are given to illustrate, highlighting the need to consider signs with negative bases. It does not include details of the specific examples or problems shown in the document.
Howard, anton cálculo ii- um novo horizonte - exercicio resolvidos v2Breno Costa
This document provides 40 examples of solving initial value problems for ordinary differential equations using separation of variables. The examples cover both first and second order linear differential equations with various forcing functions. Solutions are obtained by separating variables, integrating, and applying initial conditions to determine constants of integration. Special cases where the standard procedure fails due to singularities are also discussed.
Howard, anton calculo i- um novo horizonte - exercicio resolvidos v1cideni
This document contains exercises related to functions and graphs. Exercise set 1.1 contains word problems involving various functional relationships and graphs. Exercise set 1.2 involves evaluating and sketching functions, determining domains and ranges, and identifying piecewise functions. Exercise set 1.3 involves selecting appropriate axis ranges and scales to graph functions over specified domains.
Differential equation study guide for exam (formula sheet)Dan Al
1) The document provides an overview of topics covered in differential equation and linear algebra exams, including first and second order ordinary differential equations, systems of differential equations, and higher order linear equations.
2) Methods for solving first and second order differential equations are discussed, including separation of variables, variation of parameters, undetermined coefficients, and Euler-Cauchy formulas.
3) Solving systems of first order linear differential equations with constant coefficients is covered, including using eigenvalues and eigenvectors to find the general solution.
This document summarizes the solution to an exercise with three parts:
1) Part (a) finds the probability density function f(x) of a random variable X based on its integral from -infinity to infinity being 1. It determines that f(x) = 2 and a = 2.
2) Part (b) calculates the expected value E(x) of X by integrating x*f(x) from 0 to 1. It determines the expected value is 1/3.
3) Part (c) calculates the variance V(X) of X by finding its expected value E(X2) and subtracting the square of its expected value. It determines the variance is 1/
1. This document contains 20 multiple part questions about differential equations. The questions cover topics like determining the degree and order of differential equations, solving differential equations, identifying whether equations are homogeneous, and forming differential equations to represent families of curves with given properties.
2. The questions range from 1 to 6 marks and include both conceptual questions about differential equations as well as problems requiring solving specific equations. A variety of solution techniques are required including separating variables, homogeneous property, and identifying particular solutions given initial conditions.
3. The document tests mastery of fundamental differential equation concepts and skills like classification, solving, identifying homogeneous property, and setting up equations to model geometric situations. A solid understanding of differential equations is needed to successfully answer all
This document provides solutions to calculating the derivative functions of various given functions. It includes:
1) Finding the derivative functions of polynomials, rational functions, exponential functions, logarithmic functions, trigonometric functions, and composite functions.
2) The solutions provide the step-by-step work and final derivative function for each problem.
3) There are over 25 problems covered across multiple pages with the aim of teaching calculation of derivative functions.
1) Simultaneous equations involve two variables in two equations that are solved simultaneously to find the values of the variables.
2) To solve simultaneous equations, one first expresses one variable in terms of the other by changing the subject of one linear equation, then substitutes this into the other equation to obtain a quadratic equation.
3) This quadratic equation is then solved using factorisation or the quadratic formula to find the values of the variables that satisfy both original equations.
The document defines:
1) Powers and exponentiation, including rules for multiplying, dividing, and raising powers of numbers.
2) Examples are provided to illustrate the rules.
3) It notes that care must be taken with negative bases and even/odd exponents.
The summary provides the high level definition of powers/exponentiation and notes some key rules and examples are given to illustrate, highlighting the need to consider signs with negative bases. It does not include details of the specific examples or problems shown in the document.
Howard, anton cálculo ii- um novo horizonte - exercicio resolvidos v2Breno Costa
This document provides 40 examples of solving initial value problems for ordinary differential equations using separation of variables. The examples cover both first and second order linear differential equations with various forcing functions. Solutions are obtained by separating variables, integrating, and applying initial conditions to determine constants of integration. Special cases where the standard procedure fails due to singularities are also discussed.
Howard, anton calculo i- um novo horizonte - exercicio resolvidos v1cideni
This document contains exercises related to functions and graphs. Exercise set 1.1 contains word problems involving various functional relationships and graphs. Exercise set 1.2 involves evaluating and sketching functions, determining domains and ranges, and identifying piecewise functions. Exercise set 1.3 involves selecting appropriate axis ranges and scales to graph functions over specified domains.
Differential equation study guide for exam (formula sheet)Dan Al
1) The document provides an overview of topics covered in differential equation and linear algebra exams, including first and second order ordinary differential equations, systems of differential equations, and higher order linear equations.
2) Methods for solving first and second order differential equations are discussed, including separation of variables, variation of parameters, undetermined coefficients, and Euler-Cauchy formulas.
3) Solving systems of first order linear differential equations with constant coefficients is covered, including using eigenvalues and eigenvectors to find the general solution.
This document summarizes the solution to an exercise with three parts:
1) Part (a) finds the probability density function f(x) of a random variable X based on its integral from -infinity to infinity being 1. It determines that f(x) = 2 and a = 2.
2) Part (b) calculates the expected value E(x) of X by integrating x*f(x) from 0 to 1. It determines the expected value is 1/3.
3) Part (c) calculates the variance V(X) of X by finding its expected value E(X2) and subtracting the square of its expected value. It determines the variance is 1/
1. This document contains 20 multiple part questions about differential equations. The questions cover topics like determining the degree and order of differential equations, solving differential equations, identifying whether equations are homogeneous, and forming differential equations to represent families of curves with given properties.
2. The questions range from 1 to 6 marks and include both conceptual questions about differential equations as well as problems requiring solving specific equations. A variety of solution techniques are required including separating variables, homogeneous property, and identifying particular solutions given initial conditions.
3. The document tests mastery of fundamental differential equation concepts and skills like classification, solving, identifying homogeneous property, and setting up equations to model geometric situations. A solid understanding of differential equations is needed to successfully answer all
This document provides solutions to calculating the derivative functions of various given functions. It includes:
1) Finding the derivative functions of polynomials, rational functions, exponential functions, logarithmic functions, trigonometric functions, and composite functions.
2) The solutions provide the step-by-step work and final derivative function for each problem.
3) There are over 25 problems covered across multiple pages with the aim of teaching calculation of derivative functions.
1) Simultaneous equations involve two variables in two equations that are solved simultaneously to find the values of the variables.
2) To solve simultaneous equations, one first expresses one variable in terms of the other by changing the subject of one linear equation, then substitutes this into the other equation to obtain a quadratic equation.
3) This quadratic equation is then solved using factorisation or the quadratic formula to find the values of the variables that satisfy both original equations.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems. We started to develop ways to enhance students IQ. We started to leave an indelible mark on the students who have undergone APEX training. That is why APEX INSTITUTE is very well known of its quality of education
The document summarizes an inequality originally proven by T. Andreescu and G. Dospinescu. This inequality, presented as Theorem 1, is shown to be useful for proving several other interesting inequalities in a simple way. Applications of Theorem 1 include proving inequalities involving sums of powers and expressions divided by sums. The document concludes by listing additional inequalities that can be solved using the techniques demonstrated.
This document appears to be the table of contents and problems from Chapter 0 of a mathematics textbook. The table of contents lists 17 chapters and their corresponding page numbers. The problems cover a range of algebra topics including integers, rational numbers, properties of operations, solving equations, and rational expressions. There are over 70 problems presented without solutions for students to work through.
This document contains 20 multiple integral exercises with solutions. Some of the exercises involve calculating double integrals over specified regions, while others involve setting up approximations of double integrals using Riemann sums. Exercise 19 involves sketching solid regions in 3D space and Exercise 20 involves sketching surfaces defined by z=f(x,y).
This document contains 40 math modeling exercises involving differential equations:
1. The exercises involve setting up and solving first and second order differential equations.
2. Common methods used include separation of variables, identifying integrating factors, and determining constants from initial conditions.
3. The exercises cover a wide range of differential equation types including linear and nonlinear, homogeneous and nonhomogeneous, and those requiring advanced integrating factor techniques.
The document contains a midterm exam for an ODE class with 6 problems worth 10 points each. Problem 1 asks to find the general solution of a 7th order linear ODE using the method of undetermined coefficients. Problem 2 asks to solve a 2nd order linear ODE using either variation of parameters or undetermined coefficients. Problem 3 asks to solve a nonlinear 2nd order ODE using a substitution. Problem 4 asks to find the equation of motion for a mass attached to a spring with an external force applied. Problem 5 asks to solve an eigenvalue problem for a CE equation. Problem 6 asks to use variation of parameters to solve a 2nd order nonhomogeneous ODE.
The document provides examples of solving systems of linear equations using various methods:
1) Addition - Adding corresponding terms of equations to eliminate a variable.
2) Substitution - Solving one equation for a variable in terms of the other and substituting into the second equation.
3) Comparison - Setting corresponding terms of equations equal to each other to solve for variables.
It works through 30 examples demonstrating these methods step-by-step to solve systems with two unknown variables.
This document is a keynote presentation about summarizing different types of equations from points given on a graph. It discusses finding linear equations from two points using slope-intercept form, finding quadratic equations from three points, and finding exponential equations from two points. The presentation provides examples of predicting values using linear, quadratic, and exponential equations derived from sample point data.
This document contains a midterm exam for a differential equations course. It consists of 4 problems:
1. Find the linear/non-linear ODEs and solve the IVP for one.
2. Determine the exact ODE, solve implicitly and find an IVP solution.
3. Perform a substitution to reduce a Bernoulli equation to linear form. Also reduce a homogeneous ODE to separable form.
4. Find the orthogonal trajectories to a given family of curves.
1. The limit as x approaches 4 of x4-16 is 0. When factored, the expression becomes (x-4)(x+4)(x2+4) which equals 0 as x approaches 4.
2. The limit as x approaches infinity of x7-x2+1 is 1. When factored, the leading terms are x7 for both the top and bottom expressions, which equals 1 as x approaches infinity.
3. The limit as x approaches -1 of x2-1 is 0. When factored, the expression becomes (x+1)(x-1) which equals 0 as the factors are 0 when x is -1.
