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Unformatted text preview: 1 Operations on Matrices Definition: A matrix is a rectangular array of numbers arranged in rows and columns. An m x n matrix is a matrix with m rows and n columns and is usually denoted by a11
A= a12 a 22 a m2 a1n a 2n a mn
= a ij , I = 1,2 …m, j = 1,2 … n a 21 a m1 The entries could be complex numbers. 1 3 The matrices 1 0 and 24 1 4 01 are 3x2 and 2x3 matrices.` 2 3 A can be written in the form a ij where i = 1, 2 … m and j = 1, 2 … n A square matrix has the same number of rows and columns. A column matrix has only one column. A zero matrix has zeros for all its entries and is denoted by O. O= 0 0 0 0 An identity matrix has 1 in the main diagonal and zeros everywhere else and is denoted by I. 100 10 I= or I = 0 1 0 01 001 2 Algebra of Matrices Equality of Matrices Two matrices A and B are equal, written A=B, if they have the same dimensions and their corresponding entries are equal. Sum of Matrices Let A and B be matrices with the same dimensions. The sum of A and B, written A+B, is the matrix obtained by adding corresponding entries of A and B. Multiplication by a Scalar Let A be a matrix and c be a scalar. The scalar multiple cA is the matrix obtained by multiplying each entry of A by the scalar c. Examples 1.
1 12 Let A = 5 3 1
3 3 3A = 13 0 1 ,B= 24 23 12 47 2. Let A = A+B = Matrix multiplication Let A and B be matrices such that the number of columns of A match the number of rows of B. Suppose A is m x n and B is n x p matrix, the product AB is the m x p matrix C, defined by
n Cij =
k=1 aik bkj In other words, to obtain the entry in the ith row and jth column, we take the dot product of the ith row of A and the jth column of B. Examples 123 Let A = 3 2 1 19 13 17 23 Transpose of a Matrix AB = 4 5
B= 04 5 0 3 The transpose of the m x n matrix A, written AT or A , is the n x m matrix B whose jth column is jth row of A. Matrix Form of a Linear System A system of m linear equations in the n unknowns x1, x2 … xn can be written in the general form: a11x1 + a12 x2 + a21x1 + a22 x2 + a1nxn = b1 a2nxn = b2 am1x1 + am2x2 + … amnxn = bm This system can be written in the following matrix form: a11 a21 am1 a12 a22 am2 a1n a2n amn b1 b2 = xn bm x1 x2 AX = B Where A is a m x n matrix, X is a n x 1 matrix, and B is a m x 1 matrix. Example 1 0 2 1 x 3 2 3 4 0 1 y = 1 z 0 1 02
X= x y, z
B= 3 1 0 A= 2 3 4 0 1 1 4 The Inverse of a Matrix Let A be a square matrix. If there exists a matrix B such that AB = BA = I, then we say that A is invertible and that B is the inverse of A. We denote the inverse of A as A1. Procedure for finding the Inverse of a Matrix: By a row operation, we mean any one of the following: (a) (b) (c) Interchanging two rows of a matrix Multiplying a row of the matrix by a nonzero scalar Adding a scalar multiple of one row of the matrix to another row If the n x n matrix has an inverse, than A1 can be determined by performing row operations on the n x 2n matrix. [A:I] obtained by writing A and I side by side. In particular, we perform row operations on the matrix [A:I] until the first n rows and columns form the identity matrix; that is, the matrix is [I:B]. Then A1 = B. If this procedure fails to produce a matrix of the form [I:B], then A has no inverse. Example 1. Find the inverse of A= The matrix [A:I] = 3 I I 0 5 20 I 1 1/3  1/ 3 0 5 20 1 1 1/3  1/3 0 0 1/3  5/3 1 31 52 Divide first row by 3 Subtract 5 times the first row from the second row Multiply the second row by 3 1 1/3  1/3 0 0 1  5 3 Subtract 1/3 times the second row from the first row. A1 = 2 1 5 3 1 0  2 1 0 1 5 3 5 Example 2. Perform R2 R1 (Subtract first row from second row.) 1 1 11 0 0 1 1 1 1 0 0 1 2 3  0 1 0  0 1 2  1 1 0 0 1 10 0 1 0 1 1 0 0 1 Perform R3 R2 1 1 1 1 0 0 0 1 2  1 1 0 0 1  1 1
Perform R3 0 1 1 1 1 100 0 1 2  1 1 0 0 0 1  1 1 1 R1 R3 R2 2R3 1 1 0  2 1 1 0 1 0  1 1 2 0 0 1  1 1 1 Perform R1 – R2 1 0 0 1 0 1 R1 R2 0 1 0  1 1 2 0 0 1  1 1 1 1
A=
1 0 1 1 1 1 2 1 111 123 011
