Ma 7 Review for Systems of DEs Matrices Basic Properties Addition and subtraction: Let A a ij mn and B b ij mn.then A B a ij b ij mn 3 A 6 B 6 4 7 6 A B 6 4 3 7 6 6 7 3 Scaler Multiplication: Let k be a scalar and A a matrix of real numbers of order m n. Then ka k a ij mn 5 7 5 5 35 5 8 4 7 6 4 35 5 3 8 3 3 4 4 5 5 Some Properties of Addition and Scalar Multiplication Theorem Let A, B and C be conformable m n matrices whose entries are real numbers, and k and p arbitrary scalars. Then. A B B A.. A B C A B C 3. There is an m n matrixsuchthat A A for each A. 4. For each A there is an m n matrix A such that A A.
5. ka B ka kb 6. k pa ka pa 7. kpa kpa. 4 Note that A a ij mn A A mn Remark: We denote A by A. The Transpose of a Matrix If A is an m n matrix, the transpose of A, denoted A T,isthen m matrix whose entry a st is the same as the entry a ts in the matrix A. Thus one gets the transpose of A by interchanging the rows and the columns of A. T 4 3 3 4 9 9 Multiplication: Definition. Let A a ij mn and B b ij np be matrices. Then ABis the m p matrix C, where C c ij mp n a ik b kj k mp Remark. AB BAnecessarily. 4 3 3 4 5 3 4 5 43 44 5 4 9 9 The following occur often for matrices. 3. AB BA. AB but neither A orb
3. AB AC but B C Theorem Assume that k is an arbitrary scalar, and that A, B, C and I are matrices of sizes such that the indicated operations can be performed. Then. IA A, BI B. ABC ABC 3. AB C AB AC, AB C AB AC 4. B CA BA CA, B CA BA CA 5. kab kab AkB 6. AB T B T A T. Cramer s Rule Cramer s Rule: Let A be an n n matrix, A a ij nn and denote by A j the n n matrix formed by replacing the elements a ij of the jth column of A by the numbers k i, i,...,n. If A, the system of n linear equations in n unknowns, a x a x a n x n k a x a x a n x n k a n x a n x a nn x n k n has the unique solution x deta deta x deta deta,...x n deta n deta Example. Solve x 3y z 4x y z 5 3x 4y z 3 3
by Cramer s Rule deta 3 4 3 4 5 x 3 5 3 4 5 4 5, y 4 5 3 3 5 Systems of Equations: Elimination Using Matrices 9 5, z 3 4 5 3 4 3 5 9 5 Elementary Row Operations On Matrices I Equivalent Systems Two linear systems are equivalent if they have the same solutions. Three Elementary Operations Three basic elementary operations are used to transform systems to equivalent systems. These are:. Interchanging the order of the equations in the system.. Multiplying any equation by a nonzero constant. 3. Replacing any equation in the system by its sum with a nonzero constant multiple of any other equation in the system (elimination step). Theorem: Suppose that an elementary row operation is performed on a system of linear equations. Then the resulting system has the same set of solutions as the original, so the two systems are equivalent. Operating on the rows of a matrix is equivalent to operating on equations. The row operations that are allowed are the same as the row operations on linear systems of equations:. Interchanging the rows.. Multiplying any row by a nonzero constant. 3. Replacing any row by its sum with a nonzero constant multiple of any other row. (Add a multiple of one row to a different row.) Gaussian Elimination Definition: A matrix is said to be in row-echelon form (and will be called a row-echelon matrix) if it satisfies the following three conditions:. All zero rows (consisting entirely of zeroes) are at the bottom. 4