This document contains solutions to 100 equations of the first degree. The equations involve variables like x and y, and involve operations like addition, subtraction, multiplication and division. Each equation is presented along with its corresponding solution (e.g. x=7).
The document summarizes the method of undetermined coefficients for solving nonhomogeneous linear differential equations. It provides examples of using the method to find the particular solution by making an educated guess of the form of the solution, substituting it into the differential equation, and solving for the coefficients. It also discusses addressing duplication with the complementary function to find the appropriate trial function.
This document contains a large collection of mathematical expressions, equations, and sets. Some key points:
- It includes expressions like n(A), n(B), n[(A-B)(B-A)], and n(A × B) with various values.
- There are several equations set equal to values, such as x2 - 3x < 0, -2 < log < -1, and equations containing sums, integrals, and logarithms.
- Sets are defined containing various elements like numbers, vectors, and functions.
This document provides an overview of key concepts in differential equations including:
- A differential equation contains derivatives of dependent variables with respect to independent variables.
- The order of a differential equation is defined as the order of the highest derivative. The degree is the highest power of the highest order derivative.
- Differential equations can be formed by differentiating curves and eliminating arbitrary constants.
- Common methods for solving differential equations include variable separation, homogeneous equations, and linear equations.
The document explores different cases of conic sections defined by the general quadratic equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. It shows that a parabola occurs when either A≠0 and C=B=0, or C≠0 and A=B=0. Ellipses and hyperbolas occur when B2 - 4AC is negative or positive, respectively. Various examples of parabolas, ellipses, hyperbolas, circles and degenerate forms are worked through to demonstrate their properties.
1) Complex numbers can be represented as 2x2 matrices or as points in the Argand diagram, with the real part on the x-axis and imaginary part on the y-axis.
2) Addition and multiplication of complex numbers follows the same rules as real numbers, replacing i^2 with -1.
3) Every non-zero complex number has a square root and every polynomial with complex coefficients can be factored into linear terms.
This document provides examples of solutions to problems involving number theory, algebra, geometry, and probability. It contains the following types of problems:
- Number theory word problems involving finding missing terms in sequences
- Solving quadratic equations and determining equations with transformed roots
- Calculating areas and perimeters of geometric shapes like octagons and hexagons
- Finding equations of lines transformed by parallel shifts
- Probability problems involving counting outcomes, finding probabilities of events, and calculating combinations and permutations
The document aims to demonstrate smart solutions to a variety of mathematical problems across different topics through clear explanations and step-by-step working.
The document discusses ordinary differential equations (ODEs) and methods for solving separable differential equations. It defines ODEs, order, and degree of differential equations. It then introduces the method of separating variables for solving separable differential equations. This method involves rearranging the differential equation so that the variables are separated on each side. The document provides examples of using this method to solve three separable differential equations, finding the integral relations between the variables in each case.
Laboratorio parte i ecuaciones diferenciales 2014 ii (1)Archi Maco
This document lists 40 differential equations involving variables separated and reducible to variables separated, homogeneous and reducible to homogeneous forms. The equations cover a wide range of standard forms involving trigonometric, exponential, logarithmic and algebraic functions of variables x and y. They also include initial/boundary value problems by specifying known conditions for variables x or y at given values.
This document contains the answers to questions on a mathematics exam in Indonesia from 2006-2007. It provides the answer choices for 19 multiple choice questions on topics like algebra, geometry, trigonometry, and logic. For each question, it shows the answer choice and provides the steps taken to solve the problem.
This document contains 3 problems:
1) Showing that a sequence is increasing and less than 3.
2) Calculating the average value of a function over an interval.
3) Finding the solution to a partial differential equation of the form 2xyy' – y^2 + x^2 = 0. The solution is found to be y = √cx - x^2.
This document contains information about specialist maths exam problems from 2010-2013, including median exam scores, common student errors, and exam questions and solutions. The median exam score was a C+ and 49% of students received a B or higher. Handwriting and setting out work clearly were identified as areas of concern. Example exam questions and solutions covered topics like complex numbers, calculus, vectors, and differential equations.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems. We started to develop ways to enhance students IQ. We started to leave an indelible mark on the students who have undergone APEX training. That is why APEX INSTITUTE is very well known of its quality of education
The document summarizes an inequality originally proven by T. Andreescu and G. Dospinescu. This inequality, presented as Theorem 1, is shown to be useful for proving several other interesting inequalities in a simple way. Applications of Theorem 1 include proving inequalities involving sums of powers and expressions divided by sums. The document concludes by listing additional inequalities that can be solved using the techniques demonstrated.
This document appears to be the table of contents and problems from Chapter 0 of a mathematics textbook. The table of contents lists 17 chapters and their corresponding page numbers. The problems cover a range of algebra topics including integers, rational numbers, properties of operations, solving equations, and rational expressions. There are over 70 problems presented without solutions for students to work through.
This document contains 20 multiple integral exercises with solutions. Some of the exercises involve calculating double integrals over specified regions, while others involve setting up approximations of double integrals using Riemann sums. Exercise 19 involves sketching solid regions in 3D space and Exercise 20 involves sketching surfaces defined by z=f(x,y).
This document contains 40 math modeling exercises involving differential equations:
1. The exercises involve setting up and solving first and second order differential equations.
2. Common methods used include separation of variables, identifying integrating factors, and determining constants from initial conditions.
3. The exercises cover a wide range of differential equation types including linear and nonlinear, homogeneous and nonhomogeneous, and those requiring advanced integrating factor techniques.
The document contains a midterm exam for an ODE class with 6 problems worth 10 points each. Problem 1 asks to find the general solution of a 7th order linear ODE using the method of undetermined coefficients. Problem 2 asks to solve a 2nd order linear ODE using either variation of parameters or undetermined coefficients. Problem 3 asks to solve a nonlinear 2nd order ODE using a substitution. Problem 4 asks to find the equation of motion for a mass attached to a spring with an external force applied. Problem 5 asks to solve an eigenvalue problem for a CE equation. Problem 6 asks to use variation of parameters to solve a 2nd order nonhomogeneous ODE.
The document provides examples of solving systems of linear equations using various methods:
1) Addition - Adding corresponding terms of equations to eliminate a variable.
2) Substitution - Solving one equation for a variable in terms of the other and substituting into the second equation.
3) Comparison - Setting corresponding terms of equations equal to each other to solve for variables.
It works through 30 examples demonstrating these methods step-by-step to solve systems with two unknown variables.
This document is a keynote presentation about summarizing different types of equations from points given on a graph. It discusses finding linear equations from two points using slope-intercept form, finding quadratic equations from three points, and finding exponential equations from two points. The presentation provides examples of predicting values using linear, quadratic, and exponential equations derived from sample point data.
This document contains a midterm exam for a differential equations course. It consists of 4 problems:
1. Find the linear/non-linear ODEs and solve the IVP for one.
2. Determine the exact ODE, solve implicitly and find an IVP solution.
3. Perform a substitution to reduce a Bernoulli equation to linear form. Also reduce a homogeneous ODE to separable form.
4. Find the orthogonal trajectories to a given family of curves.
1. The limit as x approaches 4 of x4-16 is 0. When factored, the expression becomes (x-4)(x+4)(x2+4) which equals 0 as x approaches 4.
2. The limit as x approaches infinity of x7-x2+1 is 1. When factored, the leading terms are x7 for both the top and bottom expressions, which equals 1 as x approaches infinity.
3. The limit as x approaches -1 of x2-1 is 0. When factored, the expression becomes (x+1)(x-1) which equals 0 as the factors are 0 when x is -1.
This document contains solutions to 100 equations of the first degree. The equations involve variables like x and y, and involve operations like addition, subtraction, multiplication and division. Each equation is presented along with its corresponding solution (e.g. x=7).
The document summarizes the method of undetermined coefficients for solving nonhomogeneous linear differential equations. It provides examples of using the method to find the particular solution by making an educated guess of the form of the solution, substituting it into the differential equation, and solving for the coefficients. It also discusses addressing duplication with the complementary function to find the appropriate trial function.
This document contains a large collection of mathematical expressions, equations, and sets. Some key points:
- It includes expressions like n(A), n(B), n[(A-B)(B-A)], and n(A × B) with various values.
- There are several equations set equal to values, such as x2 - 3x < 0, -2 < log < -1, and equations containing sums, integrals, and logarithms.
- Sets are defined containing various elements like numbers, vectors, and functions.
This document provides an overview of key concepts in differential equations including:
- A differential equation contains derivatives of dependent variables with respect to independent variables.
- The order of a differential equation is defined as the order of the highest derivative. The degree is the highest power of the highest order derivative.
- Differential equations can be formed by differentiating curves and eliminating arbitrary constants.
- Common methods for solving differential equations include variable separation, homogeneous equations, and linear equations.
The document explores different cases of conic sections defined by the general quadratic equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. It shows that a parabola occurs when either A≠0 and C=B=0, or C≠0 and A=B=0. Ellipses and hyperbolas occur when B2 - 4AC is negative or positive, respectively. Various examples of parabolas, ellipses, hyperbolas, circles and degenerate forms are worked through to demonstrate their properties.
1) Complex numbers can be represented as 2x2 matrices or as points in the Argand diagram, with the real part on the x-axis and imaginary part on the y-axis.
2) Addition and multiplication of complex numbers follows the same rules as real numbers, replacing i^2 with -1.
3) Every non-zero complex number has a square root and every polynomial with complex coefficients can be factored into linear terms.
This document provides examples of solutions to problems involving number theory, algebra, geometry, and probability. It contains the following types of problems:
- Number theory word problems involving finding missing terms in sequences
- Solving quadratic equations and determining equations with transformed roots
- Calculating areas and perimeters of geometric shapes like octagons and hexagons
- Finding equations of lines transformed by parallel shifts
- Probability problems involving counting outcomes, finding probabilities of events, and calculating combinations and permutations
The document aims to demonstrate smart solutions to a variety of mathematical problems across different topics through clear explanations and step-by-step working.