= 1 1 0 1 2 1 1 100 0 1 0 =I 001 A A= Example 3. 1 1 1 1 6 Find the inverse of sin2t cos2t 2cos2t 2cos2t R1/sin 2t sin 2t cos 2t  1 0 2cos2t 2cos2t 0 1 ~ 1 2cos2t cot2t  csc 2t 0 2sin 2t 0 1 R2  2cos 2t R1 1 cot 2t  csc 2t 0 2csc 2t  2cot 2t 0 1 R2/ 2csc 2t 1 cot 2t  csc 2t 0 0 1 cos 2t (1/2)sin(2t) R1  cot 2t R2 1 0  sin 2t (1/2)cos(2t) 0 1  cos2t (1/2)sin(2t) A1 = sin 2t cos2t sin 2t cos 2t (1/2) cos 2t (1/2) sin 2t (1/2) cos (2t) (1/2) sin (2t) sin 2t cos 2t 10 = 2cos 2t 2sin 2t 01 A1 A = 7 Example 4. 3 5 6  1 0 0 1 2 2  0 1 0 1 1 1  0 0 1 Interchange the first row with the third row. 1 1 1  0 0 1 1 2 2  0 1 0 3 5 6  1 0 0 R2 + R1 and R3 + 31R1 1 1 1  0 0 1 0 0 1 2 10 1 1 31 0 3 R3 – 2R2 1 1 0 0 1 0 1  0 10 11 0 1 2 1 1 1 R1 + R2 1 0 00 1 2 0 1 10 1 1 0 0 1  1 2 1 R2 – R3 1 0 0 0 1 2 0 1 0  1 3 0 0 0 1  1 2 1 8 012
B = 1 3 0 is the inverse of A 1 2 1 3 5 6
Check: AB = 1 2 2 1 1 1 012 1 3 0 1 2 1
= 100 010 001
=I BA = I also. B is denoted by A1 Example 5. Find the inverse of θ cosθ 0 θ sin cos A = sin 0 0 1 θ sinθ 0  0 0 0 θ cosθ I1 cos sin 0 0 10 0 1 Do r1/cosθ, R2 + sinθ R1 1 tan 0 sec 00 0  sec 0  tan 1 0 00 10 01 R2 / secθ 1 tan 01 00 0  sec 1 0 0 0 0 0 1 0  sin cos R1 – tanθ R2 1 0 0  cos 0 1 0  sin 0 0 1 0 sin cos 0 0 0 1 9 θ scosθ 0 inθ 0 cos
Check sec 0 sinθ sinθ 0 = 1 0 0 cosθ cos 01 0 0 1 sin 0 0 1 0 0 1 001 cos
B = A = sin 0 Examples: 1. A= 12 34
1 cos 0 AT = 13 24 12
2. B= 3 4 56 BT= 135 246 Properties of Matrices and Rules of Operation Let A, B, and C be the matrices and a and b be the scalars. Assume that the dimensions of the matrices are such that each operation is defined. Then a) b) c) d) e) f) g) h) i) A+B = B+A A+(B+C) = (A+B) +C A+0 = A = 0+A A+(A) = 0 A (BC) = (AB) C AI = A = IA a(B+C) = aB + aC (A+B)T = A T + B T (AB) T = B T A T 10 Exercises 1. Let A and B be 4x5 matrices and let C, D, and E be 5x2, 4x2, and 5x4 matrices, respectively. Determine which of the following expressions are defined. For those that are defined, give the size of the resulting matrix. (a) BA 2. (b) AC+D (c) AE+B (d) E(A+B) (e) E (AC) Solve the following matrix equation for x, y, z, and t xy y+z 3t +z 2x  4t = 81 76 3. Consider the matrices 30
A = 1 2 11 Compute (a) AB (b) D+E B= 4 1 0 2 C= 142 315 152 D= 1 0 1 413 E= 613 1 1 2 413 (c) DE (d) DE (e) ED (f) –7B (g) AT (h) BTAT 11 Determinants A is a square matrix, the determinant of A is denoted by A. Let A be 2x2
a1 b1 a2 b2 det A or A is a1b2a2b1 let A be 3x3 a11 a12 a13
A = a21 a22 a23 a31 a32 a33 a11 a22 a23 a32 a33 a21 a23 a31 a33
+ a13 a12 a21 a22 a31 a32 det (A11) = a22 a23 a32 a33 det A = a11 det A11 + a12 det A12 + a13 det A13 If A is n x n matrix
n det A =
ii ( 1) i+j aij det Aij if n>2 Evaluate the following determinants 1. 12 8 = 24 – 24 = 0 32 102
2. 0 3 1 = 1 (3+2) +2 (+3) = 5 + 6 = 11 1 2 1 12 3. Determine the value of r for which det (ArI) = 0 a) A= 33 24 33 10 r 24 01 3r 3 2 4r 3r 3 = (3r)(4r) – 6 = 12 –7r + r2 – 6= r2 – 7r+6 2 4r (ArI) = = det(ArI)= det(ArI) = 0 r2 – 7r + 6 = 0 (r6)(r1) = 0 r = 6, 1 011
b) A= 1 0 1 110 0r 1 1
ArI = 1 1 r 1 1 r r(r21) – 1(rI) + 1 (1+r) = 0 r(r + 1)(r1) + (r + 1) + (1 + r) = 0 (1+r)[r(r1)+2] = 0 (1+r)(r2 + r + 2) = 0 r = 1 (r2) (r+1) = 0 r = 1, 2, 1 13 Cramer’s Rule Solve the system a11x1 + a12x2 + a13 x3 = b1 a21x1 + a22x2 + a23 x3 = b2 a31x1 + a32x2 + a33 x3 = b3 Let a11 a12 a13
Δ = a21 a22 a23 a31 a32 a33 b1 a12 a13
x1 = b2 a22 a23 b3 a32 a33
Δ a11 b1 a13
x2 = a21 b2 a23 a31 b3 a33
Δ a11 a12 b1
x3 = a21 a22 b2 a31 a32 b3
Δ 14 Examples 1. Solve 2x1 + 6x2 + 8x3 = 16 4x1+ 15x2+ 19x3 = 38 2x1 + 3x3 = 6 2 6 8 Δ = 4 15 19 203 2(45) 6(12 – 38) + 8(030) 90 + 156 – 240 = 6 16 6 8 x1 = 38 15 19 = 16(45) – 6(114 – 114) + 8(90) = (720 – 720) /6=0 603 Δ 2
x2 = 4 2 16 8 38 19 = 2(114114) – 16(12 – 38) + 8(2476) = (416 – 416) /6=0 63
Δ 2
x3 = 4 2 6 15 0 16 12 38 = 2(90) – 6(24 – 76) + 16(030) = (90 + 384  480) = 6 6 2 15 sin 2t
2. x(t) = sin 2t cos2t cos 2t e2t e2t 2e2t 6e2t 2e2t 2cos 2t 3e2t 3sin 2t 2cos2t 2sin 2t x (t) = 2cos2t 4sin 2t 6 cos 2t 2sin 2t
3. Verify that the given vector function satisfies the given system
000 x= 010x 101
0 x(t) = et 3e t 0 x= et 3e t 000 010 101
Yes it does. 0 et 3e t 0 = et 3e t =x et 0 0
4. Verify that x(t) = 0 e t e5t 0 e t e5t satisfies the matrix differential equation 100
x = 0 3 2 x 0 23 et 0 0
x= 0 e t 5e5t e t 5e5t 0 16 10 0 et 0 0e 0
t 0 e
5t et 0
= 0 5e5t
=x 0 3 2 0 2 3
