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Matrices and Their ApplicationsaAX1 bAX2 cAX3 aλ1X1 bλ2X2 cλ3X3 0, as [A – λiI]Xi 0Multiply (2) by A, again, and obtainaλ12X1 bλ22X2 cλ32X3 0Now writing (1), (2), (3) together as 1 1 1 aX1 λ λ λ bX 023 12 2 λ1 λ22 λ32 cX3 Now we see that1 B λ1λ121λ2λ2255 (2) (3) (4)1λ3 ( λ1 λ2 ) ( λ2 λ3 )( λ3 λ1 ) 0λ32Hence B–1 exists.Multiplying (4) byB–1 aX1 results in, bX2 0. cX 3 But this requires a b c 0 which is contrary to the hypothesis.Thus X1, X2, X3 are linearly independent.Theorem 2: If λ be a non-zero characteristic root (eigen value) of the non-singular n A square matrix A, thenis a charactristic polynomial of adjoint A.λProof: For non-singular n-square matrix A, the ‘characteristic polynomial’ (1)φ(λ) λI – A λn s1λn – 1 s2λn – 2 sn– 1 λ1 (–1)n A where sr (r 1, 2, , n – 1) is (–1)r times the sum of all the r-square principal minors of A.Corresponding characteristic equation is given byλn s1λn–1 s2λn–2 (–1)n A 0 (2)and on the same lines µI – Adj · A µn s1µn–1 s2µn – 2 sn – 1 µ (–1)n Adj · A (3)rwhere sr (r 1, 2, , n – 1) is (–1) times the sum of the r-square principal minors of Adj ·A.Thus by the property adj A A n–1 and definition of srwe have s1 ( 1)n sn 1n s2 ( 1) A sn 2, M Mnsn 1 ( 1) A s1; then µI – adj · A (–1)n {(–1)n µn sn– 1 µn – 1 sn – (4)2µn – 2 A s2 A n–3µ2 s1 A n –2µ A n–1}
56Engineering Mathematics through Applicationsn 1n µ µ n µ ( 1) 1 s1 sn 1 ( 1) A f (µ ) A A A nNow()() A n 11 ( 1) 1 s1 sn 1f λλλ and by equation (2), we haven 1 ( 1){n( ) A 1λ (5)n (6)} A nnnn 1λn f ( 1) λ s1 λ sn 1 λ ( 1) A 0λ Hence, A is a characteristic root of adjoint A.λTheorem 3: Eigen values (characteristic roots) of orthogonal matrix A are of absolute value 1.Proof: Let λi, Xi be characteristic roots and associated (characteristic vectors) invariant vectorsof an orthogonal matrix A, thenXi Xi X1 (A A) Xi (AXi) (AXi), since for orthogonal A, A A IXi Xi (λiXi) (λiXi) (λi Xc ) (λiXi) λiλi Xi Xi or(1 – λiλi) Xi Xi 0 impliesThus(1 – λiλi) 0, since xi xi 0 λi 1.Theorem 4: Prove if λi 1 is a characteristic root and Xi is the associated invariant vectorof an orthogonal matrix A, then Xi Xi 0.Proof: For characteristic value λi and corresponding characteristic vector Xi of the orthogonalmatrix A, we haveXi Xi Xi (A A) Xi (AXi) (AXi), (as A is given orthogonal) Xi Xi (λiXi) (λiXi) λiλiXi Xi, (1 – λiλi)Xi Xi 0 Using the transformation, AXi λiXiEither (1 – λiλi) 0 or Xi Xi 0 ButHenceλi 1Xi Xi 0.Theorem 5: For a symmetrical square matrix, show that the eigen vectors corresponding totwo unequal eigen values are orthogonal.[NIT Kurukshetra, 2004; KUK, 2004, 2006 ]Proof: Let A be any symmetric matrix i.e., A A and λ1 and λ2 two unequal eigen values,i.e., λ1 λ2Let X1 and X2 be the two corresponding eigen vectors.Now for λ1, (A – λ1I) X1 0