The document discusses ordinary differential equations (ODEs) and methods for solving separable differential equations. It defines ODEs, order, and degree of differential equations. It then introduces the method of separating variables for solving separable differential equations. This method involves rearranging the differential equation so that the variables are separated on each side. The document provides examples of using this method to solve three separable differential equations, finding the integral relations between the variables in each case.
Laboratorio parte i ecuaciones diferenciales 2014 ii (1)Archi Maco
This document lists 40 differential equations involving variables separated and reducible to variables separated, homogeneous and reducible to homogeneous forms. The equations cover a wide range of standard forms involving trigonometric, exponential, logarithmic and algebraic functions of variables x and y. They also include initial/boundary value problems by specifying known conditions for variables x or y at given values.
This document contains the answers to questions on a mathematics exam in Indonesia from 2006-2007. It provides the answer choices for 19 multiple choice questions on topics like algebra, geometry, trigonometry, and logic. For each question, it shows the answer choice and provides the steps taken to solve the problem.
This document contains 3 problems:
1) Showing that a sequence is increasing and less than 3.
2) Calculating the average value of a function over an interval.
3) Finding the solution to a partial differential equation of the form 2xyy' – y^2 + x^2 = 0. The solution is found to be y = √cx - x^2.
This document contains information about specialist maths exam problems from 2010-2013, including median exam scores, common student errors, and exam questions and solutions. The median exam score was a C+ and 49% of students received a B or higher. Handwriting and setting out work clearly were identified as areas of concern. Example exam questions and solutions covered topics like complex numbers, calculus, vectors, and differential equations.
The document contains the solutions to several calculus problems:
1) Finding the total amount spent in an economy where dollars recirculate at 90% each time.
2) Calculating the speed needed to shoot a basketball into a hoop from a given distance and height.
3) Taking derivatives and integrals to solve optimization problems.
IIT Jam math 2016 solutions BY TrajectoryeducationDev Singh
The document contains a mathematics exam question paper with 10 single mark questions (Q1-Q10) and 20 two mark questions (Q11-Q30). The questions cover topics like sequences, linear transformations, integrals, permutations, differential equations etc. Some key questions asked about the nature of a sequence involving sines, order of a permutation, evaluating a limit, checking if a differential equation is exact etc. and provided solutions to them.
This document discusses linear equations and curve fitting. It provides 18 examples of using a linear system to solve for the coefficients of linear, quadratic, and cubic polynomials that fit given data points. It also provides examples of using a linear system to solve for the coefficients of circle and central conic equations that fit given points. The linear systems are set up and solved, providing the resulting equations that fit the data in each example.
This document contains 7 exercises involving calculating integrals using techniques like integration by parts, trigonometric substitutions, and partial fraction decomposition. It also contains solutions to each exercise that demonstrate the step-by-step working to evaluate the integrals. The document is from a university course on simple integration and contains examples commonly used to teach integration techniques.
This document contains solutions to 6 math problems:
1) Finding the value of x2 - x + 1 when x = 2
2) Evaluating (a + b - c)(a - b + c) when a = 1, b = 2, c = 3
3) Simplifying (17 - 15)3 using the identity a3 - b3 = (a - b)(a2 + ab + b2)
4) Simplifying a3 - b2/(a-b) when a = 2 and b = -2
5) Simplifying a - b + b2/(1-a+b) when a = -1/2 and b = 3
This model question paper contains 55 questions divided into two parts for the subject Mathematics for IT. Part A contains one mark questions in multiple choice format covering topics like sets, relations, functions, limits, derivatives, integrals and mathematical statements. Part B contains 2 mark questions involving concepts like sets, logic, trigonometry, limits, derivatives and integrals that need to be solved. The question paper tests the understanding of core mathematical concepts required for an IT program through multiple choice and theoretical questions.
This document provides an overview of solving quadratic equations by factoring. It discusses identifying quadratic equations, rewriting them in standard form, factoring trinomials in the form x^2 + bx + c, and determining roots. Several examples of factoring trinomials and solving quadratic equations are shown. Activities include identifying quadratic equations, rewriting equations in standard form, factoring trinomials, and solving equations by factoring. The document provides resources for further learning about quadratic equations and factoring.
This document provides a review sheet with multiple math problems involving algebra, trigonometry, and calculus concepts. There are 23 problems covering topics such as simplifying rational expressions, solving equations, graphing functions, working with complex numbers, trigonometric identities, and applying trigonometric functions. Answers are provided for selected problems. The review sheet is intended to help students practice and review a wide range of mathematics topics.
This document contains the work of a student on a calculus test. It includes:
1) Solving limits, finding derivatives, and applying L'Hopital's rule.
2) Using induction to prove an identity.
3) Providing epsilon-delta proofs of limits.
4) Finding where a tangent line is parallel to a secant line.
5) Proving statements about limits of functions.
The student provides detailed solutions showing their work for each problem on the test.
This document contains the solutions to 5 questions related to calculus concepts like integration, derivatives, series approximation, and geometry of curves and surfaces. Some of the key steps include:
- Using integration to find volumes, masses, and centroids
- Finding critical points and classifying extrema
- Approximating a series to evaluate an integral
- Solving a geometric series problem to find an initial height
- Analyzing motion problems using kinematic equations
- Finding equations of planes and tangent lines to surfaces
This document provides a sample question paper for Class XII Mathematics. It consists of 3 sections - Section A has 10 one-mark questions, Section B has 12 four-mark questions, and Section C has 7 six-mark questions. The paper is for 3 hours and carries a total of 100 marks. Some questions provide internal choices. Calculators are not permitted. Sample questions include finding inverse functions, evaluating integrals, solving differential equations, and probability questions.
Pembahasan ujian nasional matematika ipa sma 2013mardiyanto83
This document contains 31 math problems and their solutions from a 2011 Indonesian national exam practice test for high school/secondary school students studying the science program. The problems cover a range of math topics including algebra, geometry, trigonometry, and statistics. The full solutions are provided for each multiple choice question, with the correct answer indicated by a letter.
The document is a sample question paper for Class XII Mathematics. It consists of 3 sections - Section A has 10 one-mark questions, Section B has 12 four-mark questions, and Section C has 7 six-mark questions. All questions are compulsory. The paper tests concepts related to matrices, trigonometry, calculus, differential equations, and vectors. Internal choices are provided in some questions. Calculators are not permitted.
Triangle ABC is given, with altitudes CD and BE from vertices C and B to opposite sides AB and AC respectively being equal. It is proved that triangle ABC must be isosceles by showing that triangles CBD and BCE are congruent by the right angle-hypotenuse (RHS) criterion, implying corresponding angles are equal, and then using corresponding parts of congruent triangles to show sides AB and AC are equal, making triangle ABC isosceles.
This document contains a midterm exam for an engineering mathematics course. It consists of 4 problems:
1. Finding the general solution to two linear ODEs.
2. Finding the complementary and particular solutions for a given non-homogeneous linear ODE, and using them to solve an IVP.
3. Repeating steps from problem 2 for another given non-homogeneous linear ODE.
4. Modeling and solving an ODE describing the motion of a damped spring-mass system subject to an external force.
The document describes the Jacobi iterative method for solving systems of linear equations. It begins with an initial estimate for the solution variables, inserts them into the equations to get updated estimates, and repeats this process iteratively until the estimates converge to the desired solution. As an example, it applies the method to a set of 3 equations in 3 unknowns, showing the estimates after each iteration getting progressively closer to the exact solution obtained using Gaussian elimination. A Fortran program implementing the Jacobi method is also presented.
This document provides a summary of lecture 2 on quadratic equations and straight lines. It covers how to factorize, complete the square, and use the quadratic formula to solve quadratic equations. It also discusses how to find the equation of a straight line given its gradient and y-intercept, or two points on the line. Additionally, it explains how to sketch lines, find the midpoint and distance between two points. Key terms defined include quadratic, surd, gradient, and intercept. Methods demonstrated include solving quadratic equations, finding lines from gradient/point and two points, and calculating midpoints and distances on a graph.
The document provides information about sets and operations on sets such as union, intersection, complement, difference, properties of these operations, counting theorems for finite sets, and the number of elements in power sets. It defines key terms like union, intersection, complement, difference of sets. It lists properties of union, intersection, and complement. It presents counting theorems for finite sets involving union, intersection. It states that the number of elements in the power set of a set with n elements is 2n and the number of proper subsets is 2n-2.
Este documento discute o Parnasianismo, Simbolismo e Pré-Modernismo na literatura brasileira. Resume os principais movimentos literários, incluindo suas características estilísticas e temáticas, e fornece exercícios sobre esses períodos.
O documento descreve a expansão burguesa e a reação operária no século XIX na Europa. Aborda os principais eventos que levaram à consolidação do capitalismo e do poder burguês, como a Revolução Industrial e as revoluções liberais de 1848, e como os trabalhadores começaram a se organizar politicamente em resposta, culminando na Comuna de Paris em 1871.
I. O documento apresenta exemplos de orações subordinadas adjetivas, adverbiais e substantivas encontradas em textos literários e músicas populares brasileiras.
II. São discutidos os tipos de orações que aparecem com mais frequência nesses contextos, como orações adverbiais condicionais e comparativas.
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1) O documento apresenta um paradoxo sobre a corrida entre Aquiles e a tartaruga. Apesar de Aquiles ser mais rápido, segundo o raciocínio inicial ele nunca alcançaria a tartaruga.
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Ensino medio livre_edicao_2012_unidade_01_gramatica
2º mat emática
1. MATEmática
Matrize s, dete rm inantes e sis tema s lineare s
1
Capítulo 1Matrizes
Conexões
56786666478886421234 ⋅⋅ 6 8 6 6 7 8 6 + + + + + + + + + +
7 6 6 4 8 8 =
+512213261218244142432812112870607440+ = î
Franciel foi aprovada
e emmat m t a,geografia biologia,masnão emhistória.
e á ic e
Exercícios complementares
13. A 28a linhacomeçarácom 28, na primeiracoluna. Assim, falta 16 elementosnessalinha: 28 +
m
16 = 44
14. At= –A
xy x z 10330203130−−0 000
z y 2 0 00 ===== ===== = = =
000 −−−− − = = ==
I.x= –x s 2x= 0 s x= 0
I 2 = –y s y= –2
I.