It satisfies 0e t e t e5t 0 e t 5e5t 5. Let A(t) = Find (a) 1 e2t 3 e2t B(t) = et e2t et e2t A(t) dt = t 3t et et e 2t /2 e2t /2 e2t / 2 e2t / 2 (b) (c) B(t) dt = d/dt [A(t) B(t)] = d/dt
et e3t 3et e3t e2t + e4t 3e2t + e4t 2e2t 4e4t 4e4t e2t + e4t 3e2t + e4t = et + 3e3t 3et +3e3t 6e2t et  e3t 3et  e3t et +e3t /3 3et + e3t /3 (d) A(t)B(t) dt = dt = e2t /2  e4t / 4 3e2t /2  e4t 17 Linear Algebraic Equations a11 x1 + a12 x2 + a13 x3 … + a1n xn = b1 a21 x1 + a22 x2 + a23 x3 … + a2n xn = b1 an1 x1 + an2 x2 + an3 x3 … + ann xn = bn Where aij’s and bi’s are given constants is called a linear system of algebraic equations in the unknown’s x1, x2, … xn. We will use GaussJordan elimination algorithm to solve the above system. The basic idea for this formulation is to use the first equation to eliminate x1 in all other equations; then use the second equation to eliminate x2 in all the others; and so on. If all goes well, the resulting system will be “uncoupled” and the values of the unknowns’ x1 , x 2 , ... x n will be apparent. Example Solve the system 2x1 + 6x 2 + 8x 3 = 16 4x1 + 15x 2 + 19x 3 = 38 2x1 + + 3x 3 = 6 Step 1: write the augmented matrix for the system 2 6 8 3 16 6 4 15 19 38 20
Step 2 (a) Obtain a leading 1 in the first row, first column. (If there is a 0 in this position, interchange the first row with a row below it so that a non zero entry appears there). Divide the first row by 2 1 2 348 036 4 15 19 38 (b) Obtain zeros in the other positions in the first column by adding appropriate multiples of the first row to the other rows. 18 Replace the second row of the sum of itself and –4 times the first. Replace the third row with the sum of itself and –2 times the first. 1 0 3 3 48 36 5 10 0 6
Step 3 (a): Obtain a leading 1 in the second row, second column (If there is a zero in this position, interchange the second row and the row below it so that a nonzero appears there. If this is not possible, go to the next column). Multiply the second row by 1/3. 1 0 3 1 4 1 8 2 0 6 5 10 (b): Obtain zeros in the second column by adding appropriate multiples of the second row to the other rows. Replace the third row with the sum of itself and 6 times the second. Replace the first row with the sum of itself and –3 times the second. 1 0 0
Step 4 0 1 0 12 12 12 (a): Obtain a leading 1 in the third row, third column (If there is a zero in this position, interchange the second row with the row below it so that a nonzero entry appears there. If this is not possible, go to the next column). There is a one in the third row, third column. (b):Obtain zeros in the third column by adding the appropriate multiples of the third row to the other rows. Replace the first row with the sum of itself and –1 times the third. Replace the second row with the sum of itself and –1 times the third. 1 0 00 0 0 1 0 00 12 19 Exercises 1. Let A (t) =
1 3 e2t e2t B (t) = et et et 3et Find the following a) A (t) b) B (t) c) d/dt A(t) B(t) d) A(t) B(t) dt 2. Verify that the following matrix function satisfies the given matrix differential equation.
100 x= 0 3 2 0 2 3 x et 0 0 x(t) = 0 e t e5t 0 e t e5t
3. Use Cramer rule to solve x1 + 2x 2 3 x3 5 2 x1 5 x2 3 x3 16 x1 8 x3 4 6 Solve the following system by GaussJordan method. x + 2y + 3z = 5 2x + 5y + 3z = 3 x + 8z = 17 20 Write the given system in the matrix form X ' = AX + f Example 1. x ' (f) = 2x(t) + sin t y ' (f) = x(t) – y(t) + 1 x '(t) y '(t) X' = 20 1 1 AX x(t) y(t) + f Example 2. dx/dt = x + y + z dy/dt = 2x – y + 3z dz/dt = x +5z sin t 1 dx/dt dy/dt dz/dt
x' = 111 = 2 1 3 1 05
Ax x y z Write the given system as a set of scalar equations Example 3
x' = 21 1 3 x + et t 1 x ' = 2x + y + te t y ' = x + 3y + e t 21 Wronskian Definition: The wronskian of n vector functions x1 col (x11 ,   x n (t) = col (x11 ,   x n1 ) is defined to be the realvalued function x11  x n1 W[x1 ,...x n ](t)= x 21  x 2n x n1  x nn If x1 , x 2 , ... x n are linearly independent then W[x1 ,...x n ] is never zero on ( , )
Examples Determine whether the following vector functions are linearly dependent or linearly independent , on the interval