Matrices and Their ApplicationsorAX1 λ1X157 (i)SimilarlyAX2 λ2X2 (ii)Taking the transpose of (ii), we get(AX2) (λ2X2) X2́ A λ2X2 X2 A λ2X2 (as λ2 is an arbitrary constant)(Since A A)X2 AX1 λ2X2 X1X2 (λ1X1) λ2X2 X1(As AX1 λ1X1)λ1X2 X1 λ2X2 X1(λ1 – λ2) X2́ X1 0 IfBut λ1 – λ2 0X2́ X1 0 x1 y1 X1 x2 and X y 22 x y 3 3 x1 XXyyy2 1 1 2 3 x2 y1x1 y2 y2 y3 y3 x3 Clearly, (y1x1 y2x2 y3x3) 0This means, the two system of co-ordinates are orthogonal. Hence the transformation is an orthogonal transformation.2 – 3 – 21 – 6 Example 33: Determine the eigen values and eigen vectors of A 2 – 1 – 20 [NIT Kurukshetra, 2008]Solution: The characteristic equation, 2 λ2 32 1 λ 6 0 1 2 λ or λ3 λ2 –21λ – 45 0The roots of above equation are 5, –3, –3.Putting λ 5, the equations to be solved for x1, x2, x3 are [A – λI]x 0i.e.–7x 2y – 3z 0, 2x – 4y – 6z 0, –x – 2y – 5z 0.Note that third equation is dependent on first two i.e. R1 2R2 ¾ R3Solving them, we get x k, y 2k, z –kSimilarly for λ –3, the equations arex 2y – 3z 0, 2x 4y – 6z 0, –x – 2y 3z 0Second and third equations are derived from the first. Therefore, only one equation isindependent in this case.
58Engineering Mathematics through ApplicationsTaking z 0, y 1, we get x –2. Again taking y 0, z 1, we get x 3. Two linearlyindependent eigen vectors are (–2, 1, 0) and (3, 0, 1). A linear combination of these viz.(–2 3k, 1, k) is also an eigen vector. 6 –22 3 – 1 .Example 34: Find Eigen values and Eigen vectors for A – 2 2 –13 Solution: The characteristic equation, A – λI 0 6 λ 22 2 3 λ 1 0 1 3 λ2–λ3 12λ2 – 36λ 32 0, λ 2, 2, 8 are the characteristic roots (latent roots).Considering [A – 8I]X 0, we may show that there exists only one linearly independentsolution 2 1 1 so that every non-zero multiple of the same is a characteristic vector for the characteristicroot 8.For the characteristic root 2, we have[A – 2I]X 0 or 4 2 2 4x – 2y 2z 0 (i)–2x y – z 0 (ii) 22 x 1 1 y 0 11 z 2x – y z 0 (iii)which are equivalent to a single equation.Thus we obtain two linearly independent solutions, may take as 1 1 0 and 2 2 0 The sub-space of V2 possessed by these two vectors is the characteristic space for the root 2.ASSIGNMENT 21. The characteristic roots of A and A are the same.2. The characteristic roots of A and A are the conjugates of the characteristic roots of A.