II
I –1 = –z s z= 1
.
∴ x+ y+ z= 0 – 2 + 1 = –1
15. O primeiromembrodaequaçãoé a somadoscemprimeirostermosdeumaPA cujoprimeirotermo
é iguala X e cujarazãoé iguala X.
SXXXX 1001002101505050= + ⋅= ⋅=().
100
Então:5.050 · X = 5050005050..l X X = 1001.0 ·
a.s 50
16. a)A vendeu16 automóveis tipo1 emmarço(a ).
do 13
B vendeu 20 automóv eisdo tipo1 emmarço(b ).13
a + b13 = 16 + 20 = 36
13
Portanto,A e B vender m,junta 36 automóv
a s, eisdo tipo1 emmarço.
b)ConcessionáriaA ConcessionáriaB
Mê 1: 12 + 15 = 27 M s 20 + 16 = 36
s ê 1:
Mê 2: 8 + 12 = 20 Mê 2: 16 + 10 = 26
s s
Mê 3: 16 + 24 = 40 M s 20 + 10 = 30
s ê 3:
Mê 4: 20 + 36 = 56 M s 24 + 26 = 50
s ê 4:
Portanto,a concessionáriaA ultra a sou concessionáriaB no volum devendas
p s a e (considerando-
sesoment os automóveis
e do tipo1 e do tipo2)nosmese demarçoe abril.
s
c)AB+ = + + + + + + + + 1 220816162020241516121024103626)n o
d)C=+=++ +
+ 1281620241512243612
D= C = 20162024241610102626
= +
+
29. d
(A+ B) = (A+ B) · (A+ B) s (A+ B) = A2 + AB + BA + B2
2 2
Se o produtodeA porB é comuta vo,(A· B = B · A), podemos
ti escrev r:
e
(A+ B) = A2 + AB + AB + B2 s
2
s (A+ B) = A2 + 2AB + B2
2
2. 30. a
I.xxx12 3121 3 − 3333 33
3 33 33333
3 333 33 − × × × 12 ,11,,10100 x3 0
23 1×
,,1
I Am× n · Bn × p – C m× p = 0m×
I. p
Observandoa ordemdasmatri e a únicapossibilidade
z s, é:
ABCDD233121212210× × × × ⋅−= × s×××− = 121210C
d⋅−0 0 1 2 0 −−o d m =00 1
x rm 1 0 1 0 x
1010011200xx oe r e 11
0
s 1200+ −++++++++++++++======= x s 1200+ −− +++++++======xxx
−− xx = s
s 1 + x– 2 = 0 s x= 1
2222 b⋅⋅
31. Aabbabbabababb 00=2b b 0 2 b + =2b b
aa = a = ab aab
A2 = A s
s abababbabb 0+2a a = b ds ababb 00+ = = = bb= Is
222 b bb be
bb r 222 ab )
(
Substituindoem( )temos:
I,
a = a s a – a = 0 s a(a– 1)= 0 s
2 2
s a = 0 (Nãoconvém.)ou a – 1 = 0 s a = 1
Logo: a = 1 e b = 0
32. e
A · At= I 12121001xy y z =0u⋅ 2 2 0 1 y 11 s
s ⋅
z x b oa 1 1 0 x =22 0
0
s 14221001 + + + + 10x yb= 0 u yx y z z
222 01 zz
yx ox z x y
3. 2
14 1 34 2 2 + = = xx ( )
s I
yxz2 0 + = s ()xzy2 2 2 = − s xzy 2 2 4 ⋅ = ( I
2 I)
y + z = 1( I)
2 2 II
Substituindo( )
Iem( I :
I)
34 4 3 2 2 2 2 ⋅ = = zyzys ( V
I )
Substituindo (IV) e m (III): a = (–1) + 1 = 1; a = (–1) + 2 = –1; a = 0
31 3 32 3 33
y y 2 3 1 + = s 43 1 2
2
Logo: A = − − − − 011101110
y= s y 34 =
2
4. b
Sendom , nije p ele en s
ij ij m to deM, N e P, resp ti a e t temos:
ec v m n e,
∴ x + y = 34 34 64
32 23 32 23 32 23 11 11 11 22 22 22 M N P mn p m n p + = + = + = i s N mnp
2 2
s 32 23 7 32 23 4 13 ⋅ + ⋅ = ⋅ + ⋅ + = 72 4 yyx() s 9 4 42 9 4 62 xyyx+ =
32 x
32 + = =
Tarefa proposta + = 4 ()()I I
2 I
Fazendo( I ( )vem:
I ) I,
–
1. c
5y– 5x= 20 s y– x= 4
M a a a a a a = ) em I I I): 11 5. e
12 21 22 31 32 p – q11 = 2 – (–3) = 5
11
p – q12 = –1 – 1 = –2
12
p – q21 = 3 – 1 = 2
21
a = 2(1 – 1) = 0 a
11 12
p – q22 = 1 – 3 = –2
22
= 2· 1+ 2= 4
Logo, a distânciaentre asmatri e p e q é 5.
z s
a = 2 · 2 + 1 = 5 a22
21
= 2(2– 2)= 0 6. c
a = 2 · 3 + 1 = 7 a32
31 O país1 exportou1,2 + 3,1 = 4,3 bilhões.
= 2· 3+ 2= 8 O país2 exportou2,1 + 2,5 = 4,6 bilhões.
Portanto: O país3 exportou0,9 + 3,2 = 4,1 bilhões.
M = ao = 8 4 5 0 7 8
n: 2 0
t O país1 importou2,1 + 0,9 = 3,0 bilhões.
O país2 importou1,2 + 3,2 = 4,4 bilhões.
O país3 importou3,1 + 2,5 = 5,6 bilhões.
2. b 7. b
A aaaa= = 0 11 12 21
− − = + + + 1 22 1 4 3 4 2 2 xxxxx
22
x+ 1 = –1 s x= –2
a = (–1) + 1 · 1 · 1 =
11 1
8. b
1 a = (–1) + 1 · 2
21 2
xa a a a = − − − + − = 22 22422482
· 1 = –2
a = (–1) + 2 · 1 · 2 =
12 1
a – 2 = 2 s a = 4 s a = –2 ou a = 2
2 2
–2 a = (–1) + 2 ·
22 2
2· 2= 4 –2a = 4 s a = –2
Logo: A = − − 12 4a = –8 s a = –2
24 –2 + a = 2 s a = 4 s a = 2 ou a = –2
2 2
Logo: a = –2
3. A a a a a a a a a a = 9. d
a a a 12 13 21 22 23
a a 11 a + a + a + … + a = (1+ 1)+ (2+ 2)+ (3+ 3)+ … + 2n =
11 22 33 nn
31 32 33 e a i ji jiji j= = 2 + 4 + 6 + … + 2n
− ≠ = + ( ) 10 , A sequ n
ê cia umaPA.
é
⋅ ()1 2 s
se, se
Então:
a = 0; a = (–1) + 2 =
11 12 1
S aan n n = +
–1; a = (–1) + 3 =
13 1
1
a = (–1) + 1 = –1; a
s S n n n = + ⋅ ()2 2 2 s S n n n n n = + ⋅ = + 2 1 2 2 ()
21 2 22
= 0; a = (–1) + 3 =
23 2
–1
5. 3
x– 1 = 2 s x= 3
y+ 2 = –3 s y= –5
3z+ 1 = 7 s z= 2
Logo: xy = 3 · (–5) · 2 = –30
z
12. xpqxyr z x y−++ −− + + ++++++++ += − + − − − 13262113ppyzqr −−−−−−− zz zz
y z x + z z zz2621
z zz
x– 1 = –x + 1 s x= 1
p = –x – 3 s p = –1 – 3 s p = – 4
q= y
x+3 = –p s 1 + 3 = –p s p = – 4
2y+ 6 = –2y – 6 s y= –3 ( )
I
De ( ),
I temos:q = –3
r = z– 2
s + + + + + + 1057152722337 − −−−−
yxx x y x −+++++ + + + + + +
y y x y −− −== = = =
–y = –q s – (–3) = –q s q= −−− − −−−−−−121031522722321
− − −−−−−−
−
–3 573371xy y+= − + = ===
x
–z + 2 = –r Resolv endoo siste a,
m temos:
z+ 1 = –z – 1 s z= –1 ( I
I) x= –2 e y= 1
De ( I temos:r = –1 – 2 =
I) , w = |x · y| = | –2 · 1| = | –2 | = 2
–3
15. e
Logo: p = – 4; q = –3 e r =
A loja L1 vendeu 30 unidades do produto P1 e 15 unidades do
–3
produtoP2, logoa somadasquantidades dosprodutosdostiposP1
e P2 vendidospelaloja 1 é 45.
L
13.
23217034212x x z zy y yz 16. b
⋅− .
xy x − + + − +
z y+
+31 31
27 42
02 =+7734322101
zx y x z −−++ + + +++++ + +
y z yz + ++ PQ − = − ==== ==
=− . d = 4 23232544123664108888=
o −1 ==
2 ====− 88
88
3x– y= 7 ( )
I
2z+ 1 = 3z– 4 s z= 5
x+ 2y= 0 ( I
I)
= − − − − − − ==== ========= =
= −−− = 46142103825125()
De ( ) e
I ( I , vem:
I)
3720xy y−=+ ====
x Logo: (P– 2Q) = −−−
t 21255
ss 6214207142xyx x
y x−
= + = (I =
)= 17. b
3A = B + C
Substituindo em ( )
I, 361243xy w w y w ww == = =++ ++++++s
z x x z w =− = =
w
ww +
temos:
3 · 2 – y= 7 s y= –1 s 333346123xyzwx y wwww + + + − + ++++++
x z ww
w=+
w
Então:
A=to y 37117401106
ã =
:
3x= x+ 4 s 2x= 4 s x= 2
Logo, o traçodamatri A
z 3y= 2 + y+ 6 s 2y= 8 s y= 4
é: 3 + 4 + 6 = 13 3w = 2w + 3 s w = 3
14. BAA=− ⋅t32 s
3z= z+ 3 – 1 s 2z= 2 s z= 1
∴ x+ y+ z+ w = 2 + 4 + 1 + 3 = 10
s yx y y x y−+
x x y x
18. b
+ + 4 6 110571527223
+ =3m
A3 × 4_ Bp × q= C3 × 5; logo: p = 4 e q = 5
37 = −−−−============ = =
−−−−
−−−1106412323216214
− 19.
