372 Chapter 7 Eigenvalues and Eigenvectors
7.4 Applications of Eigenvalues and Eigenvectors
Model population growth using an age transition matrix and an age distribution vector, and find a stable age distribution vector. Use a matrix equation to solve a system of first-order linear differential equations. Find the matrix of a quadratic form and use the Principal Axes Theorem to perform a rotation of axes for a conic and a quadric surface.
POPULATION GROWTH Matrices can be used to form models for population growth. The first step in this process is to group the population into age classes of equal duration. For instance, if the maximum life span of a member is L years, then the following n intervals represent the age classes. L 0, First age class ΄ n L 2L , Second age class ΄n n . . . ͑n Ϫ 1͒L , L nth age class ΄ n ΅ The age distribution vector x represents the number of population members in each age class, where
x 1 Number in first age class x Number in second age class x ϭ .2 ...... ΄x ΅ Number in n th age class n Over a period of L͞n years, the probability that a member of the i th age class will ϩ survive to become a member of the ͑i 1͒th age class is given by pi, where ϭ Ϫ 0 Յ pi Յ 1, i 1, 2, . . . , n 1. The average number of offspring produced by a member of the i th age class is given by
bi, where ϭ 0 Յ bi, i 1, 2, . . . , n. These numbers can be written in matrix form, as follows. . . . b1 b2 b3 bnϪ1 bn . . . p1 0 0 0 0 A ϭ 0 p 0 . . . 0 0 . .2 ...... ΄ . . . ΅ 0 0 0 pnϪ1 0 Multiplying this age transition matrix by the age distribution vector for a specific time period produces the age distribution vector for the next time period. That is, ϭ Axi xiϩ1. Example 1 illustrates this procedure. 7.4 Applications of Eigenvalues and Eigenvectors 373
A Population Growth Model
REMARK A population of rabbits has the following characteristics. If the pattern of growth in a. Half of the rabbits survive their first year. Of those, half survive their second year. Example 1 continued for The maximum life span is 3 years. another year, then the rabbit b. During the first year, the rabbits produce no offspring. The average number of population would be offspring is 6 during the second year and 8 during the third year. 168 The population now consists of 24 rabbits in the first age class, 24 in the second, ϭ ϭ and 20 in the third. How many rabbits will there be in each age class in 1 year? x3 Ax2 152 . ΄ 6΅ SOLUTION
From the age distribution The current age distribution vector is vectors x1, x2, and x3, you can 24 0 Յ age < 1 see that the percent of rabbits ϭ 1 Յ age < 2 in each of the three age classes x1 24 changes each year. To obtain ΄20 ΅ 2 Յ age Յ 3 a stable growth pattern, one and the age transition matrix is in which the percent in each age class remains the same 0 6 8 ϩ each year, the ͑n 1͒th age A ϭ 0.5 0 0 . distribution vector must be ΄ 0 0.5 0΅ a scalar multiple of the nth age distribution vector. That is, After 1 year, the age distribution vector will be ϭ ϭ xnϩ1 Axn xn. Example 2 shows how to solve this 0 6 8 24 304 0 Յ age < 1 problem. ϭ ϭ ϭ Յ < x2 Ax1 0.5 0 0 24 12 . 1 age 2 ΄ 0 0.5 0΅΄20 ΅ ΄ 12 ΅ 2 Յ age Յ 3
Finding a Stable Age Distribution Vector
Find a stable age distribution vector for the population in Example 1. SOLUTION To solve this problem, find an eigenvalue and a corresponding eigenvector x such that Ax ϭ x. The characteristic polynomial of A is ԽI Ϫ AԽ ϭ ͑ ϩ 1͒2͑ Ϫ 2͒ (check this), which implies that the eigenvalues are Ϫ1 and 2. Choosing the positive value, let ϭ 2. Verify that the corresponding eigenvectors are of the form Simulation x 16 t 16 Explore this concept further with 1 ϭ ϭ ϭ an electronic simulation available x x 2 4t t 4 . at www.cengagebrain.com . ΄ ΅ ΄ ΅ ΄ ΅ x 3 t 1 For instance, if t ϭ 2, then the initial age distribution vector would be 32 0 Յ age < 1 ϭ Յ < x1 8 1 age 2 ΄ 2΅ 2 Յ age Յ 3 and the age distribution vector for the next year would be
0 0 6 8 32 64 0 Յ age < 1 ϭ ϭ ϭ Յ < x2 Ax1 0.5 0 0 8 16 . 1 age 2 ΄ 0 0.5 0΅΄ 2΅ ΄ 4΅ 2 Յ age Յ 3 Notice that the ratio of the three age classes is still 16 : 4 : 1, and so the percent of the population in each age class remains the same. 374 Chapter 7 Eigenvalues and Eigenvectors
SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS (CALCULUS) A system of first-order linear differential equations has the form Ј ϭ ϩ ϩ . . . ϩ y1 a11 y1 a12 y2 a1nyn Ј ϭ ϩ ϩ . . . ϩ y2 a21 y1 a22 y2 a2nyn . . . Ј ϭ ϩ ϩ . . . ϩ yn an1y1 an2 y2 ann yn dy where each y is a function of t and y Ј ϭ i. If you let i i dt Ј y1 y1 y y Ј y ϭ .2 and yЈ ϭ .2 . . . . ΄y ΅ ΄y Ј΅ n n then the system can be written in matrix form as yЈ ϭ Ay.