Matrices and Their Applications593. If λ1, λ2, , λn are the characteristic roots of an n-square matrix A and if k is a scalar,then kλ1, kλ2, , kλn are the characteristic roots of kA.4. If A is a square matrix, show that the latent roots of ‘A’ are identical.1.9 LINEAR TRANSFORMATIONS AND ORTHOGONAL TRANSFORMATIONSI. Linear TransformationsLet P be a point with co-ordinates (x, y) to a set of rectangular axes in the plane-xy. If we takeanother set of rectangular axes inclined to the former at an angle θ, then the new coordinates(x , y ) referred to the new system (see the geometry) are related with x and y byx ON ON NN ( x cos θ y sin θ) y MP M P M M ( x sin θ y cos θ) (1)A more general transformation than (1) will be obtained when the new axes are rotatedthrough different angles θ and φ, and then angle does not remain a right angle.So, the most general linear transformation in two dimensions isx a1x b1y y a2 x b2 y (2)Expressed in matrix notation, thus x a1 b1 x y a2 b2 y More precisely, Y AX, where X is transformed into Y.More general, the relation Y AX,Y'Y a1 b1 k1 y1 x1 a b k2 y x , X 2 Y 2 , A 2 2 :: P yn xn an bn kn gives a linear transformation in n dimensions.y'This transformation is linear because theyrelations A (X1 X2) AX1 AX2 and A(bX) N N bAX, hold for transformation.x MθM'If the determinant value of the transformationOLmatrix is zero, i.e. A 0, the transformationxis termed as ‘Singular-transformation’,Fig. 1.1otherwise, ‘non-singular’.Non-singular transformation is also called ‘regular-transformation’. (3)X'XCorollary: If Y AX denotes the transformation of (x1, x2, x3) to (y1, y2, y3) and Z BYdenotes the transformation from (y1, y2, y3) to (z1, z2, z3), thus follows:Z BY B(AX) BAX.If 2 1 0 1 1 1 A 0 1 2 , B 1 2 3 1 2 1 3 5 1
60Engineering Mathematics through Applicationsthen the transformation of (x1, x2, x3) to (z1, z2, z3) is given by Z (BA)X, where 1 1 1 2 1 0 1 4 1 BA 1 2 3 0 1 2 1 9 1 1 3 5 1 21 7 12 1 Observations: It is seen that every square matrix defines a linear transformation. Further more, it is possibleto write the inverse transformation X A–1Y for only non-singular matrix A.II. Orthogonal TransformationsA transformation from one set of rectangular coordinates to another set of rectangularcoordinates is called an ‘orthogonal transformation’ or in other words, the lineartransformation Y AX is said to be orthogonal, if matrix A is orthogonal, i.e. AA' I A'A.Thus, an important property of this transformation is carried out only if transformationmatrix is orthogonal or vice versa.We have x1 x X X x1 x2 xn 2 x12 x22 xn2: xn y1 y Similarly,Y Y y1 y2 yn 2 y12 y22 yn2: yn If Y AX is an orthogonal transformation, thenX'X Y'Y (AX)' AX X'A'AX X'(A'A)Xwhich is possible only if A'A I AA' and A–1 A'.Hence a square matrix ‘A’ is said to be orthogonal if AA A'A and A–1 A'.Observations:(i)(ii)(iii)(iv)(v)A linear transformation preserves length if and only if its matrix is orthogonal.The column vectors (row vectors) of an orthogonal matrix are mutually orthogonal unit vectors.The product of two or more orthogonal matrices is orthogonal.The determinant of an orthogonal matrix is 1.If the real n-square matrix A is orthogonal, its column vector (row-vectors) are an orthogonal basis ofVn R (n-dimensional vector space in field of real) and conversely.α, η x sin α y cos α, write the matrix A of transformationExample 35: If ξ x cos α – y sinαand prove that A–1 A'. Hence write the inverse transformation.}ξ x cos α y sin αSolution: Given η x sin α y cos α (1)We can write the above system of equations in matrix notation as: ξ cos α sin α x η sin α cos α y (2)