. The first nonzero entry from the left in each nonzero row is a, called the leading for that row. 3. Each leading is to the right of all leading s in the rows above it. Definition: A row-echelon matrix is said to be in reduced row-echelon form (and will be called a reduced row-echelon matrix) if it satisfies the following condition: 4. Each leading is the only nonzero entry in its column. Reduce the matrix 4 3 7 8 to row-reduced echelon form. 4 3 7 8 RR3; RR4 4 3 4 8 3 R; R3 4 3 4 8 3 R3 R 4 4 8 3 3 R3R4 4 4 8 3 4 RR 4 8 3 Solve the system AX C, where 5
3 A, X 7 From the previous example 4 x x x 3 x 3 3, row echelon form: 7 8 4 and C 8 4 4 8 3 Thus the solution is: x 3 3x 4,x x 4 4,x 4x 4 8,x 4 x 4 Inverse of a Matrix Definition: If A is a square n n matrix, a matrix A is called the inverse of A if and only if AA I A A A matrix A that has an inverse is called an invertible or nonsingular matrix. Find A for A 7 4 3.Weform 7 4 3 7 4 3 RR; RR3 4 3 RR3; 4 3 4RR; R 3 4 7 3 R; R R 3 3 3 3R3R; R3R 3 3 3 6
Thus A 3 3 3 Eigenvalues Definition: The values of such that deta I are called eigenvalues. The vector X corresponding to an eigenvalue is called an eigenvector of the matrix A. Example. Find all eigenvalues and eigenvectors of the matrix A Solution:. deta I det 3 Thus deta I eigenvalues,,. A IX Thus x,x x 3 or x x x 3 x x x 3 x x x 3 x x x 3 x 3x x 3 x x x 3. Similarly, 3 A, eigenvectors: 3,, Matrix Methods for Linear Systems of Differential Equations 7
Linear Systems in Normal Form Asystemofn linear differential equations is in normal form if it is expressed as x t Atxt ft where xt and ft are n column vectors and At a ij t nn. A system is called homogeneous if ft ; otherwise it is called nonhomogeneous. When the elements of A are constants, the system is said to have constant coefficients. Express the equation in normal form Solution: Defining we have y 6y y 6ty cost x t Atxt ft x y,x y,x 3 y x x Thus x x 3 x 3 6tx x 6x 3 cost x x x At and ft x 3 6t 6 cost Solving Normal Systems. To determine a general solution to the n n homogeneous system x Ax : a. Find a fundamental solution set x,...,x n that consists of n linearly independent solutions to the homogeneous equation. b. Form the linear combination x Xc c x c n x n where c colc,...,c n is any constant vector and X x,...,x n is the fundamental matrix, to obtain a general solution. Theorem Suppose the n n constant matrix A has n linearly independent eigenvectors u,u,...,u n.letr i be the eigenvalue corresponding to the u i.then 8
e rt u,e rt u,...,e rnt u n is a fundamental solution set on, for the homogeneous system x Ax. Hence the general solution of x Ax is xt c e rt u c n e rnt u n where c,...,c n are arbitrary constants. Example Find a general solution of x 5 4 x 5 4, eigenvectors:, 4 4 Thus xt c e t Thus the solution is c e 4t 4 x t c e t 4c e 4t x t c e t c e 4t Find a fundamental matrix for the system x t 3 7 xt Solution: 9
3 7, eigenvectors: 3,, 8 7, 3 Hence the four linearly independent solutions are e t 3,e t,e 7t 8,e 3t Therefore a fundamental matrix is e t e t e 7t e 3t 3e t e 7t e 7t e 3t 8e 7t Solve the initial value problem x t 3 7 xt x We know from above that the solution general solution to the system is
xt c e t 3 c e t c 3 e 7t c 4 e 3t 8 c e t c e t c 3 e 7t c 4 e 3t xt 3c e t c 3 e 7t c 3 e 7t c 4 e 3t 8c 3 e 7t c c c 3 c 4 Then x 3c c 3 c 3 c 4. Therefore we must solve the system 8c 3 c c c 3 c 4 3c c 3 c 3 c 4 8c 3, Solution is: c 3,c 4,c 3,c 3,and xt 3 e t 3 et e 3t e t e 3t Complex Eigenvalues Consider x t Axt in the case where A is a real matrix and the eigenvalues are complex. Denoting the eigenvalues by i, letz a ib, wherea and b are real vectors, be an eigenvector corresponding to the eigenvector r.then x t e t costa sintb x t e t sin ta costb are two real lineraly independent solutions of the system. Find the general solution of x t 4 xt Solution: This is problem on page 573 of our DEs text and was assigned for homework. Eigenvalues:
