# Systems of Algebraic Equations and Systems of Differential Equations

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1 Systems of Algebraic Equations and Systems of Differential Equations Topics: 2 by 2 systems of linear equations Matrix expression; Ax = b Solving 2 by 2 homogeneous systems Functions defined on matrices Determinant of a 2 by 2 matrix Trace of a 2 by 2 matrix Eigenvalues/eigenvectors of a 2 by 2 matrix Characteristic equation Systems of autonomous homogeneous with constant coefficients DEs; two dependent variables Case real eigenvalues Revised 2/22/2016

2 Matrices and Systems of Algebraic Equations A 2 2 system of linear equations is easily represented using matrices. We adopt the following matrix notation. For system ax + bx =k 1 cx + dx =k 2 we identify three matrices associated with this pair of equations. a b x1 k1 A, x, k c d x k 2 2 coefficient matrix column of unknowns rightside column ax 1 + bx 2=k1 In place of the system we use the matrix expression Ax = k. cx + dx =k 2 Writing Ax in matrix algebra indicates a product of matrix A times matrix x. The result of this matrix product using dot products of rows of A with the column x is ax 1 + bx2 Ax cx 1 + dx 2 This is a product of 2 2 matrix A time 2 1 matrix x resulting in a 2 1 matrix which is the left side of the linear system.

3 Example: Linear system 2x - 3x = -6 x - 2x = -5 in matrix form is denoted Ax = k, where A 2 3 x1 6, x x, k 2 5 The Determinant of a 2 by 2 Matrix Definition: The determinant of a 2 2 matrix computed by the expression det(a) = ad bc.. A a b c d is denoted det(a) and is An easy way to recall this computation is from the diagram. We take the product of the diagonal entries and add to it the negative of the product of the reverse diagonal. Examples: 4 1 x det = -8-3 =-11, det =x + 6, det = 4-4 = 0, λ 2 2 det =(1 - λ)(2 - λ) + 6 = λ - 3λ λ Greek letter lambda; λ.

4 Trace of a 2 by 2 matrix Definition: The trace of a 2 2 matrix the expression tr(a) = a + d. A a b c d is denoted tr(a) and is computed by The trace function computes the sum of the diagonal entries of the matrix. Examples: tr = 5 + (-2) = 3, tr = 4 + (-4) = A Special Matrix 1 0 I 0 1 The identity matrix is an example of a diagonal matrix; that is, all entries are zero except possibly those on the diagonal. The letter I is reserved to denote the identity matrix. The next example shows the results of applying our matrix functions to the 2 2 identity matrix. Example: det(i) = 1 and tr(i) = 2. Definition: The 2 2 identity matrix is denoted by the letter I and is defined to be The identity matrix plays an important role in matrix multiplication. If A is any 2 by 2 matrix then by applying matrix multiplication we have that AI = IA = A.

5 Homogeneous 2 by 2 Systems of Algebraic Equations A homogeneous system of 2 equations in 2 unknowns has the form ax + bx = 0 cx + dx = 0 There are just two types of solutions to such systems: 0 (i) A single solution x 1 = 0, x 2 = 0 which we will express as the column vector x. 0 (ii) Infinitely many solutions in which one of the unknowns, either x 1 or x 2, can be chosen arbitrarily. The form of the solution of a homogeneous system is determined by properties of the coefficient matrix A a b c d

6 Omitting the proof we state the following for solutions of system which can be expressed as the matrix equation Ax = 0 ax + bx = 0 0 (i) System cx 1 + dx 2= 0 has only the solution x if and only if 0 (ii) System ax + bx = 0 cx + dx = 0 det(a) det a b ad-bc 0 c d has infinitely many solutions if and only if det(a) det a b ad-bc = 0 c d ax + bx = 0 cx + dx = 0 0 x 0 The column of all zeros is called the trivial solution. In upcoming work we are only interested in the second case. The next example illustrates this case.