Matrices and Their Applications61cos α sin α ξ x or more precisely, Y AX, where Y , A and X , representing η sin αcos α y linear transformation with A as the matrix of transformation.Now,Find, cos α sin α A' sin α cos α (3) cos α sin α cos α sin α 1 0 AA 1cos α sin α cos α 0 1 sin αAlsoA'A I. Hence A is an orthogonal matrix.But if A is an orthogonal, then A' A–1.Thus, for the transformation Y AX, we can write the inverse transformation cos α sin α X A–1Y, where A 1 A . sin α cos α 2Example 36: Is the matrix 4 – 3 – 3 1 3 1 orthogonal? If not, can it be converted into an1 9 orthogonal matrix?[KUK, 2005]Solution: Let the given matrix be A. Then to check its orthogonality, find AA'Thus 2 3 1 2 4 3 3 1 3 31 AA 4 3 1 9 1 1 9 8 9 1 6 3 9 14 0 0 4 9 1 8 9 116 9 1 12 3 9 0 26 0 9 1 81 0 0 91 6 3 9 12 3 9As AA' I, hence A is not an orthogonal matrix.However, it can be made an orthogonal by nromalization, i.e. on dividing every elementof a row by the square root of the sum of squares of each element of the respective row sothat product of resultant matrix (normalization) with its transpose would be a unit matrix. Hence, the orthogonal form of the matrix A is 214426391 3143261911 14 1 .26 9 91
62Engineering Mathematics through Applications0 n l m 0 00 – 1 Example 37: Prove that is orthogonal, when l 2 , m 3 , n 6 .0 n l –m777 ––l0mn Solution: If we denote the given matrix by ‘A’ then it implies that (l, m, n) must have( 72 , 73 , 76 ) is their one of the values that makes A as an orthogonal matrix. In other words,deduce that AA' I is possible with l 236,m ,n .777n 0 l 0n m l m 0 00 1 m 0ln AA n l m 0 n 0 m l 00 m n l 0 0 1Now l2 m2 n2 0 nl ml mn lm mn nl 0100AA 222 nm ln lm nl ml nm 0 n m l222 ml nm ln 0 mn nl ml m n l For matrix A to be rothogonal, AA' 1i.e.nl ml – nm 0l2and m2 n2 1 (1) (2) l lFrom (1), we have, m n 1ll k , then (1 k )nmLetl 23l or m 1m32Again suppose k , then 3l 1 or n 3l n 3 (3) (4)Then using (4) in (2), we get949 2l2 m2 n2 l2 l2 9l2 l 1 44 Either l 27orl 2736Taking l 2 , we get m and n 777Hence with(l, m, n) ( 72 , 73 , 76 ) , A is orthogonal. (5)
Matrices and Their Applications63Theorem 1: Prove that both ‘the inverse and transpose’ of an orthogonal matrix are alsoorthogonal.Solution: As we know that for an orthogonal matrix say A,AA' I A'A and A' A–1LetA–1 BCase I: Then for B to be an orthogonal, we are to prove thatBB' B'B I BB' (A–1) (A–1) A–1(A')–1 A–1(A–1)–1 A–1A ISimilarly,B'B (A–1)' A–1 (A')–1A–1 (A–1)–1A–1 AA–1 IHence inverse of an orthogonal matrix is also an orthogonal.Case II: Let A' B. For B to be orthogonal, we need to prove thatBB I B B BB A'(A')' A'A I;AlsoB'B (A')'A' AA' IHence transpose of an orthogonal matrix is also orthogonal.Theorem 2: A linear transformation preserves length if and only if its matrix is orthogonal.Solution: Let Y1, Y2 be the respective images of X1, X2 under the linear transformationY AXSuppose A is orthogonal, then AA' I A'ANow,Y1 · Y2 Y1'Y2 (AX1)'(AX2) X1 (A A)X2 X1 · X2 inner product.Hence the transformation preserves length.For vice versa, suppose lengths (i.e., inner products) are preserved.Then,Y1 · Y2 Y1' Y2 (AX1) (AX2) X1 (A A) X2But, Y1 · Y2 X1 · X2 (given) i.e., X1 (A A)X2 must be equal to X1 · X2 which is only possiblewhenA A IHence A is orthogonal.2 1 33 1For example, the linear transformation Y AX 233 22 33 is orthogonal.a 2b 2c 2a b 2cThe image of X [a b c]' is Y 33 3 3 3 3and both vectors are of lengtha2 b2 c2 .2 3 2 X31 3 2a 2b c 33 3