Solving a System of Linear Differential Equations Solve the system of linear differential equations. Ј ϭ y1 4y1 Ј ϭ Ϫ y2 y2 Ј ϭ y3 2y3 SOLUTION From calculus, you know that the solution of the differential equation yЈ ϭ ky is y ϭ Ce kt . So, the solution of the system is ϭ 4t y1 C1e ϭ Ϫt y2 C2e ϭ 2t y3 C3e .
The matrix form of the system of linear differential equations in Example 3 is yЈ ϭ Ay, or Ј y1 4 0 0 y1 Ј ϭ Ϫ y2 0 1 0 y2 . ΄ Ј ΅ ΄ ΅΄ ΅ y3 0 0 2 y3 ϭ it So, the coefficients of t in the solutions yi Ci e are given by the eigenvalues of the matrix A. If A is a diagonal matrix, then the solution of yЈ ϭ Ay can be obtained immediately, as in Example 3. If A is not diagonal, then the solution requires more work. First, attempt to find a matrix P that diagonalizes A. Then, the change of variables y ϭ Pw and yЈ ϭ PwЈ produces PwЈ ϭ yЈ ϭ Ay ϭ AP w wЈ ϭ P Ϫ1AP w where P Ϫ1AP is a diagonal matrix. Example 4 demonstrates this procedure. 7.4 Applications of Eigenvalues and Eigenvectors 375
Solving a System of Linear Differential Equations Solve the system of linear differential equations. Ј ϭ ϩ y1 3y1 2y2 Ј ϭ Ϫ y2 6y1 y2 SOLUTION 3 2 First, find a matrix P that diagonalizes A ϭ . The eigenvalues of A are ΄6 Ϫ1΅ ϭ Ϫ ϭ ϭ Ϫ T 1 3 and 2 5, with corresponding eigenvectors p1 ͓1 3͔ and ϭ T p2 ͓1 1 ͔ . Diagonalize A using the matrix P whose columns consist of p1 and p2 to obtain
1 Ϫ1 1 1 4 4 Ϫ3 0 P ϭ , PϪ1 ϭ , and P Ϫ1AP ϭ . ΄Ϫ3 1΅ 3 1 ΄ 0 5΅ ΄4 4΅ The system wЈ ϭ P Ϫ1AP w has the following form. w Ј Ϫ3 0 w w Ј ϭ Ϫ 3w 1 ϭ 1 1 1 Ј Ј ϭ ΄w2 ΅ ΄ 0 5΅΄w2΅ w2 5w2 The solution of this system of equations is ϭ Ϫ3t w1 C1e ϭ 5t w2 C2e . ϭ To return to the original variables y1 and y2 , use the substitution y Pw and write y 1 1 w 1 ϭ 1 Ϫ ΄ y2΅ ΄ 3 1΅΄w2΅ which implies that the solution is ϭ ϩ ϭ Ϫ3t ϩ 5t y1 w1 w2 C1e C2e ϭ Ϫ ϩ ϭ Ϫ Ϫ3t ϩ 5t y2 3w1 w2 3C1e C2e .
