Advanced Higher Mathematics for INFOTECH Final exam 28. March 2012

Transcription

1 Advanced Higher Mathematics for INFOTECH Final exam 8. March Problem. (8 points) We have to solve the linear system of equations Ax =. We bring the system (A ) into echelon form: We first interchange the first and second row: A = We add multiples of the first row to the second and third row: A = We add a multiple of the second row to the third row: A = 3. (3 points) We can choose two variables in an arbitrary way, x 3 = µ, x = λ with µ, λ R ( point). Then x 3µ + λ =, hence x = 3µ λ ( point). Finally, x (3µ λ) + µ λ =, hence x = 5µ ( point). The general solution is x = 5µ 3µ λ µ λ It is easy to check that x is indeed a solution of the original system. A basis for the kernel is given by 5 v = 3, v =. ( points) By definition, v and v span the kernel. It is also clear that they are linearly independent, because they are not parallel.. Problem. ( points) We first solve the homogeneous equation ẍ + ẋ + x =.

2 The ansatz x = e λt leads to the characteristic equation The solutions are λ + λ + =. λ, = ± = ± 3i. ( points) Hence the general complex solution of the homogeneous equation is x(t) = e t (C e 3it + C e 3it ). The real solution is x h (t) = e t (A cos(3t) + A sin(3t)). ( points) To solve the inhomogeneous equation we first complexify the right hand side ẍ + ẋ + x = 6e it and make the ansatz x(t) = ce it. We get ẋ(t) = ice it ẍ(t) = ce it. Hence This implies ( + i + )c = 6. c = 6 6(6 i) 5(3 i) = = = 3 i. 6 + i Hence the complex solution of the inhomogeneous equation is The real part is x(t) = (3 i)e it. x p (t) = 3 cos(t) + sin(t). (5 points) The general solution of the inhomogeneous equation is therefore x(t) = x h (t) + x p (t) = e t (A cos(3t) + A sin(3t)) + 3 cos(t) + sin(t), with A, A R ( point). Problem 3. ( points) We use the method of Lagrange. The surface S is given by h () for the function h(x, y, z) = x + y z. ( point) The function f that we want to minimize is f(x, y, z) = (x ( ) + (y 5) + z. )

3 ( point) We get the system of equations x + y z = x (x ) + (y 5) + ( z ) + λx = y 5 (x ) + (y 5) + ( z ) + λy = z (x ) + (y 5) + ( z ) λ =. ( points) The second and last equation imply x ( = x z ), hence = xz. The third and last equation imply y 5 = y ( z ), hence The first equation implies 5 = yz. 6x z + 6y z 8z 3 =. With the two equations above this yields + 5 = 8z 3 (5 points for the correct equations) hence z = 3 x = 6 ( point) and y = 5 6 ( point). ( point). This implies Problem. ( points) We use the theorem of Gauss F n da = div F dv. M ( point) The divergence of F is given by div F = x z x + 3z = 3x + z. ( points) The set M in cylindrical coordinates is given by r, φ π, z. M

4 ( points) The volume element in cylindrical coordinates is given by dv = rdrdφdz. ( point) We have to calculate the integral π (3r cos φ + z )rdφdrdz. The integral π cos φdφ is zero ( point). We get π z rdrdz. ( point) The integral π rdr is equal to π[r ] = 3π. ( point) The integral 3π z dz is equal to π[z 3 ] = 8π. ( point) This is the result of the surface integral. Problem 5. ( = points) a) The polynomial X 3 a of degree 3 is irreducible if and only if it has no zero. We make a table of all values of X 3 for X Z 7 : X X We see that there is a zero if and only if a =, or 6. Hence the polynomial is irreducible precisely for a =, 3,, 5 (3 points). b) Every polynomial of degree is of the form g = X + ax + b where a, b Z. The polynomial is irreducible if and only if it has no zero. is a zero of g if and only if b =. is a zero of g if and only if + a + b =. Hence there are no zeroes if and only if b = and + a + b =, i.e. b = and a =. The unique irreducible polynomial of degree is X + X + ( points). c) We have to prove that f = X + X 3 + X + X + is irreducible ( point). If f is reducible it is either a product of polynomials of degree and 3 (i.e. it has a zero) or a product of the irreducible polynomial of degree with itself. The polynomial f has no zero because f() = and f() =. The product of X + X + with itself is X + X + which is not equal to f ( points). d) The set {, X, X, X 3 } are a basis for K over Z, hence K has 6 elements ( points). In K we have X 5 = : We have X 5 = Xf + f + =. Hence the inverse of X is X and of X it is X 3 ( points).

5 Problem 6. (3 + 5 = 8 points) a) The rows of the generator matrix are a basis for the code. If we perform elementary row operations we still get a generator matrix for the same code. Adding the third to the first row we get G = ( points) This is now in standard form. We can read off the parity-check matrix: H =. ( point) b) The codewords are given by the linear combinations of the rows of G. Hence they are: (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ). ( point) The cosets are given by y + C where y is an arbitrary element in F 7. One coset is given by the code itself. It has coset leader (,,,,,, ) and syndrome (,,, ) T. A second coset is (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ). It has coset leader (,,,,,, ) and syndrome (,,, ) T. A third coset is (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ), (,,,,,, ). It has coset leader (,,,,,, ) and syndrome (,,, ) T ( for the leaders, for the syndromes, for the cosets = points).. Total: 6 points