1. s in t , cos t 1 0, 1 t 0, t s in 2 t cos 2t t2 0 t2 2. S o lu tio n s 1 . w [x 1 , x 2 ] s in t s in 2 t cos t cos 2t = s in t c o s t 2 t  c o s t s in 2 t = s in t ( 2 c o s 2 t  1 )  2 c o s t s in t c o s t =  s in t = 0 W hen t = 0 If t 0 , s in t 0 0 , x2 1 0 1 W hen t = 0 x 1 = S o w e h a v e o n ly o n e v e c to r fu n c tio n . So t 0 a n d x 1 , x 2 a r e lin e a rly in d e p e n d e n t 1 t t2 0 0 0 =0 1t t2 2 . w [x 1 , x 2 , x 1 ] N o t lin e a rly in d e p e n d e n t 22 Theorem 1: Let x1 , x 2 ... x n be n linearly independent solutions to homogeneous system
x' = A(t) x(t)  (1) on the interval I, where A(t) is an n x n matrix function continuous on I. Then every solution to (1) can be expressed in the form x(t) = c1 x1 (t) + c2 x 2 (t) +  cn x n (t) where c1 , c 2 ,  c n are constants. Definition: A set of solutions { x1 , ... x n } that are linearly independent on I or equivalently, whose Wronskian does not vanish on I, is called a fundamental solution set for (1) on I. Theorem 2: Let xp be a particular solution to the nonhomogeneous system x ' (t) = A(t) x(t) +f(t)     on the interval I, and let { x1 , x 2 ... x n } be a fundamental solution set on I for the corresponding homogeneous system x ' (t) = A(t) x(t). Then every solution to on I can be expressed in the form
x(t) = x p (t) + c1 x1 (t) + c 2 x 2 (t) +  c n x n (t) where c1 , c 2 ,  c n are constants. Example Verify that the vector functions e3t x1 0 e3t , x2 e3t e3t , x 3 0 e3t e3t e3t are solutions to the homogeneous system 1 x ' = Ax = 2 2
on ( , ), and that 2 1 2 2 2x 1 23 5t + 1 xp 2t 4t + 2
is a particular solution to x ' = Ax +f(t), where 9t f(t) 0 18t
Find the general solution to x ' = Ax + f(t) Solution Let us check if x1 , x 2 ,x 3 are solutions to the homogeneous system x ' = Ax. 3e3t x1 ' 0 3e3t , x '2 3e3t 3e3t , x '3 0 3e3t 3e3t 3e3t x1 is a solution to the system if x1 ' = Ax1 x2 is a solution to the system if x2 ' = Ax2 x3 is a solution to the system if x3 ' = Ax3
1 2 1 2 2 2 1 e3t 0 e3t e3t 0 e3t x '1 Ax1 = 2 2 So x1 is a solution to the system x1 ' = Ax 1
Ax2 = 2 2 2 1 2 2 2 1 e3t 0 3e3t = x '2 0 e3t = 3e3t So x2 is a solution to the system x2 ' = Ax 1
Ax3 = 2 2 2 1 2 2 2 1 e3t e3t 3e3t 3e3t e3t = 3e3t =x '3 24 So x3 is a solution to the system x3 ' = Ax
1 AX p 2 2 22 1 2 2 1 5t 2t 4t 2 1 5t 1 4t 8t 4 10t 2 2t 8t 4 10t 2 4t 4t 2 5 X 'p 2 4 9t 5 9t AX p f (t ) 5 2 4
So Xp is a particular solution to X ' AX f (t ) 9t 5 2 18t 4 20 18t 4 18t So the general solution to x ' = Ax is
x(t) = x p + c1 x1 c2 x 2 c3 x 3 25 1. Exercises Write the following system in the matrix form. x = Ax x (t) = 3x – 5y y (t) = 4x + 8y Write the following system in the form x = Ax+f dx/dt = 3x + 4y + et sin 2t dy/dt = 5x + 9y + 4et cos 2t Determine whether the following vector functions in exercises 3 and 4 are linearly dependent or linearly independent on the interval ( , ) 1 3 et , et 5 15 2. 3. 1
4. 1 e
2t 0
2t e 2t 0, 5 1 ,e 1 1 0 5. Verify that the vector functions et et x1 and x2 = et 3e t Are solutions to the homogeneous system 21 x AX x 32 On –(
, ) and that xp 3 tet 2 tet 1 et 4 3et t 2t 0 1
AX +f(t) where f(t) = col Is a particular solution to the nonhomogeneous system x ' et , t . Find a general solution to x AX f (t ) . 6. Verify that the given vector functions form a fundamental solution set to the system x' t Ax t
e x1 2e e
t t t et x2 0 x3 et e 3t e 3t 2e3t 26 Approach to Solving Normal Systems 1. To obtain a general solution to the n x n homogeneous system a) Find a fundamental solution set {x1,  xn} that consists of n linearly independent solutions to the homogeneous system. b) A general solution is x = xc = c1x1 +  + cnxn where c = col(c1,  cn) is a constant vector and x = [x1,  xn] is the fundamental matrix whose columns are the vectors in the fundamental solution set 2. To obtain a general solution to the nonhomogeneous system x = Ax + f a) Find a particular solution xp is the nonhomogeneous system. b) A general solution to the nonhomogeneous system is x = xp + xc where xc = c1x1 +  + cn xn Homogeneous Linear Systems with Constant Coefficients X (t) = AX(t) where A is a real constant n x n matrix. The general solution we seek will be defined for all t because the elements of A are just constant functions, which are continuous on ( , ) let x(t) = er t u then r e r t u = A e r t u (A  r I )u = 0 where r I denotes the diagonal matrix with r’s along its main diagonal. x(t) = ert u is a solution to if r and u satisfy . We Since u = 0 is the trivial case it is of no help in finding linearly independent solutions to equate u 0, such vectors are given a special name. They are called eigen vectors. 