If A has eigenvalues with multiplicity greater than 1 or if A has complex eigenvalues, then the technique for solving the system must be modified. 1. Eigenvalues with multiplicity greater than 1: The coefficient matrix of the system Ј ϭ 0 1 y1 y2 ϭ Ј ϭ Ϫ ϩ is A . y2 4y1 4y2 ΄Ϫ4 4΅ The only eigenvalue of A is ϭ 2, and the solution of the system is ϭ 2t ϩ 2t y1 C1e C2te ϭ ϩ 2t ϩ 2t y2 ͑2C1 C2͒e 2C2te . 2. Complex eigenvalues: The coefficient matrix of the system Ј ϭ Ϫ 0 Ϫ1 y1 y2 ϭ Ј ϭ is A . y2 y1 ΄1 0΅ ϭ ϭ Ϫ The eigenvalues of A are 1 i and 2 i, and the solution of the system is ϭ ϩ y1 C1 cos t C2 sin t ϭ Ϫ ϩ y2 C2 cos t C1 sin t. Try checking these solutions by differentiating and substituting into the original systems of equations. 376 Chapter 7 Eigenvalues and Eigenvectors
QUADRATIC FORMS Eigenvalues and eigenvectors can be used to solve the rotation of axes problem introduced in Section 4.8. Recall that classifying the graph of the quadratic equation ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 0 Quadratic equation is fairly straightforward as long as the equation has no xy -term (that is,b ϭ 0 ). If the equation has an xy -term, however, then the classification is accomplished most easily by first performing a rotation of axes that eliminates the xy -term. The resulting equation (relative to the new xЈyЈ -axes) will then be of the form aЈ͑xЈ͒2 ϩ cЈ͑yЈ͒2 ϩ dЈxЈ ϩ eЈyЈ ϩ fЈ ϭ 0. You will see that the coefficients aЈ and cЈ are eigenvalues of the matrix a b͞2 A ϭ . ΄b͞2 c΅ The expression ax 2 ϩ bxy ϩ cy 2 Quadratic form is called the quadratic form associated with the quadratic equation ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 0 and the matrix A is called the matrix of the quadratic form. Note that the matrix A is symmetric . Moreover, the matrix A will be diagonal if and only if its corresponding quadratic form has no xy -term, as illustrated in Example 5.
y Finding the Matrix of a Quadratic Form
3 Find the matrix of the quadratic form associated with each quadratic equation. a.4x 2 ϩ 9y2 Ϫ 36 ϭ 0 b. 13 x 2 Ϫ 10 xy ϩ 13 y2 Ϫ 72 ϭ 0 1 SOLUTION x ϭ ϭ ϭ −2 −1 1 2 a. Because a 4, b 0, and c 9, the matrix is − 1 4 0 A ϭ . Diagonal matrix (no xy -term) ΄0 9΅ x2 y2 −3 + = 1 ϭ ϭ Ϫ ϭ 32 22 b. Because a 13, b 10, and c 13, the matrix is 13 Ϫ5 Figure 7.3 A ϭ . Nondiagonal matrix (xy -term) ΄Ϫ5 13 ΅
In standard form, the equation 4x 2 ϩ 9y2 Ϫ 36 ϭ 0 is (x′)2 (y′)2 y 2 + 2 = 1 x 2 y2 3 2 ϩ ϭ 1 32 22 y′ 3 x′ which is the equation of the ellipse shown in Figure 7.3. Although it is not apparent by inspection, the graph of the equation 13 x 2 Ϫ 10 xy ϩ 13 y 2 Ϫ 72 ϭ 0 is similar. 1 Њ 45 ° In fact, when you rotate the x - and y -axes counterclockwise 45 to form a new x xЈyЈ-coordinate system, this equation takes the form −3 −1 1 3 xЈ 2 yЈ 2 ͑ ͒ ϩ ͑ ͒ ϭ 2 2 1 −2 3 2
−3 which is the equation of the ellipse shown in Figure 7.4. To see how to use the matrix of a quadratic form to perform a rotation of axes, let 13 x2 − 10 xy + 13 y2 − 72 = 0 x X ϭ . Figure 7.4 ΄ y΅ 7.4 Applications of Eigenvalues and Eigenvectors 377
Then the quadratic expression ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f can be written in matrix form as follows. a b͞2 x x X TAX ϩ ͓d e͔X ϩ f ϭ ͓x y͔ ϩ ͓d e͔ ϩ f ΄b͞2 c ΅΄ y΅ ΄ y΅ ϭ ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f If b ϭ 0, then no rotation is necessary. But if b Þ 0, then because A is symmetric, you can apply Theorem 7.10 to conclude that there exists an orthogonal matrix P such that P TAP ϭ D is diagonal. So, if you let xЈ P T X ϭ XЈ ϭ ΄ yЈ΅ then it follows that X ϭ PX Ј, and X TAX ϭ ͑PX Ј͒TA͑PX Ј͒ ϭ ͑XЈ͒TP TAPX Ј ϭ ͑XЈ͒TDX Ј. The choice of the matrix P must be made with care. Because P is orthogonal, its determinant will be ±1. It can be shown (see Exercise 65) that if P is chosen so that ԽPԽ ϭ 1, then P will be of the form cos Ϫsin P ϭ ΄ sin cos ΅ where gives the angle of rotation of the conic measured from the positive x -axis to the positive xЈ -axis. This leads to the Principal Axes Theorem.