0⋅−9
3012432
121
23s
syxxy y x y−+
x y x 102321512310502100−0 0 .
00
0 g :
o p = = + − − + + 1154136031203+ +
+ 14 322
2 23 1
121 10571527223
371122
4=
− + + +++++ −=== =
+= ==
= −−−−============ = =
−−−−
−−110641232329332
6920323323017
292552
33172s
6. 20. c
ABA⋅= − ====⋅⋅
== BA = − ====⋅=
=010000010100sBBO ≠
BABA⋅=AA 0 B 0 0 B 0000101000000s== O
0⋅−A A 0 =AA 0
B0 0⋅
=⋅−
A2010001000000= −=== == = 1 1 = =1 0
0 0 0 1
BBO 22000100010001= ⋅⋅ 0 1 0
0 0 =======≠ s
ABA+ = − ==== =+++ + += −=== =
= ++ = =+010000010101sBBO ≠
21. e
A4 × 7 · B7 × 9 = C 4 × 9 (quatrolinhase novecolunas)
c63 é o ele e to
m n dalinha6 e coluna3, logoc63 não existe.
22. b
ABAmnABABAn1212112× × × ⋅ ⋅⋅× × ()[()ln
] i ×Amn1221 se
2121211121 1⋅⋅ 1 1 1
1 2 2 ⋅−1 1 = − ==== s
− 11 2 2 bb
s 221211121 b⋅−1 1 =
bb−bb 2 1 − ==== s
b
s 4222111211121bbbb−−−− 2 − − 2 s
1= 1()
1 1
s 4b11 – 2b21 = 2 s 2b11 – b21 = 1 s
s b21 = –1 + 2b11
23. a
(A· B) · C = ()ABnD342 × × ⋅× · C m×
32 2
n= 4
Então:D3 × 2 · C m× 2
Logo: m= 2
∴ m= 2 e n = 4
24. c
7. PQRSxy ⋅=
z ⋅⋅ Q S y Q S y Rx⋅ s1111111111111t
R x ⋅⋅ R x =QS z
y s
sxxyy z t+++1t = + + +
t z 1 1121211
I.x+ 1 = 2 s x= 1
I z+ 1 = 2 s z= 1
I.
I I + 1 = t+ 1 s 1 · y+ 1 = t+ 1 s y= t
I xy
.
IV.y+ 1 = z+ ts y+ 1 = 1 + ts y= t
25. b
I.(V)A3 × 2 · B2 × 1 = C 3 × 1
I (F)A5 × 4 · B5 × 2 (Nãoexist o produto.)
I . e
I I A2 × 3 · B3 × 2 = C 2 × 2
I (V)
.
26. a)A2 = A · A =
1 1⋅−−A
= 123012111123012111−−1 111
111
1 · A = e ==
= 103223343002012022101211+ − + + + − + − + + + − − + + − + − − 3321+ ++++++ + +
+ + +=
= −−− 274230020
1 1⋅−−
b)A · At= 123012111101211321−−1 111
111
1 A · A e =
=
= 1490261230260140121230121++ + + − + − + + + + + − − + − + − + + + ++++++ + +=11
++
= 1482851213−−−−33 33
3 33
3 33
c)2A + 3At=
= 246024222303633963−−3 333
3 33 ++++++ + + +
333 + + + +== −− = = = 543657785
==== = = = −
27. A2 × 2 · X = B2 × 2
Paraquea igualdad sej possíve a matri X deve
e a l, z serdo tipo2 × 2.
SejaXmnpq=ig l
ud
a
Como A · X = B, temos:
= ⋅⋅
21031020 X B 1 3 0 01mnpqs 22331020mpnqpq++tm:=j p
0 1 =130
2 e s ao
o s
2122312131623mp m p+=+ = = = = psívsss
m m o e
202000nqnnq+= = = = m ss
mp
Logo: X=n + = = m
n = = m160230
q p
( ) )) x x x e e t ⋅ssen( cos ) xs o
28. a)cos()() )cos( cos( ( x x x s ns nsenaiz
r ) ( x oc
( =
)
= cos()( ( cos ) ) os 2xxxx x+ + ⋅sensensenseensensen()cos() )cos( cos( ( x x x⋅+ ⋅+)22 o =
)) ((c ( ) x ⋅ 2 2 ( ) )) x x x )s
c(
4
8. = 112sen2sen ) ( ( co ( x x⋅⋅⋅
( cos ) ) s ) x x ⋅ 1 2 =
1 s
= 1221sensen ) ) x11
( ( x= 2
2
b)A(x) A(x) A(x)
· = s 32. c
s 1221sens ens s n ) ) s ) )
en e ( ( co ( (
AB⋅= − − − − ==== =
⋅−−.
()cox x xs e
x x ns
e =ss( x s(
n ) o)
()
= y 1 1 3 211011111s
y 1 1 z
• cos(x) 1 s x= 0 ou x= 2π ( )
= I
• sen(2x) sen(x)
= s
s 2sen(x) cos(x) sen(x)
· = s
s 2sen(x) cos(x) sen(x) 0 s
· – =
AB⋅= − ⋅−+ ⋅−+ − ⋅⋅−+ − ⋅−+2111110111()() ) ) ) ) −
s sen(x) [2cos(x) 1] = 0
· –
i) sen(x) 0 s x= 0 ou x= π ou x= 2π ( I
= I) s ( ( (((
ii)2cos(x – 1 = 0 s
)
s cos(x) 12 s
=
− ⋅⋅ ⋅ 1 1 11)s
s x= π3 ou x= 53π (III)
S = ( ) [( I ( I ) {0; 2π}
I% I ) I I=
5 ] 0 1
29. b
AB⋅= − ====⋅−B
s AB⋅= − − + − ++++++211011s
== . ) )= + + +
++++3014211060302410– 1 = 06361
= =
BA⋅= − ====⋅−A + 3 1 + − − +
== + 0 = s AB⋅=B 1 00
) 11
+++211030146104300000
−− 0 ====== = 7430
0 ==
33. b
ABBA− ==== = −−−=== = 63617430=
== == − 1791 XABXC− = + + 23 s
s 3(X– A)= 2(B + X)+ 6C s 3X – 3A = 2B + 2X + 6C
y⋅⋅
s
s X = 3A + 2B + 6C s
30. 1123411xy−yy 0 .
yy
y 3 = − ======s
0 sX= −======+++++++++ −++++++63932420246126 s
−
sX= − + + − + + − + + ++++++62243469212306s
s 3413811+− +++++++= −====== x s
() y sX== 3 +
A281233
s 3441421238139+ − = = = + = − = − = − xxxyyy
34. c A=3 +2
A 8 2081512B=66 =
C 32Xxy=1
C “
33ss3
ssss 01
A · B = Xs
8⋅
⋅
31. b
s 208151232 21
05 0 1 1 85xy
8 5 =011
2 s
22⋅⋅
2130112−2 2
22 . 3 1 2 = +. 31 z xs
0 4 1 0 xy y z
4
⋅
20382153122 + ⋅⋅+ ⋅⋅ ⋅ 3 2 =⋅ 32xy
8 1 81 s5
s 232xyz z x − +++++++= +++++++s
y y z++
s 76692x =0 5xy
1y 8 1
5 12
s 232xyz y x
y z z++= − + = + +++++
35. d
x= –y
–2y + y+ 3z= 2y
3z= 3ys z= y
xy z x++=
y z − ++ − = − + − = − yyy y 1111
y y
9. AA⋅= − − ==== =
⋅−
=
−A > == + t10101210011210++ +
− + − + + +++++ −− === = 10020020142225
+= ==
(A· At–3I)· X = B
2225300312−−2 2 −2 222
2 2 2222
22 2222
2 ⋅⋅ 3 ) · =3xy· X
I I
s
s −−
−−− − − − − ⋅⋅ I y = I
x · xy· 122212xy
s −−− + + + + + + + = = = = = = = xy y22212s
x
s −− = − += = = = xy y21222+
x s –3x = 3 s x= –1
e y= 0
Logo: x+ y= –1 + 0 = –1
5
10. 36. −n
Annnn= − − − − − − − − =111111111111 ()( (
)) Bpppp
nnnn 123123123123
=n 11122223333 1332Û
11 313
()((
)) Û 11 3
1 3
a)Paran par:
2110010xx > 2 – x – x> 0 s x + x– 2 < 0
xs 2 2
S = –1 + 1 – 1 + 1 – 1 … + 1 = 0
Paran ímpar:
S = –1 + 1 – 1 + 1 … – 1 = –1 1– 2
b)An × n · Bn × p = C n × p S = ]–2; 1[
c42 = a · b12 + a · b22 + … + a · bn2 s
41 42 4n 16. Aplicando o teoremade Laplacena quinta
coluna,temos:
s c42 = 1 · 21 + 1 · 22 + 1 · 23 + … 1 · 2n s
2132111023403210252321110234032 ⋅−⋅−= ⋅−+
s cnn42232222= + + +
+ +SomadosprimeirostermosdaPGs 55
s caqqn42111= −−() s
s 409421212.()=−− n s ()11025
s – 4.094 = 2 – 2n + 1 s Aplicando o teore a de Laplac na segunda
m e
coluna,temos:
s 2n + 1 = 4.096 s
221123432125 ⋅⋅−⋅−= + ()
s 2n + 1 = 212 s n = 11 linhas
12
Capítulo 2Determinantes
Conexões
A área podeserobtidapor:
A
= – 4 · (15– 4 + 24 + 9 – 4 – 40)= – 4 · 0 = 0
A = AABC + ACDA s
29. a
s A = DD1222+ O segundo deter minan e a combinaçãolinear
t é
D1 = 011201131−− e D2 131111011− do primeiro (L = –L 1 + L2). Portan
2 to, os
determinan e sãoiguais.
t s
∴ A = 9262+ = 7,5 u.a.