7 Example: Let 2 1 A, then solve Ax = 0. Computing det(a) = 4 4 = 0. This implies 4 2 2x1 x2 0 there are infinitely many solutions. The linear system can be written in the form for x 1 in each equation giving the system. that 1 x1 x2 x 2 x 2 x2 4x 2x 0. In this system we can solve This pair of equations implies Thus we can compute every solution of the system Ax = 0 by choosing x 2 arbitrarily and so a column represents the set of all possible solutions. x x x x Then we call the general solution of Ax = 0. Often the general solution is represented by letting x 2 = r, any arbitrary value, so then. Note that if r = 0 we get the trivial solution, for r = 1 we get, for r = 2 we get, and for r = -8 we get 4 x 8 1 x x 2 x2 2 1 x 2 1. There are infinitely many nontrivial solutions. 1 r x 2 r 1 x 2

8 Eigenvalues and Eigenvectors of 2 2 matrices Matrices are useful in a wide variety of applications. So you might suspect that we can construct functions involving matrices. As with functions you met in calculus matrix functions have a domain (input information) and a range (output information). One of the simplest matrix functions can be expressed as y = f(x) = Ax where A is a 2 2 matrix, x, an input, is a 2 1 column matrix. This function requires that we perform a matrix product A times x in order to produce an output y, which is also a 2 1 column matrix. An important feature of a matrix function y = f(x) = Ax involves determining input columns x 0 so that the output y is a multiple of x. That is, so that we have Ax = λx, where λ is some number (often called a parameter). (We can think of 2 1 column matrices as vectors in a Cartesian plane, so here we want the output vector y to be parallel to the input vector x.)

9 In the previous example we illustrated that if Ax = λx, then for any number r, A(rx) = λ(rx). Hence once we have one column x so that the output of function y = Ax is parallel to the input x there are infinitely many such columns. From our earlier discussion involving 2 2 homogeneous linear equations Ax = 0 with det(a) = 0 we showed there were infinitely many solutions. So it is likely that the two situations are related. We show there is such a relationship and how to determine columns (vectors) so that when they are multiplied by matrix A the result is a column representing a parallel column (vector). For a 2 2 matrix A we seek 2 1 columns x 0 so that Ax = λx. This matrix expression is called the eigen-equation. This equation is actually a nonlinear equation since both column x and number λ are unknown. Definition: Let A be a 2 2 matrix. The number λ is called an eigenvalue of matrix A if there exists a 2 1 column x 0 so that Ax = λx. Every 2 1 column x that satisfies the eigen-equation is called an eigenvector of A associated with eigenvalue λ. (Note: by definition an eigenvector of A cannot be a column of all zeros.)

10 Here we can rephrase a previous statement using eigen terminology: If x is an eigenvector associated with eigenvalue, then so is rx for any number r 0. How do we compute eigenvalues and eigenvectors of a 2 2 matrix? We start with the eigen-equation Ax = λx and restructure it into a homogeneous system of 1 0 I w w w 2 equations. We previously defined the 2 2 identity matrix. If is any 2 1 matrix it follows that Iw w. (Matrix multiplication is required.) So we can replace x on the right side of the eigen-equation to obtain Ax = λix. Applying some matrix algebra we rearrange this expression as Ax λix = 0. Recall that 0 represents the column system 0 0.) Then factoring we have the homogeneous linear (A λi)x = 0.

11 In our previous discussion of solutions of homogeneous systems, we stated that a homogeneous system had a nontrivial solution (that is, 0 x 0 ) only if the determinant of the coefficient matrix is zero. Since an eigenvector cannot be the zero column it must be that for linear system (A λi)x = 0 to have a nontrivial solution that the determinant of the coefficient matrix is zero; that is, det(a λi) = 0. (Since matrices A and I are known it is the value of eigenvalue λ that controls when the determinant is zero.) Definition: The determinant det(a λi) is called the characteristic polynomial of A and the equation det(a λi) = 0 is called the characteristic equation of A. Thus we have a compact expression for the characteristic equation in terms of the trace and determinant of matrix A. det(a λi) = λ 2 tr(a)λ + det(a) = 0

12 For the 2 2 matrix A the characteristic equation is a quadratic and its roots are the eigenvalues of A. Thus we know there are exactly two eigenvalues for 2 2 matrix A. Once we have computed the eigenvalues we return to the homogeneous linear system (A λi)x = 0 to find an associated eigenvector for each of the eigenvalues. We illustrate this in the next example.