64Engineering Mathematics through Applications a b c Example 38: Given that A b c a , where a, b, c are roots of x3 x2 k 0 (k is a c a b constant). Prove that A is orthogonal.Solution: a, b, c are the roots of the cubic x3 x2 k 0 impliesS1 Sum of the roots taken one at a timeco–eff. of x2 –1co–eff. of x3a b c ( 1) (1)S2 Sum of the roots taken two at a timeab bc ca ( 1)2co–eff. of x 0co–eff. of x3S3 Sum of the roots taken three at a timeabc ( 1)3 (2) (3)constant term kco–efficient of x3Now, to check whether A is orthogonal, find the product AA Here a b c a b c AA b c a b c a c a b c a b a2 b2 c2 ab bc ca ca ab bc ab bc ca b2 c2 a2 bc ca ab 222 ca ab bc bc ca ab c a b (4)On using the values of S1 and S2, i.e. a b c –1 and ab bc ca 0we see that (a b c)2 (a2 b2 c2) 2(ab bc ca) results in a2 b2 c2 1. (5) 1 0 0 On using (1), (2), (3) and (5) AA 0 1 0 1 0 0 1 Hence with a, b, c as the roots of the given cubic, the matrix A is an orthogonal. l1 m1 n1 Example 39: If l2 m2 n2 defines an orthogonal transformation, then show that l m n 33 3lilj mimj ninj 0(i j); 1(i j); i, j 1, 2, 3.
Matrices and Their Applications65Solution: We know that for an orthogonal matrix A, AA I A A and A A–1 l1AA l2 l 3 m1m2m3n1 l1 l2n2 m1 m2n3 n1 n2l3 m3 , for given A.n3 l12 m12 n12 l1l2 m1m2 n1n2 l22 m22 n22 l2 l1 m2 m1 n2 n1 l3l1 m3m1 n3 n1 l3l2 m3 m2 n3n2l1l3 m1m3 n1n3 l2 l3 m2 m3 n2 n3 l32 m32 n32 For A to be an orthogonal, AA I which is possible only if,(l12 m12 n12) (l22 m22 n22) (l32 m32 n32) 1and(l1l2 m1m2 n1n2) (l2l3 m2m3 n2n3) (l3l1 m3m1 n3n1) 0.ASSIGNMENT 31. Prove that the product of two orthogonal matrix is orthogonal. cos θ 0 sin θ 10 is an orthogonal matrix.2. Prove that the matrix 0 sin θ 0 cos θ a b c 3. Given that A b c a , where a, b, c are the roots of x3 x2 k 0 c a b (where k is a constant). Prove that ‘A’ is orthogonal.4. Show that the modulus of an orthogonal transformation is either 1 or –1.[Hint: Since AA' I, then A A' 1 ]1.10 DIAGONALISATION OF MATRICES, THEIR QUADRATIC AND CANONICAL FORMS1. Diagonalization: If a square matrix A of order n has n linearly independent eigenvalues, then a matrix P can be found such that P–1AP, called a matrix of transformation.We prove this theorem for a square matrix of order n 3 as follows:Let λ1, λ2, λ3 be the three eigen values of the square matrix A. Let X1, X2, X3 be the x1 x2 x3 corresponding eigen vectors, where X1 y1 , X2 y2 , X3 y3 z z z 1 2 3 Let a square matrix whose elements are three column matrices X1, X2, X3 be denotedby P or more precisely,then x1 x2P X1 X2 X3 y1 y2 z z 1 2AP A[X1 X2 X3] [AX1x3 y3 .z3 AX2 AX3] [λ1X1 λ2X2 λ3X3]
66Engineering Mathematics through Applicationsλ3 x3 x1 x2 x3 λ1 0 0 λ3 y3 y1 y2 y3 0 λ2 0 λ3 z3 z1 z2 z3 0 0 λ3 PD, where D is the diagonal matrix such that P–1 AP D.The resulting diagonal matrix D, contains the eigen values on its diagonal.This transformation of a square matrix A by a non-singular matrix P to P–1AP istermed as Similarity Transformation. The matrix P which diagonalizes thetransformation matrix A is called the Modal Matrix and the matrix D, so obtained bythe process of diagonalization is termed as Spectral Matrix. λ1x1 λ1y1 λ z 11λ2 x2λ2 y2λ2 z2Observations: The diagonalizing matrix for matrix An n may contain complex elements because thezeros of the characteristics equation of An n will be either real or in conjugate pairs. Further, diagonalizing matrix is not unique because its form depends on the order in which the eigen values of An n aretaken.2. Quadratic Forms: A homogeneous expression of second degree in several variables iscalled