30. Trata- edeumdeterminante Vandermonde.
s de
Suaresoluçãoé dadapor:
(3– 2)· (– 4 – 2)· (– 4 – 3)= 1 · (– 6)· (–7) = 42
31. Trata- edeumdeterminante Vandermonde.
s de
Então:
(k – k)· (k – k)· (k – k ) · (1 – k)· (1 – k ) · (1
2 –1 –1 2 2
Exercícios complementares – k )= 0
–1
I.k – k = 0
2
13. d
k(k– 1)= 0 s k = 0 ou k = 1
I.ad– bc = 0 s ad= bc
I abdcadbc001022=+
I. I k – k = 0 s 101 kkkk−= −
I –1
. 2 s = 0 s k2 = 1 s k = ±1
I I –1 – k = 0 s 101 kkkk−= −
Ik. 2 23 s= 0 sk = 1 sk= 1
3
Substituindo I em I , concluímos que o
I
IV.1 – k = 0 s k = 1
deter minan e
t vale:
2bc + bc = 3bc V. 1 – k = 0 s k = 1 s k = ±1
2 2
VI.1 – k = 0 s 1 –101kkk=− s= 0 s k = 1
–1
14. a
De I,I I I V e VI,temos:
I I IV,
, ,
00100cos( s n ) e ( cos ) x y=
) e (s n) (x y S = {–1; 0; 1}
= cos(x) cos(y – sen(x) sen(y =
· ) · ) 6
= + = = cos()cosxyπ312
15. a
11. 32. Trata se
- de um deter minan e
t de s c = –3 ou c = 5
Vander monde.
10. a
Temos,então:
xxx1110010101010−−− = s 1 – x = 0 s x= ±1
2
(x– 2)· (x– 3 – 2)· (x– 3 – x)· (1– 2)· (1– x)·
(1– x+ 3)< 0 s
s (x– 2)· (x– 5)· (–3)· (–1) · (1– x)· (4– x)< 0 s 11. c
s (x– 2)· (x– 5)· (1– x)· (4– x)· 3 < 0 s 2x· log x · 3 – 8x· log x· 3 = 0 s
2 2 2
s (x– 2)· (x– 5)· (1– x)· (4– x)< 0 s 2x· 2 · log x– 8x· log x= 0 s
2 2
Tratase de uma inequação- roduto,
- p cuja s 2x+ 1 · log x– 23x· log x= 0 s
2 2
resoluçãoé dadapor:
s log x· ()22
2 13xx+− = 0 s
I.x– 2 = 0 s x= 2
s log x= 0 s x= 1 ou ()22
2 13xx+− = 0 s
I x– 5 = 0 s x= 5
I.
s x+ 1 – 3x= 0 s x= 12
I I – x= 0 s x= 1
I1.
11232+ =
IV.4 – x= 0 s x= 4
12. b
xxx x x − + =11311113110s
x x x+
––+ + + – – – + – + – – + + + – – – – + + – – + 1 1 224455IIIIIIIVI _ II _ III _ IV
S = {x3 ® | 1 < x< 2 ou 4 < x< 5}
s (x – 1)· x+ 3x+ x– x – 3(x– 1)– (x+ 1)= 0 s
2 3
s x – x+ 3x+ x– x – 3x+ 3 – x– 1 = 0 s
3 3
Tarefa proposta s x= 2
1. e 7
xaax011011=
x+ ax– ax = 1 s
2
s ax – (a+ 1)x+ 1 = 0
2
(a+ 1) – 4a = 0 s
2
s a2 + 2a + 1 – 4a = 0 s a – 2a + 1 = 0 s
2
s (a– 1) = 0 s a = 1
2
2. a)2 · 5 – 4 · 1 = 10 – 4 = 6
b)5 · 0 – 3(–1) = 0 – (–3) = 3
3. a)sen 20° – (–cos 20°)=
2 2
= sen 20° + cos 20° = 1
2 2
b)sen75° · cos75° + sen75° · cos75° =
= 2sen75° · cos75° = sen(2· 75°) =
= sen150° = 12
4. e
Maaaa=Ma
aa
a=
11122122s M=− − 11
22
1142
detM=−1142 = 1 · 2 – 4 · (–1) = 2 + 4 = 6
5. a
a – b + (–a2 + b2)= a – b2 – a + b2 = 0
2 2 2 2
6. d
Aaaaa=2 ° =
2 11122122s A=12 2
22
1 2345
detA=2345 = 2 · 5 – 4 · 3 s detA= 10 – 12 s
detA= –2
7. 121142101121410−− =
= 4– 4+ 0+ 4+ 0– 2= 4– 2= 2
8. 124221511122524xxxx=s
s 2 + 4x+ 20x– 8 – 10 – 2x = 24 s
2
s –2x2 + 24x– 16 – 24 = 0 s
s 2x – 24x+ 40 = 0 s
2
s x – 12x+ 20 = 0 s
2
s x= 2 ou x= 10
S = {2; 10}
9. d
111191311119ccc = 0 – 2
0 =
= 27 + c + c – 9 – c2 – 3 = 0 s
s c2 – 2c – 15 = 0 s
12. 13. (–1) · (–1) + 2 · 131123142−− + (1)· (–1) + 4 ·
4 4
123112114−−−− =
= –(– 4 + 9 – 4 + 2 + 6 – 12) + (4 + 3 + 4 – 3 – 8 – 2) 22. e
= bc – (b – 4ac)= bc s –(b2 – 4ac)= 0 s
2
= –(–3) + (–2) = 3 – 2 = 1 s b2 – 4ac= 0 (Δ = 0)
14. xxxx x x 4444440= s x + 4x + 16x– 4x
x x x 3 2 2 Portanto, o gráfico da função tangencia o eixo
Ox.
– 4x – 4x = 0 s
2 2
23. a
ab x2222
xxx 22− = ++++++ −−++ + += −− eeee
+ ++
s x – 8x + 16x= 0 s x(x – 8x+ 16)= 0 s
3 2 2
s x= 0 ou x – 8x+ 16 = 0 s Soma= 8
2
= + + −− + = −−eeeeee 2022022424 x
xxx
A somadasraízes 0 + 8 = 8.
é
= + + − + − = −−eeee 224xxxx
2222
15. 11213130xx= x + 6 – 13 – 3x= x – 3x–
2 2
= = 441
7
24. b
⋅⋅ene 2232 s
3023xx=
11213130302xx s x – 3x– 7 = 3xs
x= 2 dxx= n
s x – 6x– 7 = 0 s x= 7 ou x= –1
2
ssdxx= ⋅⋅ene 2322
16. b
n
AB⋅= − ====⋅−−.
== 7 o = − ======21343122401711
u
s d = 2x· 2 · 23xs d = 24x+ 1
log d = log 24x+
2 2 1 = (4x+ 1)· log 2 =
2
= 4x+ 1
⋅
25. Aplicando o teorema de Laplace na última
coluna,temos:
det )AB= − = 40171144
( (–1) · 2100021190047502181110−
10
Observandoa quarta coluna,paraa aplicaçãode
Laplace, podemos concluir que o determinan e
t
val zero.