27 Eigen Values and Eigen Vectors Definition: Let A = [aij] be a n x n constant matrix. The eigen values of A are those (real or complex) numbers r for which (Ar I) u = 0 has at least one nontrivial solution u. The corresponding nontrivial solutions u are called the eigen vectors of A associated with r. We know that a linear homogeneous system of n algebraic equations in n unknowns has a nontrivial solution if the determinant of its coefficients is zero. Hence a necessary and sufficient condition for (A–r I) u = 0 to have a nontrivial solution is that  A – r I  = 0 Expanding the determinate in , we find that it is the nth degree polynomial in l, that is,  A – r I  = p(r) Finding the eigen values of the matrix A is equivalent to finding the zeros of the polynomial p(r). Equation 4 is called the Characteristic Equation of A, and p(r) in 5 is the Characteristic Polynomial of S. The Characteristic Equation plays a role for systems similar to the role played by the auxiliary equation for scalar linear equations. Examples Find the general solution of x (t) = Ax(t) where A = Step 1: Find eigen values The Characteristic Equation for A is A – rI = o 6r 2 3 =0 1 r 6 2 3 1 (6r)(1r) + 6 = 0 6 – r –6r + r2 +6 = 0 r2 – 7r + 12 = 0 (r – 4)(r – 3) = 0 r = 4, 3 28 Step 2: Find eigen vectors corresponding to r = 3, 4 Solve (A3I) u1 = 0 3 2  3 u1  2 u2 =0 3 u1  3 u2 = 0 2 u1  2 u2 = 0 u1 = u2 so u1 = s 1 1 solve (A  4I) = 0 2 2  3 u1 =0  3 u2 2 u1  3 u2 = 0 2 u1  3 u2 = 0 2u1 = 3u2 u2 = u1 3 3 = u1 = s 2/3 u1 2 2 So the solutions are x1(t) = e3t x2(t) = e4t 1 e 3t = 3t 1 e 3 3e 4t = 2 2e 4t So the general solution is x(t) = c1
e 3t e 3t + c2 3e 4t 2e 4t 29 1
2. A = 2 2 2 1 2 2 2 1 Step 1: Find eigen values 1r
A – rI = 2 1r 2 2 2
=0 2 2 1r (1r)[(1r) 2 – 4] + 2 [2 (1r) + 4] + 2 [4 – 2 (1r)] = 0 (1r)3 – 4 (1r) – 4 (1r) + 8 +8  4 (1r) = 0 (1r)3 – 12(1r) + 16 = 0 let 1 – r = u u3 – 12 u +16 = 0 u = 2 8  24 + 16 (u2)(u2 + 2u – 8) = 0 (u2)(u+4)(u2) = 0 u= 2, 2, 4 1r=2 r = 1 r = 1,1 r = 1, 1, 5 1 – r = 4 r = 5 r=5 30 Step 2: Find eigen vectors r = 1 2 2 2 2 2 2 2 2 2 u1 u2 = 0 u3 2 u1  2 u2 + 2 u3 = 0 2 u1 + 2 u2  2 u3 = 0 2 u1  2 u2 + 2 u3 = 0 u1  u2 + u3 = 0 Let u2 =s and u3 = t u1 = s –t st
u1 = s t 1
=s 1 +t 0 1 0 1 let s = 1, t = 0 and let s = 0 , t = 1, we get the two linearly independent 1 1 eigen vectors u1 = 1 , u2 = 0 0 1 for r = 5 4 2 2 2 4 2 2 2 4 u1 0 u2 = 0 0 u3 4 u1  2 u2 + 2 u3 = 0 2 u1  4 u2  2 u3 = 0 2 u1  2 u2  4 u3 = 0 Add the first two 6 u1  6 u2 = 0 Add the last two  6 u1 = 6 u2 u1 =  u2 u2 =  u1 31 6 u2  6 u3 = 0 u2 =  u3 u3 =  u2 = u1 u1
u3 =  u 1 = s 1 1 1 u1 let s = 1 1
u3 = 1 1 Since the eigen vectors u1, u2, u3 are linearly independent, a general solution to x (t) =A x(t) is 1
x(t) = c1e
t 1
t 1
t 1 + c2e 0 0 + c3e 1 1 1 If A is not symmetric, it is possible for A to have a repeated eigen value but not to have two linearly independent corresponding eigenvectors. 3. 5 3 x(t) 3 1 Step1 Find the eigen values Solve x' (t) = Solve A  rI = 0 5r 3 3 =0 1 r (5r) (1r) + 9 = 0 5 + r –5r + r 2 + 9 = 0 r 2  4r + 4 = 0 (r 2) 2 r = 2, 2 32 Step 2. Find the eigen vector for r = 2 3 3 3 3 u1 u2 = 6 0 3 u1  3 u 2 = 0 1 =s u1 1 u1 = u 2 So the solution is x1(t) = e 2 t 1 1 To find a second linearly independent solution to x' = Ax, try x 2 (t ) te 2t u1 e 2t u2 as a solution. We substitute x 2 in the system x' =Ax So x' 2 = A x 2 e 2t u 1 2 t e 2t u 1 2e 2t u 2 2t e 2t u 1 Ate 2t u 1 (A2I) u 1 = 0 u 1 + 2 u 2 =A u 2 u 1 = (A2I) u 2 A(t e 2t u 1 e 2t u 2 ) Since u 1 must be an eigen vector, set u 1 = 1 3 3 = 1 3 3 1 and solve for u 2 1 u1 u2 3 u1  3 u 2 = 1 3 u 1 = 1 + 3 u 2 or 3 u 2 = 3 u 1  1 u1 u2 = u1 (3u 1 1) / 3 u1 1 1 0  1/3 let u 1 =1 = 1 3 or 2/3 2 1 3 3e 2t + e 2t = 1 2 2e 2t
te 2t te 2t x 2 (t) = t e 2t General solution is x(t) = c1 x 1 ( t) c 2 x 2 ( t) 33 4. Find the general solution of 3