58 Engineering Mathematics through Applications Taking z 0, y 1, we get x –2. Again taking y 0, z 1, we get x 3. Two linearly independent eigen vectors are (–2, 1, 0) and (3, 0, 1). A linear combination of these viz. (–2 3k, 1, k) is also an eigen vector. Example 34: Find Eigen values and Eigen vectors for
3.1 EGR 101: Introductory Mathematics for Engineering Applications The WSU model begins with the development of EGR 101, "Introductory Mathematics for Engineering Applications," a novel freshman-level engineering mathematics course. The goal of EGR 101 is to address only the salient mathematics topics actually used in the primary core
IBDP MATHEMATICS: ANALYSIS AND APPROACHES SYLLABUS SL 1.1 11 General SL 1.2 11 Mathematics SL 1.3 11 Mathematics SL 1.4 11 General 11 Mathematics 12 General SL 1.5 11 Mathematics SL 1.6 11 Mathematic12 Specialist SL 1.7 11 Mathematic* Not change of base SL 1.8 11 Mathematics SL 1.9 11 Mathematics AHL 1.10 11 Mathematic* only partially AHL 1.11 Not covered AHL 1.12 11 Mathematics AHL 1.13 12 .
as HSC Year courses: (in increasing order of difficulty) Mathematics General 1 (CEC), Mathematics General 2, Mathematics (‘2 Unit’), Mathematics Extension 1, and Mathematics Extension 2. Students of the two Mathematics General pathways study the preliminary course, Preliminary Mathematics General, followed by either the HSC Mathematics .
2. 3-4 Philosophy of Mathematics 1. Ontology of mathematics 2. Epistemology of mathematics 3. Axiology of mathematics 3. 5-6 The Foundation of Mathematics 1. Ontological foundation of mathematics 2. Epistemological foundation of mathematics 4. 7-8 Ideology of Mathematics Education 1. Industrial Trainer 2. Technological Pragmatics 3.
Materials Science and Engineering, Mechanical Engineering, Production Engineering, Chemical Engineering, Textile Engineering, Nuclear Engineering, Electrical Engineering, Civil Engineering, other related Engineering discipline Energy Resources Engineering (ERE) The students’ academic background should be: Mechanical Power Engineering, Energy .
Advanced Engineering Mathematics Dr. Elisabeth Brown c 2019 1. Mathematics 2of37 Fundamentals of Engineering (FE) Other Disciplines Computer-Based Test (CBT) Exam Specifications. Mathematics 3of37 1. What is the value of x in the equation given by log 3 2x 4 log 3 x2 1? (a) 10 (b) 1(c)3(d)5 E. Brown . Mathematics 4of37 2. Consider the sets X and Y given by X {5, 7,9} and Y { ,} and the .
2. Further mathematics is designed for students with an enthusiasm for mathematics, many of whom will go on to degrees in mathematics, engineering, the sciences and economics. 3. The qualification is both deeper and broader than A level mathematics. AS and A level further mathematics build from GCSE level and AS and A level mathematics.
1) Grewal B.S. - Higher Engineering Mathematics, 40/e, Khanna Publishers. 2) Kreyszig E.K. - Advanced Engineering Mathematics, John Wiley. 3) Ramana B.V. - Higher Engineering Mathematics, (TMH) 4) Singh R.R. & Bhatt M. - Engineering Mathematics, (TMH) I A 2 ENGINEERING PHYSICS Aim : To enabl