e
17. ax– x = 0 s x – ax= 0 s x(x– a)= 0 s x = 0
2 2
26. c
ou x= a M – k· I −
= − 30451001k=
Paraduasraíze reaise iguais,temos:a = 0
s
18. Aplicando Laplace na terceira linha, = − − 304500kk=
temos:
= − + − ++++ + ( 3045kk
+ () )
detA= 3 · (–1) · 10011011a− = 3 · (–1) ·
7
(1+ 1)s
det – k · I = 0 s –(3 + k)· (5– k)= 0 s
(M )
s (3+ k)· (5– k)= 0 s k = –3 ou k = 5
s detA= –3 · 2 = – 6
27. FazendoC 1 = C 1 – C 2; C 2 = C 2 – C 3; C 3 = C 3 – C 4,
19. p(x) 6 + 2x+ 2 = 2x+ 8
= temos:
a)P(5)= 2 · 5 + 8 = 18 kg
abbabbab a b −−− = − ⋅000000000 ()
b)30 = 2x+ 8 s 2x= 22 s x= 11 anos
20. a b a a 3
0310324330043330 xxx
xxx = s8 · 3x – 4 · 3x =
0 s 4 · 3x= 0 s 3x= 0
∅
28. Por setratar determinante umamatriz
do de de
Vander monde,temos:
S=
(x– 2)· (3– 2)· (3– x)· (1– 2)· (1– x)· (1– 3)= 0
s
s (x– 2)· (3– x)· (1– x)= 0 s
21. 21341102411012−−− =nnnn s –2n + n2 –
s x= 2 ou x= 3 ou x= 1
n + 3n – 4n = 12 s
S = {1; 2; 3}
s n2 – 4n – 12 = 0 s n = 6 ou n = –2
8
S = {–2; 6}
13. xxx ( ⋅−⋅−−−= + 312277077000 ()
29. d
AplicandoLapl e
ac na segundacoluna,temos:
x− ) 11 s
3 · (–1) · 13014142121205141421 +
3 4 ⋅−⋅()=
ssxxxx− ) ( ⋅=3270()
( ⋅−⋅− )
= (–3) · 11 + 2 · (–66)= –165
30. AplicandoLaplace primeira
na linha,temos:
x· (–1) · xx120300216= s
2
s x=3 ou x= 2 ou x= 7 ou x= 0
Logo, o triângulo é retângulo, pois:
222 )
732 ( = )
(+
s x· 2x = 16 s
2
sx = 8 s
∴ A = 2323 =
⋅
3
s x= 2
Ou seja:
α = 2 s α2 = 22 = 4
31. Por setratardo determinante umamatriz
de
deVandermonde,temos:
34. d
(5– 7)· (x– 7)· (x– 5)= 0 s Por Vander monde:
s (–2) · (x– 7)· (x– 5)= 0 s (log 20 – log 2) · (log 200 – log 2) · (log
s (x– 7)· (x– 5)= 0 s 200 – log 20) · · (log 2.000 – log 2)( log
s x= 7 ou x= 5 2.000 – log20)(log2.000 – log200)=
=.0 ⋅⋅ . 0 0 ⋅⋅ . 0 0 ⋅logloglog20220022002
S = {5; 7}
32. b
Aaaaa=; 7=
} 11122122
0–
0 0 0
a 421= ⋅+++++ += =sen(11)s ππ
0
11 + en
⋅⋅ g o ⋅⋅
l g g o ⋅logloglog20002200020200
l g
axx 12=
x ⋅−[] − = − sen(12)sen() en(
= s ) 02...000g0 =
20
02
= (log10)· (log100)· (log10)· (log1.000)
· (log100)· (log10)=
= 1 · 2 · 1 · 3 · 2 · 1 = 12
⋅−[]
35. Por Jacobi,vem:
axx =
21 =sen(21)s (
en )
1111111+ b11111+ a1111+
c111000b0001a000c + + + × × × – 1 – 1 – 1
=
Aplicando o teore ade Lapl ena primeira
m ac
a 40= ⋅+ + = ( =sen(22)s ππ
coluna,temos:
det( ⋅−⋅= + 11000000 abcabc
22 02n = en
)= 11
Logo: Axx=−− s(
=en
10sen( s n )
) e (
36. p(x) (3– x)· (a– x)· (1– x)+ 4 · (3– x)= 0
=
101414− = =sen()s ( s n ) e 2xxx ss
en ) e ( s n( x ))=±12 s
s p(x) (3– x)· [(a– x)· (1– x)+ 4] = 0 s
=
∴ S = −−−− 1167656665676116ππππππππ;;;;;;;;;;;;;
s (3– x)= 0 s
s x= 3 (únicaraizreal) ou
(a– x)· (1– x)+ 4 = 0 s
s a – ax– x+ x + 4 = 0 s
2
s x – (a+ 1)x+ (a+ 4)= 0
2
33. b Devemos ter Δ < 0, para que não exist m
a
FazendoC 1 = C 2 – C 1, C 2 = C 2 – C 3 e C 3 = C 3 – C 4,
outrasraíze reais.Assim:
s
temos:
Δ = (a+ 1) – 4 · (a+ 4)< 0 s
2
xxx −−−−−− =3322770227700770000
x s a + 2a + 1 – 4a – 16 < 0 s
2
Aplicando Laplace na primeira coluna, s a – 2a – 15 < 0
2
temos:
5– 3
15. Capítulo 3Complementos de Assim:M= −− === = 91111
==
|M| = – 9 + 11 = 2
determinantes M− = 112
Conexões 30. e
detA= –2 + 2 + 3 = 3
a)A = 3152e exs detA= 6 – 5 = 1
sd det(B ) = det(2A) 1detB= 23 · detAs 1detB=
–1 s
A–1 = 2153−−3 3
33
33 24 s
b)B = 5101−1 1 s detB= 5
11
11 s detB=124
B–1 = 15150555 5 1 Â = 151501
= “ 31. a
Aplicandoo teore am deBinet(P.8):
det(A· B–1 )= det(4A) det –1 )( )
P.8
· (B I
c)C = −− 2346s detC= –12 + 12 = 0 Como A é uma matri de ordem n, pode- e
z s
colocaro 4 emevid nê cia (P.4).
Não exist C –1 (a matri C não é invertív l
e z e ), n vez s
e emdet(4A),logo:
poisdetC= 0.
Exercícios complementares det(4 d t fatores nn=
) e d AA ⋅⋅⋅⋅⋅= ⋅44444 eet
4
13. e
B = k· A
Aa = ⋅4
det t = detB= det( · A)s
(B) k
s k3 · detA= 96 s k3 · 1,5 = 96 s n
, sss kkk 9615644== =
33
14. detA= det(A)
t
e
Aaaaa=A a = a=
aa aa
a== a 111221224278 det(d tBBb= =
e − 111)
Substituindoessesvalore em( )temos:
s I,
det(4414 ABabab ⋅= ⋅⋅= ⋅− )
|A| = |A t = 32 – 14 = 18
|
15. e
1 nn
det (3A· 2B)= det (3A) det(2B) 32 · detA· 22
· =
· detB=
= 9 · 2 · 4 · 3 = 216
16. d 32. c
Pabcd=1 d M · P = I s
4o
in e 2
det(2A· At)= 4k s det(2A) det t)= 4k s
· (A
1301711001de · P ⋅
⋅
s 23 · detA· detA= 4k s 8 · d · d = 4k s 2d2 = k
det (3B) 162 s 3d · detB= 162 s 3d · 2 = 162 s
=
s oM = 3 1 1 07abcd
0 7 =31 1
1 s
s 3d = 81 s 3d = 34 s d = 4
∴ 2 · 42 = k s k = 32
abacbd33771001++ · P I e ()
= e= I
s ,
I.aaaccc3137037037== + = + = = − ===== =sss
=
I bbbdd300711== + = = = I ss
I. es
Logo: k + d = 36
29. b
M · A – 2B = 0 s M · A = 2B s Logo: P= −=== = = = 30371
= ===
⋅⋅
A somados ele en sda diagonalprincipa é
m to l
abcdA 2
=B b d =bd =3412161422
c A cA 3 + 1 = 4.
Tarefa proposta
I.316421431627abab a
ab b+= + = t2 = + = 3
(+ s 1. c
SejadetA= d. Dividindo umafileiraporx(x≠ 0)
∴ a = 9 eb = –11
e multiplicandoumafileirapory(y≠ 0), temos:
det=⋅dxy dyx s dxy
s ⋅
I 324223221cdcdcdcd+= + = +=t = + = +=
I. + s
∴ c = 1 ed = –1 10
16. 2. a detB= – 6 (triangul r
a inferior)
Aaaaaa a a a a 11121314212223243132333441
a a a a a= det · B) = detA· detB= 36
(A
4224344aa24222
44124
3423comoa = 2i – j,vem:
ij 8. c
AA=− detA= 2sen (x) 2cos (x)
2 + 2 s
−−−− − − − − −
− − − − − −=10123210543276541edet s detA= 2[sen (x) 2cos (x)] 2 · 1 = 2
2 + 2 =
0012321054327654−−
t 1414 ⋅⋅ 2 d t 5e⋅=AA)
Por Jacobi,vem:
1– 1– 230532170246540– 104– 283–
detde ( 525 5 e =2d (
t
41 120246 6= + + × × 1 – 2
5–
Aplicando o teore a de Laplac na
m e
= ⋅⋅⋅= 1165detdetAAx
primeiralinha:
detA= (–1) · A13 s
x
s detA = (–1) · (–1) ·
4 42284412660−−
−= e
P.5
= ⋅= ⋅= = 11611622 ()detde AA
detA= detA = 0
t
55 t
3. a
Aabc = = b= d 23426 det t)= 2 · detB
t a e
c t 5 s (A
9. d
Como det(A)= detA,temos:
t I.(F)Um contra x m l
e e p o:
detA= 2 detB 23460=
4. d
I .
I (V)
mbncp=⋅4111 detBmanbpc=3111
aaaaaa a nnnnn1112122231122
a a 00000 ⋅⋅ nn
a⋅
detAa
·
I I 2121211+ ( ⋅−( = − =
I (V)
. ) )
Como detA= 2, temos:
24111= ⋅ambncp
detB= 2121+ ( ⋅−( ⋅detAs
s 11112ambncp= s
) )
manbpc11112= −
∴ detB= 31232 e B = = −
⋅−d t
s detB= 1 · detA
10. c
det 3)= det I s
(M (82)
s (detM3 = 82 · det( )s
) I
2
5. a s (detM3 = 64 · 1 s
)
PG(a;b; c; d),então: s detM=643 s
b = aq; c = aq e d = aq
2 3
s detM= 4
detMabcda aqaqaq== 23
Por P.5: detM= 0 11. d
Multiplicandoa primeiralinhapor a, a segundalinha
6. d
porb e a terceiralinhaporc, temos:
y x z=−⋅= − ss(P.4)
111= ⋅
123691212312323412x z y
abcaaabcbbabcccabcaabbcc
232323232323
ss12323442341234xy x z=− = (P.3)
z y
11
7. d
detA= – 6 (triangul r
a superior)
17. 12. b s x + 8 – 3 – 6x– 2x+ 2 = 0 s
2
[detM] = 25 s
2 s x – 8x+ 7 = 0 s x= 1 ou x= 7
2
s detM= ±5 ( )
I 23. c
detM= 3x+ 12 + 12 – 27 – 4x– 4 = –x – 7 ( I
I)