deta ri Eigenvectors: r i : i 4 i r 4 r u u R says iu 4u u then u s r 4 r i i, so, i s s i 4 i 4 i u i u. Let u s; is. Let s : u i a ib. So the general solution is xt c e t cost e t sin t c e t sint e t cost c cost cost sin tt c sint sint cost Nonhomogeneous Systems Undetermined Coefficients Consider the nonhomogeneous constant coefficient system Find the general solution of x t Axt ft e t x t xt 4e t e t Solution: We first find the homogeneous solution., eigenvectors: 3,, 3 Since these eigenvectors are linearly independent, then
x h t c e 3t c e 3t c 3 e 3t We seek a particular solution of the form x p t e t a a a 3 Then x p t e t a a Ax p t e t a a e t 4e t a 3 a 3 e t e t a a a 3 a a a 3 a a a 3 4 Thus a a a a 3 a a a a 3 4 a 3 a a a 3, Solution is: a,a,a 3 Therefore and x p t e t xt x h t x p t c e 3t c e 3t c 3 e 3t e t Note: the Method of Undetermined Coefficients works only for constant coefficient systems. The Matrix Exponential Function Definition: Let A be a constant n n matrix. The we define 3
e At I At A t! tn An n! A n n! tn n This is an n n matrix. Remark: If D is a diagonal matrix, then the computation of e At is straightforward. Example Let D.Then D 4, D 3 8,... D n n n Therefore e Dt D n n! tn n n n! tn n n n tn n! e t e t In general if D is an n n diagonal matrix with r,r,...,r n down its main diagonal, then e Dt is the diagonal matrix with e rt,e rt,...,e rnt down its main diagonal. In general it is not easy to calculate e At for any matrix A. Remark: In one case it is relatively easy to find e At. It can be shown that if a matrix A has n linearly independent eigenvectors, then P AP is a diagonal matrix, where P is formed from the n linearly independent eigenvectors of A. Thus P AP D where D is a diagonal matrix. In fact, D has the eigenvalues of A along its diagonal. Now implies that when A has n linearly independent eigenvalues we have A PDP so that 4
e At e PDP t I PDP t PDP tpdp t I PDP t PDP PDP t I PDP t PD P t P I Dt Dt P Pe Dt P We saw above that A 3 7, has eigenvectors: 3,, 7, 3 8 Let P 3 8 Then P 3 4 3 3 8 4 so P AP 7 3 D Thus 5
e Dt e 7 3 t e Dt e t e t e 7t e 3t Hence e At Pe Dt P P e t e t e 7t e 3t P e t 3 e t 3 et e t e 3t 4 e t 3 et 8 e7t 4 e3t e t 8 e t 8 e7t e 3t 4 e7t 4 e3t e 7t Calculating e At for Nilpotent Matrices Definition: An nxn matrix A matrix is nilpotent if for some positive integer k A k. Since e At I At A t! tn An n! A n n! tn n we see that if A is nilpotent, then the infinite series has only a finite number of terms since A k A k and in this case e At I At A t! Ak tk k! This may be taken further. The Cayley-Hamilton Theorem says that a matrix satisfies its own characteristic equation, that is, pa. Therefore, if the characteristic polynomial for A has the form pr n r r n,thatisa has only one multiple eigenvalue r,thenpa n A r I n. Hence A r I is nilpotent and e At e ria rit e rit e A rit e rt n tn I A r It A r I n! 6
Example Find the fundamental matrix e At for the system x t Axt where A Solution: The characteristic polynomial for A is pr det r r r r 3 3r 3r r 3 Hence r isaneigenvalueofa with multiplicity 3. By the Cayley-Hamilton Theorem A I 3 and e At e t e A It e t I A It A I t! Thus A I e At e t te t and A I e t te t te t te t te t e t te t te t te t te t e t e t 7