x (t) = Ax(t) where A = 0 4 1 3 8 0 1 2 Step 1: Find the eigen values The Characteristic Equation for A is A – rI = o 3r 1 0 4 3r 8 0 1 =0 2r (3r)[(3r)(2r) + 8] + 4 = 0 3r2 – r3 – 3r – r2 – 6 –2r +4 = 0  r3 – 4 r2 –5r + 2 = 0 r3 + 4 r2 +5r + 2 = 0 (r + 1)( r2 –3r + 2) = 0 (r +1)(r + 1)(r +2) = 0 r = 1, 1, 2 Step 2: Find the eigen vectors corresponding to r = 1, 1, 2 r = 1 Solve (A + I) u1 =0 2 1 0 1 3 u1 u2 = 0 u3
=0 u2 = 2 u1 u3 = 2u2 = 4u1 0 2 4 8
2 u1 + u2  2 u2 + u3 = 0 4 u1 – 8 u2 + 3 u3 = 0 4 u1 –16 u1 + 12 u1 = 0 34 u1 1 u1 = 2u1 = s 2 4u1 4 r = 2 Solve (A + 2I) u1 = 0 1 1 0 1 4 8 0 1 4 u1 u2 = 0 u3
u2 = u1 u3 = u2 = u1  u1 + u2 =0  u2 + u3 = 0 4 u1 – 8 u2 + 4 u3 = 0 u1 1 u2 = u2 = s 1 u3 1 Solutions are 1 e t x1(t) = et 2 = 2e  t 4 4e  t 1 e 2t x3(t) = e2t 1 = e 2t 1 e 2t Notice r = 1 is a repeated root. So the second linearly independent solution to x =Ax can be x2(t) = tetu1 + et u2 substitute x2 into the system x =Ax x2 =A x2 tetu1 + et u1  tetu2 =A (tetu1 + et u2) tetu1 = A tetu1 (A +I) u1 = 0 et u1  tetu2 =A (et u2) (A +I) u2 = u1 35 2 0 4 1 2 8 0 v1 1 2 4
v2 = 1 + 2 v1 v3 = 2 + 2 v2 = 2 + 2(1 + 2 v1) = 4 +4 v1 1 v2 3 v3
=1 2v1 + v2 2v2 + 3v3 = 2 4v1 8 v2 + 3v3 = 1 v1 v1 1 v1 3 8
let v1 = 1 v2 = 1 2v2 v3 4 4v3
So the solution is 1 1 te t
t t e t x1(t) = t et 2 + et 3 = 2te 4 8 4te The general solution is x(t) = c1 x1 t 3e 8e t t + c2 x2 t + c3 x3 t Solve the initial value problem. 0
5. X= 1 1 1 0 1 1 1 x(t) 0
x(0) = 4 0 1 Step 1: Find the eigen values A – rI  = 0 r 1 1 1 1 r 1 1 =0 r r(r2 – 1) – 1( r – 1) +1(1 + r) = 0 r(r + 1)(r  1) + (r + 1) +(1 + r) = 0 (1 + r)[r (r – 1) + 2] = 0 (1 – r) (r2 + r + 2) = 0 36 (1 – r)(r – 2)(r + 1) = 0 r = 1, 1, 2 Step 2: Find the eigen vectors r=1 111 111 111 u1 u2 u3 0
=0 0 u1 + u2 + u3 = 0 u3 =  u1  u2 u1
u1 = 1 u2 0 u2  u1 = u1 0 + u2 1 1 1 let u1 = 1, u2 = 0, and u1 = 0, u2 = 1 We get two linearly independent eigen vectors. 1
u1 = 0
u2 = 1 1 0, 1 let r = 2 2 1 1 11 2 1 1 2 u1 u2 u3 0
=0 0 1 2 3 u1 = u2 u3 = u2 or u3 = u1 2 u1 + u2 + u3 = 0 u1  2 u2 + 4 u3 = 0 u1 + u2  2 u3 = 0 Subtract 3 from 1 3 u1 + 3u2 = 0 Subtract 3 from 2 3 u2 + 3u3 = 0 37 eigen vector is u1 u3 = let u1 = 1 1 u 1 = u1 = 1 1 u1 1
so u3 = 1 1 let r = 1 1 1 1 1 1 1 1 1 1 v1 0 v2 = 0 0 v3 + v1 + v2 + v3 = 0 v1 + v2 + v3 = 0 v1 + v2 + v3 = 0 v3 =  v1  v2 eigen vector corresponding to r = 1 is v1 v2 = v3 v1 v2 v1 v2 1 0 = v1 0 + v2 1 1 1 let v1 = 1 v2 = 1 1
Eigen vectors are 0 1 1 0 and 1 38 General solution is 1
X(t) = c1 e
t 0
+ c2 e
t 1
2t 0 1 1 + c3 e 1 1 1 1
X(0) = 4 0 1 4 0
= c1 1 0 0 1 c1 c2 c1 c3 c3 c2 c3 + c2 1 + c3 1 = 1 1 1 c1 + c3 = 1 c2 + c3 = 4 c1  c2 + c3 = 0 Subtract from c1  c2 = 5 Add and 2 c2 + c3 = 5 Subtract 3 c2 = 9 from from from c2 = 3 c1 = 5 + 3 = 2 c3 = 4  c2 = 4 – 3 = 1 Solution of the initial value problem is 1
X (t) = 2 e
t 0
+3e
t 1
2t 0 1
2e
t 1 +e 1 1 1 e 2t e 2t
t 2e
= t t e 2t e 2t e 2t = 3e 2e
t t 3e e
t 3e e 2t 39 6. Find a general solution to the system 3 X '(t ) 2 21 1 1 X (t ) 441 Step 1: Find the eigen values 3r 2 4 2 1r 4 1 1 1r 0 (3 r )[( 1 r )(1 r ) 4] 2[2(1 r ) 4] [8 4(1 r )] (3 r )( 1 r r r 2 4) 4(1 r ) 8 8 4(1 r ) 0 (3 r )(r 2 5) 4 4r 16 4 4r 3r
2 0 r 3 15 5r 8r 16 0 r 3 3r 2 3r 1 0 r 3 3r 2 3r 1 0 (r 1)3 0 r 1,1,1 Let r = 1 2 2 21 21 u1 u2 0 4 4 0 u3 2u1  2 u2 + u3 = 0 2u1  2 u2 + u3 = 0 4u1 + 4u2 = 0 u1 = u2 From 2 u1 – 2 u1 + u3 = 0 u3 = 0 So the eigen vector corresponding to r =1 is 1 u1 1 0
1 et et 0 One solution is X1(t) = et 1 0 To obtain a second linearly independent solution, we need to find eigen vector u2 such that ( A I )u2 u1 40 2 2 4 21 21 4 0 v1 v2 v3 1 1 0 2v1  2v2 + v3 = 1 2v1  2v2 + v3 = 1 4v1 + 4v2 = 0 From 2v1 2v1 + v3 = 1 v3 = 1 v1 = v2 1
So u2 0 0 1 1 0 e0 1
t t 1 or u2 1 So the solution is x2 (t ) te 1 0 To obtain a third linearly independent solution we need to find u3 such that ( A I )u3 u2 2 2 4 21 21 4 0 u1 u2 u3 0 0 1 2u1  2 u2 + u3 = 0 2u1  2 u2 + u3 = 0 4u1 + 4u2 =1 4u2 = 1 + 4u1 From
2u1 2 4u1 1 4u1 u3 0 4 1 4u1 2u3 0 2u3 1 u3 1 2 u1 u3 1 u1 4 1 2 41 Let u1 = 1 4 1 4 u3 0 1 2 So the solution is
X 3 (t ) t2 t e u1 tet u2 et u3 2 1 t2 t e1 2 0 0 te 0 1
t et 1 4 0 1 2 So the general solution is X (t ) c1 x1 (t ) c2 x2 (t ) c3 x3 (t ) 42 Exercises Find the general solution of the following systems.