101120154364251025485−− = − − + = − + = −
De ( I ( ),
I ) I vem:
e
–x – 7 = 5 s –x = 12 s x= –12 ou
–x – 7 = –5 s –x = 2 s x= –2 Logo, o determinan e
t dainversaserá:−548
–12 + (–2) = –14 24. detA≠ 0
13. A= − ======1111detA= 2 3x– 6x ≠ 0 s x≠ 0 e x≠ 12
2
det 2)= (detA2 = 22 = 4
(A )
25. Sendoabcdtm inversa,temos:
ein
ra
14. b
10011001ce ⋅ 0 1 0 01abcd
⋅
det 2 · B2)= det 2)· det 2)s
(A (A (B
s det 2 · B2)= (detA2 · (de
(A ) tB) s
2
tm 0 1 =01 0
dr 0
s det 2 · B2)= (–1) · (–1) = 1
(A 2 2
15. b
a = 1; b = 2; x= 3 e y= 4
A== 1 b
; 1234e detA= –2 abcd1a = r 1001
0b m
0c in
det(AB) detA· detB= detA· detA s
= t 26. A = 2153−−3 3 s detA= –1 (existe –1 )
33
33 A
s det(AB) (de ) s det(AB) 4
= tA2 =
Aplicando o dispositivo prático (página 41),
16. d temos:
1 · 2 · 3 · 4 · 5 + (–1) · (1· 2 · 3 · 4 · 5 · 6)=
3152−−2 2
22
22
= 120 + (–1) · (720)=
= 120 – 720 = – 600
27. c
17. d a)(F)detB= 4
Q 3 = –2Q 2 s det(Q )= det
3 (–2Q2)s Pelodispositivoprático(página41):
s (detQ) = (–2) · (de
3 4 tQ) , dividindo ambos os
2 BA− = − ==== = = = ==1134014
= = == ≠
membros por(detQ) , temos:detQ = 16
2 b)(F)detA= 2; detB= 4
∴ detA≠ detB
18. d
sen()cos ) e ( cos ) e ( s2xxx x
(s n) (s n) e x 1011=nn()cos()
⋅=1110111
cos()xxx
2
c)(V)AB⋅=) ( )⋅⋅
V ) (V =) ( )110213041708
) V
= sen(x) [cos(x) 1 – cos(x) cos (x)]
· + – 2 =
= sen(x) [1 – cos (x)]
· 2 =
= sen(x) sen (x) sen (x)
· =
BA⋅=A( )⋅⋅
2 3
19. b
A2 = –2A ts V A (V =A( )130411021708
) V
s det 2)= det
(A (–2A )s (detA2 = (–2) · det(A)s
t ) 3 t
s (detA2 = –8 · detAs detA= –8
)
∴ AB = BA
20. d
A–1 · B · A = D s
s det –1 · B · A)= detDs
(A
s det –1 )· detB· detA= detDs
(A
⋅⋅
d)(F)det(A· B) = detA· detB= 2 · 4 ≠ 0
e) (F)
s 15detde d t
t e ABA = s
BBBA213041304115016= ⋅=34 0
01⋅⋅
3 3 4 3 01
0 1 =34
s detB= 5
21. detA≠ 3
=≠
=
6 – 12 – 4x≠ 0 s x≠ −32
22. e 12
detA= 0 s
18. mapa a a ⋅= − ⋅+ ⋅=cos()s ( s n )
m p ⋅+
28. d
⋅⋅ b d b ca⋅⋅ b d b
en ) e ( cos ) = oe
( 10cs
abcdabcd c a =bd c c a 12011201s
b
naqanaqa+ ⋅= − ⋅+ ⋅=cos()sen ) e ( co ( 01c
⋅
s aabccdacbdcd2222++ =+ + o
( s n ) s ) =o s
Temos:a = a + 2c s c = 0
29. a
I.–log x · logx– 1 – 3logx≠ 0
2
Resolv endo ambos os siste a
m s,
–2[log x] – 3logx– 1 ≠ 0
2
encontra mo s:
Δ = 9– 8= 1
m= cos(a);n = –sen (a);p = sen(a) q = cos(a)
e
I logxx
I. x≠−≠ ≠ − 12101010 ss
12 Assim,substi uindoemI,vem:
t
s ⋅
e logxxx≠−≠ ≠ − 110110 ss
1
30. AXBC –1 = B Xaaaa=−− s, cos() en ) e ( cos ) s b b b s n
im s (s n) ( co ( b b ) e
Multiplicandoà esquerdaporA–1 e à direitapor
C, temos:
A–1 · A · X · B · C –1 · C = A–1 · B · C s ()sen )
( cos )−o )
( − ss
(s
s I X · B · I A–1 · B · C
· =
s
Sendo I matri identidadede mesm ordemque
a z a
Xababb=⋅+ ⋅⋅cos()cos()sen( sen ) en( coss( s n( cos( sen( c
A, B e C.
X · B = A–1 · B · C ) (s ) ) e ) ) )
Multip licandoà direitaporB –1 , temos:
X · B · B–1 = A–1 · B · C · B –1 s
s ( a a ⋅⋅− ( cos ) e ( s n ⋅⋅+ ⋅⋅ ⋅ a
s X · I A–1 · B · C · B –1 s X = A–1 · B · C · B –1
=
os() en a b b− bbaabab cos )
) ( cos ) (s n) e ( ) a
31. d
⋅ ⋅ as
det(2A) det 2)s
= (A
s 22 · detA= (de )
tA2
Dividindo ambos os membros por detA, temos: s Xabababab=−−−−− − a
acos() en ) e ( cos c
s (s n) (o
)s
detA= 4
Observação: A informaçãosen (a)· cos (a)≠
32. 0 pode ser excluída do enunciado do
SeAaaaaB=−− . A
2 = =cos()sen ) e ( co ( e
( s n ) s ) ccos )
(s
proble a, mas, nesse caso, a resolução
m
en() en )
s ( cos ) b b−c( temos:
( b b − o)
s ,
implicariaumadiscussãomuitolonga.
A · X= B
33. d
Multip licando à esquerda por A–1 (que existe, A · A–1 = I s A · B = I s
2 2
isso⋅
uã⋅
poisdetA≠ 0), temos:
A–1 · A · X = A–1 · B s I X = A–1 · B s X = A–1 · B (I)
·
130143101001pq dcs m 3 1 3 0 3
0 4 =31 1
4sss
Calcul mo A–1 :
e s
s
Amnpq =mq1 e A · A–1 = I
− nl
pe
cos() en ) e ( cos ) a a n q− I ⋅⋅
113041001+0 qiss m pqq
1 dcs u
p u =o it
I.qq414= = s
s (s n) (a a m p − · I · = =I
I 1301340+ = + = pqp ss413112pp=− = − s
I.
∴ s qp− = − − ======= + = 12412112415
= 1001 s
s
mapanaqa ⋅+
m ⋅⋅+ ⋅−⋅cos()sen( cos( s n ) enn( c
) ) e (s ) 34. a)(A+ B) · (A– B)= A2 – AB + BA – B2
b)O produtoAB deveserigualao produto
BA.
( a a aqa+⋅−⋅+
os()s ( cos ) p n
en ) ⋅⋅ s()s =
e c)detde ( d t ) e2AAAA− = − ⋅=(1
t) e 1d t
d)detdetBA=1
= ( s 1001
)en
Para que a igualdad se verifique, é preciso
e 35. b
que: log [det
3 (2A )]= log (detA )s
–1 27 –1
s log [25 · det(2A )]= log 3 [det –1 · A–1 )]s
3 –1 3 (2
19. s
loglogde 33132113132
t( ⋅⋅ 2 1 1 ⋅⋅ 2 1 1
1 3 = 1 3 −
det)AAs
s logdetlogdet 321321AA 22 ⋅⋅ 2 3 1s
3313 33
11
= 1 2
s 32132 detdetAA=d s
13 etÛ
s 212155detdetAA= s
3
s 220 · detA= det A s 220 = det A s
3 2
s detA= 1.024 = 210
36. a)AB = BA s AB · B–1 = BA · B–1 s A · I B =
· A · B–1 s
s A = B · A · B–1 s B–1 · A = B–1 · B · A · B–1 s
s B–1 · A = I A · B–1 s B–1 · A = A · B–1
·
∴ A · B–1 = B–1 · A (c.q.d.)
13
20. b)A2 + 2AB – B = 0 s B = A2 + 2AB s B = A(A+ 2B)s
s det(B) det[A(A+ 2B)]s
=
s det(B) det(A) det(A+ 2B)
= ·
Do enunciado,sabemosqueB é inversíve Logo, det
l. (B) 0.
≠
Assim:det ) det(A+ 2B)≠ 0 s det ) 0 e det + 2B)≠ 0
(A · (A ≠ (A
Então,sedet ) 0, A é inversíve
(A ≠ l.
(c.q.d.)
Capítulo 4 sistemas lineares
Conexões
1.
– 5 05 – 2 1 2 y x r t s
2. S1(r;s):yxyx= + = − 52
S2(r;t):yxyx= + = − 52
S3(t;s):yxyx= − = − 22
3. S1(r;s):yxyx= + = − 52s
s x+ 5 = –2x s 3x= –5 s x= − 53
∴ S= − 53 10 3 ; (Asretassãoconcorren e )
t s.
S2(r;t):yxyx= + = − 52 s
s x+ 5 = x– 2 s 0x= –7 (F)
∴ S = ∅ (Asretassãopara e a distinta )
l l s s.
S3(t;s):yxyx= − = − 22s
s x– 2 = –2x s 3x= 2 s x= 23
∴ S = 23 43 ; − (Asretassãoconcorren e )
t s.
Exercícios complementares
13. D a a a a a = − = − − cos()sen()sen()cos()cos()s2 en() aD s = −1
2
⋅
D a a a a x= − − = = − sen(2)sen()cos(2)cos()sen(2)cos()sen()cos()aaaa + ⋅ 2s
s Dx= –sen (a)
D a a a a a y= − = − cos()sen(2)sen()cos(2)cos(2)cos()sen2sen()⋅ − ⋅ aaa
Dy= – cos(a)
xDD a a x= = − − = sen()sen()1
yDD a a y= = − − = cos()cos()1
21. ∴ S = {(sen(a);cos(a))}
14. a)SPD (sistemaescalonadodo primeirotipo)
a bbb + = = = a 5 2 12 6 s
e
Substitui eb naprimeira
s
- igualdade:
a + 6 = 5 s a = –1
∴ S = {(–1; 6)}
b)SPD (sist m escalonadodo primeirotipo)
e a
2z= 4 s z= 2
Substitui eznasegundaigualdad
s
- e:
3y+ 4 · 2 = 14 s 3y= 6 s y= 2
Substitui eye zna primeira
s
- igualdade:
x+ 2 · 2 – 2 = 0 s x= –2
∴ S = {(–2; 2; 2)}
c)SP (sist m escalonadodo segundotipo)
I e a
p é a variáv l
e livre.
n + 2p= 5 s n = 5 – 2p
Substitui eo valorden naprimeira
s
- equação.
–m + 2 · (5– 2p)+ 3p= 4 s
s –m + 10 – 4p+ 3p= 4 s
s –m – p = – 6 s
s m= 6 – p
∴ S = {(6– p; 5 – 2p; p),comp 3 ® }
15. d
Sendoy= 0, subs tuímos
ti essevalornasegundaequaçãoe obtemos = 2.
x
Substituindonaprimeiraequação,temos:
(λ + 1)· 2 + 0 = 0 s λ + 1 = 0 s λ = –1
16. d
Antes Hoje Depois
Eu y 2x a
Tu x y 2x