Principal Axes Theorem For a conic whose equation is ax 2 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 0, the rotation given by X ϭ PX Ј eliminates the xy -term when P is an orthogonal ϭ REMARK matrix, with ԽPԽ 1, that diagonalizes A. That is, Note that the matrix product 0 PTAP ϭ 1 d e PX Ј has the form ͓ ͔ ΄ 0 2΅ d cos ϩ e sin xЈ ͑ ͒ where 1 and 2 are eigenvalues of A. The equation of the rotated conic is ϩ ͑Ϫd sin ϩ e cos ͒yЈ. given by Ј 2 ϩ Ј 2 ϩ Ј ϩ ϭ 1͑x ͒ 2͑y ͒ ͓d e͔PX f 0.
Rotation of a Conic
Perform a rotation of axes to eliminate the xy -term in the quadratic equation 13 x 2 Ϫ 10 xy ϩ 13 y 2 Ϫ 72 ϭ 0. SOLUTION The matrix of the quadratic form associated with this equation is 13 Ϫ5 A ϭ . ΄Ϫ5 13 ΅ Because the characteristic polynomial of A is ͑ Ϫ 8͒͑ Ϫ 18 ͒ (check this), it follows ϭ ϭ that the eigenvalues of A are 1 8 and 2 18. So, the equation of the rotated conic is 8͑xЈ͒2 ϩ 18 ͑yЈ͒2 Ϫ 72 ϭ 0 which, when written in the standard form ͑xЈ͒2 ͑yЈ͒2 ϩ ϭ 1 32 22 is the equation of an ellipse. (See Figure 7.4.) 378 Chapter 7 Eigenvalues and Eigenvectors
In Example 6, the eigenvectors of the matrix A are 1 Ϫ1 x ϭ and x ϭ 1 ΄1΅ 2 ΄ 1΅ which you can normalize to form the columns of P, as follows.
1 Ϫ 1 Ί2 Ί2 cos Ϫsin P ϭ ϭ 1 1 ΄ sin cos ΅ ΄Ί2 Ί2΅ Note first that ԽPԽ ϭ 1, which implies that P is a rotation. Moreover, because cos 45 Њ ϭ 1͞Ί2 ϭ sin 45 Њ , the angle of rotation is 45 Њ, as shown in Figure 7.4. The orthogonal matrix P specified in the Principal Axes Theorem is not unique. Its entries depend on the ordering of the eigenvalues 1 and 2 and on the subsequent choice of eigenvectors x1 and x2. For instance, in the solution of Example 6, any of the following choices of P would have worked.
x1 x2 x1 x2 x1 x2 Ϫ 1 1 Ϫ 1 Ϫ 1 1 1 Ί2 Ί2 Ί2 Ί2 Ί2 Ί2 Ϫ 1 Ϫ 1 1 Ϫ 1 Ϫ 1 1 ΄ Ί2 Ί2΅ ΄ Ί2 Ί2΅ ΄ Ί2 Ί2΅ ϭ ϭ ϭ ϭ ϭ ϭ 1 8, 2 18 1 18, 2 8 1 18, 2 8 ϭ 225 Њ ϭ 135 Њ ϭ 315 Њ For any of these choices of P, the graph of the rotated conic will, of course, be the same. (See Figure 7.5.)