X AX Where: 1. A= 15 1 3 011
2. A= 1 0 1 110 4 1 1
3. A= 1 5 1 0 1 3 216 4. A= 025 002 43 Complex Eigen Values We know that the homogeneous system x' (t) = Ax(t), where A is a constant n x n matrix, has a solution of the form x(t) = e rt u if and only if r is an eigen value of A and u is a corresponding eigenvector. We know how to solve the above system if the eigen values are real. So if the real matrix A has complex conjugate eigen values + i with corresponding eigen vectors z = a + ib and aib. Then two linearly independent real vector solutions to
x' (t) = Ax(t) are x1 (t) = eαtcosβt a  eαtsinβt b x 2 (t) = eαtsinβt a + eαtcosβt b because we know that we have the solution w(t) = eγtz = e(α+iβ)t(a+ib) = eαtei βt(a+ib) = eαt (cosβt + isinβt)(a+ib) = (eαtsinβt a  eαtsinβt b) + i(eαtsinβt a + eαtcosβt b) 44 Examples: 1. Find a general solution of the system x (t) = Ax(t) where A = 2 1 5 2 Step 1 Find the eigen values. Solve  ArI  = 0 2r 5 =0 1 2r (2r)(2r)+4 = 0 r= Step2 4 2r + 2r + r2 + 5 = 0 i r2 + 1 = 0 Find eigen vectors 2r 5 1 2i (2i)z1 – 5z2 = 0 z1 + (2– i) z2 = 0 Do (1) – (2i) * (2) (2i)z1 – 5z2 = 0 – (2i) z1 – (2i)(2– i) z2 = 0 0=0 (2+i)z1 = 5z2 z2 =
(2+i)z1 5 z1 z2 0 0 ……. (1) ……..(2) z1 z2 z1 2i z1 5 z1 1 2 5 0 11 5 Let z1 = 5 45 5 2 So the solutions are =
X1(t) i 5 2 0 1  e0t sint 0 5cos t = 1 2 cos t sin t = e0t cost X2(t) = e0tsint 5 2 + e0tcost 0 5sin t = 1 2sin t cos t General solution is
X(t) = C1 X1(t) + C2 X2(t) 2. Find the solution to the system x (t) = Ax(t) that satisfies the given initial condition. 10 1 x (t) = 0 2 0 x t 101 2
x(0) = 2 1 Step 1 Find the eigen values of A
A rI 0 0 1r 0 1 0 2r0 1 0 1r (1r) [(2r)(1r)] – 1[(2r)] = 0 (1r) (2r)(1r) + (2r) = 0 (2r)[(1r)2 + 1] = 0 r = 2,
2 r2 – 2r +2 = 0
48 2 2 2i 2 Therefore, r = 1i 46 Step 2 Find the eigen vectors Let r = 2 1 0 1 0 0 0 1 0 1r u1 u2 u3 0 0 0 u1 – u3 = 0 u1 – u3 = 0 u3 = 0, u1= 0
u1 u2 u3 0 u2 0 0 s1 0 Let s = 1 0
u1 = 1 0 Let r = 1 + i i 0 1 0 1i 0 1 0 i z1 z2 z3 0 0 0 iz1  z3 = 0 (1i)z 2 = 0 z1  iz3 = 0 z1 = iz3 z2 = 0 z1 z2 z3
Let z 3 = 1 iz 3 0 z3 i z3 0 1 47 0 0 1 0 a 0, 1 b 1 i0 0 1 0 0 Solutions are 0
X2(t) 1  e t sint 0 e t cost = et cost 0  et sint 0 = 1 0 0
X3(t) 1 e t cos t 0 e sint
t = etsint 0 + etcost 0 = 1 0 General solution is 0
X(t)  e t sint e t cos t = C1 1 0 + C2
t 0 e cost + C3
t 0 e sint 2
X(0) = 2 1 2 2 1 0 C1 1 0 1 C2 0 1 1 C3 0 0 C3 = 2 C1 = 2 C2 = 1 C3 = 2  C2 = 2+1 = 1 C1 = 2, C2 = 1, C3 = 2 48 So the solution is 0
X(t) e t cos t  e t sint e t cos t + e t sint =21 0 = 1 0 e t cost 2 2 0 e t sint e t cos t + e t sint 2e t sint  2e t cost e cost  2e t sint 3e t cos t  e t sint
t =
t 2 e cost  2e t sint 49 Exercises Find a general solution of the following system x (t) = Ax(t) where 1. A= 6 1 54 324 2. A = 2 0 2 423 3. Find the solution to the system x (t) = Ax(t) that satisfies the given condition x (t) = 24 1 6 1 6 x(t) x(0) = 50 Nonhomogenous Linear System Earlier we discussed the method of variation of parameters for general constant coefficient linear differential equations. The idea was that if a general solution to the second order homogenous has the form x(t ) c1 x1 (t ) c 2 x 2 (t ), where x1 (t ) and x 2 (t ) are linearly independent solutions to the homogenous equation, then a particular solution to the nonhomogenous equation would have a form x p (t ) v1 (t ) x1 (t ) v 2 (t ) x 2 (t ) where v1 (t ) and v 2 (t ) are certain functions of t . We found v1 (t ) and v 2 (t ) by using variation of parameters method. A similar idea can be used for systems. Let x(t ) be a fundamental matrix for the homogenous system x '(t ) = A(t ) x(t )......(1) Where the entries of A maybe any continuous function of t . The general solution of (1) is x(t )c where c is a column nx1 vector, we seek a particular solution to the nonhomogenous system x '(t ) of the form x p (t ) Where v(t )
A(t ) x f (t )......(2) x(t )v(t )......(3) col (v1 (t ),......vn(t )) is a vector function of t to be determined. To derive a formula for v(t ) , we first differentiate (3) using the matrix version of the product rule. We obtain xp(t) = x(t)v (t) + x (t)v(t)……(4) Since x p (t ) is a solution of (2)
x p (t ) A(t ) x p (t ) f (t )......(5) Substitute for x p (t ) from (4) into (5)
x(t )v '(t ) x '(t )v(t ) A(t )[ x(t )v(t )] f (t ) A(t ) x(t )v(t ) f (t ) Replace A(t ) x(t ) by x ' (t) because of (1)
x(t)v'(t)+x'(t)v(t)=x'(t)v(t)+f(t) x(t) v ' (t) = f(t) Multiply both sides by x 1 (t) v '(t ) x 1 (t ) f (t ) 51 Integrating v(t ) x 1 (t ) f (t )dt x(t )v(t ) x(t ) x 1 (t ) f (t )dt......(6) Hence a particular solution to (2) is x p (t ) Combining (6) with the solution x(t ) to the homogenous system yields the following general solution for (2)
x '(t ) x(t )c x(t ) x 1 (t ) f (t )dt Examples: Use the variation of parameters formula to find a general solution of the system x '(t ) Ax(t ) f (t ) where A and f (t ) are given. (1) A 11 , 41 f(t) t 1 4t 2 Step 1 Find x(t ), the fundamental matrix for the homogenous system
