Problem 1 (1): The determinant is 12. (2): For instance, the following matrix works:
Problem 2 (1): The matrix is given by
Problem 3 (1): t3−t2. (2): The Jordan canonical form is given by
Problem 4 (1): Use the definition of unitary matrices. (2): No student received a positive point.
Problem 5: No student received a positive point.
Your points may be reduced in case you have not explained your answer by sentences. It is also the case when you have used logical symbols in bad ways.
Problem 1 (1): The determinant is 1.
Although the determinant is easily calculated by elementary row or column operations, many students have strangely chosen the much harder way of calculating it by Laplace expansion.
Some students created and used a wrong formula for determinants of 4 by 4 matrices imitating the Sarrus rule for determinants of 3 by 3 matrices. Creating own wrong formulas is a typical symptom of weak students.
Problem 1 (2): The signature is (2,1) as readily seen by the square completion
Q(x, y, z) = (x + 2y + 2z) 2 - 4 (y + 5z/4) 2 + z2/4,
where the transformation of variables is invertible.
Some students erroneously calculated the signature by using a sum of more than 3 square terms ignoring the condition on the transformation of variables occuring in the square completion. Calculating without thinking of conditions is also a typical symptom of weak students.
The characteristic polynomial of the representation matrix is calculated as chA(t)=t3+t2-11t+1. This polynomial has 3 real roots but they do not have simple expressions by rational numbers or square roots of rational numbers. So a solution given in terms of roots with simple explicit expressions is incorrect. Note that one can find the signs of roots by considering the graph of the function y = x3+x2-11x+1. But no student seems to have tried this method.
Problem 2 (1): The characteristic polynomial is calculated as chA(t) = t3-9t = t (t-3) (t+3). Thus the eigenvalues are λ = 0, 3, -3. The matrix is diagonalized by a real orthogonal matrix as Q-1AQ = diag(0,3,-3) with Q given for instance by
Problem 2 (2): The Jordan canonical form is given by J = J3(2) ⊕ J2(2) ⊕ J1(2).
You must explain the reason why the Jordan canonical form is determined as above from the conditions given in the question. It is possible that no point will be given if the explanation is incorrect or not accurate even though the Jordan canonical form is correctly given as above.
Note that we eventually obtain dim Ker (A-2E)2 = 5 and dim Ker (A-2E)3 = 6. Some students have erroneously evaluated these dimensions first and then tried to determine the Jordan canonical form.
Problem 3 (1): A basis is given, for example, by 1, et, e-t.
Your points may be reduced in case you have not explained your answer by sentences.
Problem 3 (2): The exponential is given by
Problem 4 (1): The Jordan canonical form is given by J2(-1) ⊕ J2(-1). Indeed, by calculating A2, one easily finds A2=-2A-E, and this relation together with the fact that A is not a scalar multiple of E implies that the minimial polynomial of A is mA(t)=(t+1)2. Now by checking dim Ker (A+E), one obtains sufficient data to determine the Jordan canonical form.
Many students have gone into the extremely hard calculation of the characteristic polynomial by Laplace expansion. Sticking to a specific method is again a typical symptom of weak students.
Problem 4 (2): For instance, the following matrices work fine:
Problem 5: No student but one received a positive point.
Problem 1 (2) line 2: t-2 should be t-1.
Problem 2 (2) Notes ii) line 5 and 6: Ker A should be Ker (A-E).
Problem 2 (2) Notes ii) line 6: a complement should be the complement.
The schedule may change.
The following topics will be covered by the course.
Tutorial by assistant professors will take place as shown below.
Date: As marked by pink in the tables below.
Time: 5th period for Monday and Wednesday, and 11:15-13:00 for Friday.
Place: Room 109 of ¶ð¾ì¹ñºÝ¶µ°é¸¦µæÅï