13

14 Continuing with Step 2.

15 Summary: For our eigenvalue/eigenvector computations we use the following two-step procedure. Step 1. Determine the characteristic equation is λ 2 tr(a)λ + det(a) = 0 and solve for the eigenvalues. Step 2. For each eigenvalue λ determine a nontrivial solution to the homogeneous linear system (A λi)x = 0. The nontrivial solution will provide the set of eigenvectors associated with the eigenvalue λ.

16 For 2 2 matrices the characteristic equation is a quadratic so there are three types of roots, hence eigenvalues. They are real and distinct, λ 1 λ 2, real but repeated, λ 1 = λ 2, and complex λ ± µi, µ 0. In the preceding example that matrix had real distinct eigenvalues λ 1 = 4 and λ 2 = -1. A 2 2 matrix A with real entries can have complex eigenvalues. In such a case the corresponding eigenvectors also have complex entries. For now we focus on the case where the eigenvalues are real and distinct.

17 Solving systems of two autonomous homogeneous linear differential equations with constant coefficients A system of two autonomous homogeneous linear differential equations with constant coefficients in independent variable t and dependent variables u 1 (t) and u 2 (t) has the form u '(t) = a u (t) + a u (t) u '(t) = a u (t) + a u (t) Often the inclusion of the independent variable t is omitted to provide a more compact expression, then the system appears in the form u ' = a u + a u u ' = a u + a u

18 Our goal is to show how the eigenvalues and eigenvectors play a role in determining the general solution, the set of all solutions, to this system of differential equations. The proof of the following solution procedure requires some matrix theory beyond the scope of this introduction. But it is reminiscent of the technique for solving a homogeneous second order linear DE with constant coefficients. Procedure for obtaining the general solution to u' = Au when the eigenvalues of the a c b d 2 2 coefficient matrix A are real and distinct λ 1 λ 2. Step 1. Find the eigenvalues of matrix A. Call them λ 1 and λ 2. 2 Use the quadratic tr(a) det(a) 0 find the roots. Step 2. Compute the associated eigenvectors for each eigenvalue. Call them v and w. To find v compute a nonzero solution to the system a- λ v + bv = 0 1 c v + d- λ v = 0 1 similarly for w find a nonzero solution to Step 3. The general solution of u' = Au is given by the equation u 1(t) 1t 2t u(t) C1e v C2e w u 2(t) where C 1 and C 2 are arbitrary constants. a- λ w + b w = 0 2 Note the exponential functions that appear here. c w + d- λ w = 0 2

19 We illustrate the use of this procedure with examples. Example: Let s return to our example involving matrix A where now we have the system of differential equations u 1'(t) 2 3u 1(t) u'. Previously we u 2 '(t) Αυ 2 1 u2( t) computed the eigen information for this coefficient matrix as Using the expression for the general solution u (t) t t u(t) C e v C e v u (t) we get Now let s consider initial conditions for this system of differential equations so we can determine the arbitrary constants.

20

21 Another example: Determine the solution of the IVP given by u '(t) - 141u (t) - 54u (t) 1 u '(t) 360 u (t) 138u (t) 2 The coefficient matrix is A= tr(a) = -3 and det(a) = So the characteristic polynomial is The roots are the eigenvalues : λ 1= 3, λ 2 = λ - tr(a)λ +det(a) = λ - (-3)λ - 18 = λ +3λ - 18 To find the eigenvectors we solve the homogeneous linear systems (A - λ 1 x) = 0 and (A λ 2 x) = 0. We find the eigenvectors which can be chosen to have integer entries are respectively Then the solution of the system of DEs is -3-2 v 1 = and v 2 = 8 5 u1() t 3t 3 6t 2 u(t) C1e C2e u2() t 8 5

22 If we had initial conditions u 1 (0) = 2 and u 2 (0) = 1, then we must solve the system C We find C 1 = 12 and C 2 = -19. The solution of the IVP C u '(t) = u (t) - 54u (t) 1 u '(t) = 360 u (t) + 138u (t) 2 u 1(0) = 2, u 2(0) = 1 u () t 3 2 u(t) 12e 19e u2() t 8 5 is 1 3t 6t

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