x '(t ) Ax(t ) Find the eigenvalues and eigenvectors of A Solve: A rI =0 1r 1 =0 4 1r (1r) 2 4 = 0 = 0
r2 2r 3 = 0 0 (r 3)(r 1) r = 3, 1 52 Eigenvector for r = 3 2 4 1 2 u1 u2 = 0 0 2 u1 4 u1 u2 2u 2
2u1
u1 2u1 0 0 u2
So u1 Let So u1 u2 u1 1 2 u1
u1 1
1 2 Eigen vector for r = 1 21 42 u1 u2 = 0 0 2u1 u2 u2 0 2u1
u1 1 = u1 2u1 2 u2 Let u1 = 1 53 So, solutions are x1 (t ) e3t 1 2 1 2 e 3t 2e3t e
t t x 2 (t ) e t 2e So the fundamental matrix for the homogenous system is e 3t 2e 3t
Step 2 x 1 (t ) = e t t 2e We need x 1 (t ) 1 4e 2t 2e t e e 3t t 1/ 2e 3t
1/ 4e 3t 2e3t 1/ 2et 1/ 4et v(t ) x 1 (t ) f (t )dt = 1/ 2e 3t 1/ 4e 3t t1 4t 2 1/ 2et 3 / 2te
3t 1/ 4et e
3t dt = 1/ 2tet
1/ 2te
3t dt = 1/ 2e 3t 1/ 2tet 1/ 2et So a particular solution is x p (t )
t 2 e 3t 2e 3t e t t 1 / 2te 3t 1 / 2e 1 / 2e t 3t 2e 1 / 2te t = 54 General solution is x(t ) c1 e 3t 2e
3t c2 e t t t 2 2e Example (2) A = 01 10 f (t ) 8 sin t 0 Find the eigenvalues and eigenvectors of A Solve A rI 0
r 1 1 r 0 r= i r2 1 0 Solutions are x1(t) = e0t cost a  e0t sint b x2(t) = e0t sint a  e0t cost b Let r = i i 1 1 i z1 z2 = 0 0 iz1 + z2 = 0 z2 = iz1 z1  iz2 = 0 iz2 = z1 z2 = (1/i) z1 = iz1 z1 z2 z1 iz2 1 0 i 0 1 55 Let z1 = 1 So a= 1 0 ,b= 0 1 1 0  sint = 0 1 1 0  cost = 0 1 cos t sin t sin t cos t x1(t) = cost x2(t) = sint Fundamental matrix is x(t) = We need x1 (t) cos t x1(t) = sin t
v(t ) x 1 (t) f(t)dt cos t sin t sin t cos t dt cos t sin t sin t cos t sin t cos t 8sin t 0 dt 8cos t sin t 8sin 2 t 8sin 2 t 2 sin 2t 4t 2 4sin 2 t 4t 2sin 2t So, a particular solution is xp(t) = x(t) v(t) cos t sin t sin t cos t 4sin 2 t 4t 2sin 2t = = 4sin 2 t cos t 4t sin t  2 sin t sin 2t 4sin 3 t 4t cos t  2 cos t sin 2t 56 General solution is x(t) = C1 cos t sin t C2 sin t cos t
4sin 2 t cost 4t sin t 2sin t sin 2t 4sin 3 t 4t cos t 2 cos t sin 2t The formula for the particular solution can be troublesome because of the requirement of finding the inverse of the fundamental matrix. The computation is even worse when the fundamental matrix X(t) is 3 by 3. Fortunately, there is a way to avoid these difficulties if we use the right fundamental matrix and follow the following process. If A is a square matrix and t is a scalar, we define the matrix exponential exp (At) by its Taylor series. exp(At) = 1 nn A t =I+ n=0 n! 1 nn At n=1 n! This is the unique solution of the equation: dX = AX where X(0) = I dt I is the identity matrix. Thus, as long as we choose the fundamental matrix X(t) so that X(0) = I, we can exploit the properties of the exponential function. Let’s start with the equation on the bottom of page 46: V (t) = X1(t) f(t) Take the definite integral of both sides. V(t) – V(0) =
t o X 1 (s) f(s) ds If we let C = V(0), we get the solution: Xp(t) = X(t) V(t) = X(t) C + X(t)
t 0 X 1 (s)f(s) ds Here is where we exploit the properties of the exponential: X(t)
t 0 X 1 (s)f(s) ds = eAt
t 0 t 0 eAs f(s) ds = t 0 e A(ts) f(s) ds = t 0 X(t s) f(s) ds It is much simpler to use X(t s) f(s) ds than it is to use X(t) X 1 (t) f(t) dt . There are fewer matrix multiplications and we don’t need the inverse of the fundamental matrix. Let’s see how this would work out in the examples we used earlier. 57 Take a look at example 2, beginning on page 49. A = the fundamental matrix. X(t) = cos t sin t sin t cos t 01 8 sin t and f(t) = . We produced 1 0 0 which, by happy coincidence, has the property that X(0) = I. This means that we can go right to the formula: Xp(t) = X(t) C + cos t sin t
t 0 X(t s) f(s) ds
t 0 = sin t + cos t cos(ts) sin(ts) sin(ts) cos(ts) 8 sin s ds 0 The integration is no more difficult than it was before and it’s less matrix algebra to get to the final answer. We can always select the fundamental matrix so that it equals the identity matrix at t = 0. For example, consider example 1, beginning on page 47. We use the fundamental matrix: X(t) =
e3t et 2e3t 2et This matrix does not equal I at t = 0, so it is not exp(At). However, we can solve for the matrix K so that X(0) K = I and then use X(t)K as our fundamental matrix instead. The result is: 1 3t e 2 e3t 1 t e 2 et 1 3t 1 t e e 4 4 1 3t 1 t e e 2 2 use this as our new X(t). it may look like a more complicated matrix, but it now permits the use of the formula
t 0 X(t s) f(s) ds to obtain a particular solution. 58 Exercises Use the variation of parameters formula to find a general solution of the system x ' (t) = A x(t) + f(t) where A and f(t) are given. 1. A= 01 1 0 21 3 2 f(t) = 1 0
2e t 4e t e 2t e3t 2. A= f(t) = 3. 11 A= 01 f(t) = ...
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This note was uploaded on 03/20/2011 for the course MA 243 taught by Professor Poon during the Spring '09 term at EmbryRiddle FL/AZ.
 Spring '09
 POON
 Calculus, Matrices

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