Exam Study Questions for PS0- (*=solutions given in the back of the textbook) p 59, Problem p 59 Problem 3 (a)*, 3(b) 3(c) p 55, Problem
p547, verify the solutions Eq (8) to the Marcov Processes being described Explain why the eigenvalue of = corresponds to a steady state population To verify (8), one would simply verify that Ae =e, Ae =e, Ae 3 =e 3 But you should be sure to note that a negative population is not possible, therefore the only acceptable solution is e with being a negative constant The reason e represents a steady-state population is that the product of A with that population represents the new population after a year And since that new population Ae equals the starting population e then the population always stays the same year to year, which is describes a steady-state situation p 55, Problems 3a*, 3i*, 3n*
3 Find the eigenvalues and eigenspaces, as well as a basis for each eigenspace 0 0 3 (n) 3 7 The eigenvalues/vectors are determined by asserting that ( A I) x =0 has a non-trivial solution For that to be true the matrix ( AI) must have LD columns; and that is ensured if det( AI) 0 FIRST, I find the values for that make det( AI) 0 0 0 3 0 0 0 0 3 det( AI) 0 0 0 0 0 3 7 3 7 3 det( AI) (7 ) ( 3) 03 0 ( ) 7 33 0 7 0 0; 7 0 0; 7 4940 73 One root is 0 The other two are = 0,, and 5 NEXT, for each, I find the vectors that solve ( A I) x=0 0 0 3 3 7 3 0 3 0: 0 0 0 e 3 7 0 0 3 0 3/ 0 3/ 0 3/ 3/ 3 : 0 0 / 0 / 0 / / e 3 5 0 5/3 0 0 /9 5 0 3 0 3/5 0 3/5 0 3/5 3/5 3 5: 0 5 0 /5 0 /5 0 /5 e3 /5 3 0 3 7/5 0 0 0/5 5 3 3 T T T 0; e 3,, 0 ; span 3,, 0 ; basis is 3,, 0 T T T 0; e 3,, ; span 3,, ; basis is 3,, T T T 0; e 3,, 5 ; span 3,, 5 ; basis is 3,, 5 p 55, Problem 5(a)*, 5(b)* Use FreeMAT to compute A*x If cross(a*x,x)=0 (ie, the cross product of Ax and x=0), then yes, x is parallel to Ax, and it is an eigenvector
p 565, Problem (a)*, (d)* Solve the eigenvalue problem (see my example above for 55 Problem 3n) Since the matrices are symmetric the eigenvectors should be orthogonal The orthogonal basis is then the two eigenvectors with the scalar multiples (eg, and ) set to any specified value Greenberg sets these two multiples both to p 580, Problem (a)*, (d)*, (g) 3 ( a) The rows and columns are LD One way I know this is any row or column that is all 0 0 zeros is LD on the other rows and columns Another way I know this is that I can see that column is -3/ times column Because the columns are LD, this matrix is not diagonalizable This is because the columns of matrix Q are the eigenvalues and if they are LD, then the matrix is singular, detq 0, and Q does not exist ( d) Columns and are LD and so this matrix is not diagonalizable (actually none of the matrices in problem are diagonalizable!) p 580, Problem (a)* x' x4y Here how to solve (b) y' x y Since the coupled set of ODE's is homogeneous and have constant coefficients rt rt rt I know the solutions are e, and thus have the form xqe and y qe Plugging those forms into the equation I get the matrix equation 4q rt q rt 4q q, or simply e r e r r q q q q Ax = x We then solve the eigenvalue problem (see example above for p 55) rt The eigenvalues r give (in general) two LD exponential function of e, and the eigenvectors give the coefficients ( q, q ) for each function p 580, Problem 3 A singular matrix may or may not be diagonalizable A singular matrix is defined as a matrix with zero determinant Correspondingly, the rank is less than the number of rows, n, which also means the matrix has LD columns and rows But these characteristics do not determine whether the eigenvectors are or are not LI, which is the key requirement for a matrix to be diagonalizable p 687, Problem
p 687, Problem 7 p 69, Problem (a)*, (h)*, (k)* (k) With the tails of u, v, and w all meeting at the origin, their ends will be three points in 3D, space that form the vertices of a triangle The area of the triangle will be half the area of the parallelogram having two sides formed by by any two sides of the triangle The magnitude of the cross product of two vectors on the sides of the triangle give the area of the parallelogram (see p 686) The vectors forming the three sides are ( wv), ( wu), and ( vu) The answer is thus ( wv) ( wu) ( wv) ( vu) ( wu) ( vu) p 69, problem 4(a)*, 4(d)* Note that the magnitude of a cross product between vectors that point away from any vertex is the area of the parallelogram bounded by two side coinciding with the vectors The area of the triangle is half the area of the parallelogram The two vectors would be represented by the difference in the x, y, and z coordinates of the 3 sides involved eg, for 4(a), lets pick point (,4,3) as one vertex with two vectors point out from that point to (,0,-) and (0,0,5)
p 765, Problem (a)*, (d)* (d) v xyz( ˆi ˆj kˆ ) x y z At P (3,, 4), v (4) 3(4) 3( ) 5 C v v v v x y z yz xz xy means that the st partial derivatives of v are continuous If a function is continuous then the derivative of that function must exist The non-existence of a derivative would mean the slope is infinite, which corresponds to a discontinuous function This means that second partial derivatives of v must exist And indeed they do