y y y
3 x′ 3 3 y′
225 ° 135 ° 315 ° x x x −31 3 −31 3 −3 1 3
−2 −2 −2 ′ ′ ′ x′ −3 y y −3 −3 x (x′)2 (y′)2 (x′)2 (y′)2 (x′)2 (y′)2 + = 1 + = 1 2 + 2 = 1 32 22 22 32 2 3
Figure 7.5 The following summarizes the steps used to apply the Principal Axes Theorem. 1. Form the matrix A and find its eigenvalues 1 and 2. 2. Find eigenvectors corresponding to 1 and 2. Normalize these eigenvectors to form the columns of P. 3. If ԽPԽ ϭ Ϫ 1, then multiply one of the columns of P by Ϫ1 to obtain a matrix of the form cos Ϫsin P ϭ . ΄ sin cos ΅ 4. The angle represents the angle of rotation of the conic. Ј 2 ϩ Ј 2 ϩ Ј ϩ ϭ 5. The equation of the rotated conic is 1͑x ͒ 2͑y ͒ ͓d e͔ PX f 0. 7.4 Applications of Eigenvalues and Eigenvectors 379
Example 7 shows how to apply the Principal Axes Theorem to rotate a conic whose center has been translated away from the origin.
Rotation of a Conic
Perform a rotation of axes to eliminate the xy -term in the quadratic equation 3x 2 Ϫ 10 xy ϩ 3y 2 ϩ 16 Ί2x Ϫ 32 ϭ 0. SOLUTION The matrix of the quadratic form associated with this equation is 3 Ϫ5 A ϭ . ΄Ϫ5 3΅ The eigenvalues of A are ϭ ϭ Ϫ 1 8 and 2 2 with corresponding eigenvectors of ϭ Ϫ ϭ Ϫ Ϫ x1 ͑ 1, 1 ͒ and x2 ͑ 1, 1͒. This implies that the matrix P is
Ϫ 1 Ϫ 1 Ί2 Ί2 P ϭ 1 Ϫ 1 ΄ Ί2 Ί2΅ cos Ϫsin ϭ , where P ϭ 1. ΄ sin cos ΅ Խ Խ Because cos 135 Њ ϭ Ϫ 1͞Ί2 and sin 135 Њ ϭ 1͞Ί2, the angle of rotation is 135 Њ. y Finally, from the matrix product 10 Ϫ 1 Ϫ 1 8 Ί2 Ί2 xЈ ͓ d e ͔PX Ј ϭ 16 Ί2 0͔ ͓ 1 1 ΄yЈ΅ 6 Ϫ ′ ΄ Ί2 Ί2΅ x 4 ϭ Ϫ 16 xЈ Ϫ 16 yЈ 135 ° the equation of the rotated conic is x −4 −22 4 6 8 8͑xЈ͒2 Ϫ 2͑yЈ͒2 Ϫ 16 xЈ Ϫ 16 yЈ Ϫ 32 ϭ 0. −2 y′ In standard form, the equation ′ − 2′ 2 (x 1)− ( y + 4) ͑xЈ Ϫ 1͒2 ͑yЈ ϩ 4͒2 2 2 = 1 Ϫ ϭ 1 1 2 12 22 Figure 7.6 is the equation of a hyperbola. Its graph is shown in Figure 7.6.
Quadratic forms can also be used to analyze equations of quadric surfaces in R3, which are the three-dimensional analogs of conic sections. The equation of a quadric surface in R3 is a second-degree polynomial of the form ax 2 ϩ by 2 ϩ cz 2 ϩ dxy ϩ exz ϩ fyz ϩ gx ϩ hy ϩ iz ϩ j ϭ 0. There are six basic types of quadric surfaces: ellipsoids, hyperboloids of one sheet, hyperboloids of two sheets, elliptic cones, elliptic paraboloids, and hyperbolic paraboloids. The intersection of a surface with a plane, called the trace of the surface in the plane, is useful to help visualize the graph of the surface in R3. The six basic types of quadric surfaces, together with their traces, are shown on the next two pages. 380 Chapter 7 Eigenvalues and Eigenvectors
Ellipsoid z x 2 y2 z2 z ϩ ϩ ϭ 1 a2 b2 c2 xz -trace yz -trace
Trace Plane Ellipse Parallel to xy -plane Ellipse Parallel to xz -plane Ellipse Parallel to yz -plane y y x The surface is a sphere when x a ϭ b ϭ c Þ 0. xy -trace
z z Hyperboloid of One Sheet x 2 y2 z2 ϩ Ϫ ϭ 1 a2 b2 c2
Trace Plane Ellipse Parallel to xy -plane xy -trace y y Hyperbola Parallel to xz -plane Hyperbola Parallel to yz -plane x x The axis of the hyperboloid corresponds to the variable whose coefficient is negative.
xz -trace yz -trace
z Hyperboloid of Two Sheets yz -trace z xz -trace z2 x 2 y2 Ϫ Ϫ ϭ 1 c2 a2 b2
Trace Plane Ellipse Parallel to xy -plane parallel to Hyperbola Parallel to xz -plane xy -plane no xy -trace x y Hyperbola Parallel to yz -plane x y The axis of the hyperboloid corresponds to the variable whose coefficient is positive. There is no trace in the coordinate plane perpendicular to this axis. 7.4 Applications of Eigenvalues and Eigenvectors 381
z z xz -trace Elliptic Cone x 2 y2 z2 ϩ Ϫ ϭ 0 a2 b2 c2
Trace Plane Ellipse Parallel to xy -plane xy -trace Hyperbola Parallel to xz -plane (one point) y Hyperbola Parallel to yz -plane y x x The axis of the cone corresponds parallel to to the variable whose coefficient xy -plane is negative. The traces in the coordinate planes parallel to this axis are intersecting lines. yz -trace
z z Elliptic Paraboloid yz -trace xz -trace x 2 y2 z ϭ ϩ a2 b2
Trace Plane Ellipse Parallel to xy -plane Parabola Parallel to xz -plane Parabola Parallel to yz -plane parallel to xy -plane The axis of the paraboloid
corresponds to the variable raised y x y to the first power. xy -trace x (one point )
z z Hyperbolic Paraboloid yz -trace y2 x 2 z ϭ Ϫ b2 a2
Trace Plane y y Hyperbola Parallel to xy -plane x x Parabola Parallel to xz -plane Parabola Parallel to yz -plane The axis of the paraboloid corresponds to the variable raised parallel to xy -plane to the first power. xz -trace 382 Chapter 7 Eigenvalues and Eigenvectors
LINEAR Some of the world’s most unusual architecture makes use ALGEBRA of quadric surfaces. For instance, Catedral Metropolitana APPLIED Nossa Senhora Aparecida, a cathedral located in Brasilia, Brazil, is in the shape of a hyperboloid of one sheet. It was designed by Pritzker Prize winning architect Oscar Niemeyer, and dedicated in 1970. The sixteen identical curved steel columns, weighing 90 tons each, are intended to represent two hands reaching up to the sky. Pieced together between the columns, in the 10-meter-wide and 30-meter-high triangular gaps formed by the columns, is semitransparent stained glass, which allows light inside for nearly the entire height of the columns.
The quadratic form of the equation ax 2 ϩ by 2 ϩ cz 2 ϩ dxy ϩ exz ϩ fyz ϩ gx ϩ hy ϩ iz ϩ j ϭ 0 Quadric surface is defined as ax 2 ϩ by 2 ϩ cz 2 ϩ dxy ϩ exz ϩ fyz . Quadratic form The corresponding matrix is d e a 2 2 d f A ϭ b . 2 2 e f c ΄2 2 ΅ In its three-dimensional version, the Principal Axes Theorem relates the eigenvalues and eigenvectors of A to the equation of the rotated surface, as shown in Example 8.
Rotation of a Quadric Surface
Perform a rotation of axes to eliminate the xz -term in the quadratic equation 5x 2 ϩ 4y2 ϩ 5z2 ϩ 8xz Ϫ 36 ϭ 0. z SOLUTION
6 The matrix A associated with this quadratic equation is z' 4 5 0 4 2 A ϭ 0 4 0 ΄4 0 5΅ 2 2 ϭ ϭ ϭ Ј Ј Ј 6 4 which has eigenvalues of 1 1, 2 4, and 3 9. So, in the rotated x y z -system, 2 2 2 x 6 the quadratic equation is ͑xЈ͒ ϩ 4͑yЈ͒ ϩ 9͑zЈ͒ Ϫ 36 ϭ 0, which in standard form is y xЈ 2 yЈ 2 zЈ 2 ͑ ͒ ϩ ͑ ͒ ϩ ͑ ͒ ϭ 2 2 2 1. x' 6 3 2 Figure 7.7 The graph of this equation is an ellipsoid. As shown in Figure 7.7, the xЈyЈzЈ -axes represent a counterclockwise rotation of 45 Њ about the y -axis. Moreover, the orthogonal matrix 1 1 0 Ί2 Ί2 P ϭ 0 1 0 1 1 Ϫ 0 ΄ Ί2 Ί2 ΅ whose columns are the eigenvectors of A, has the property that P TAP is diagonal. ostill, 2010/Used under license from Shutterstock.com