141B 1
MATH Instructor: Dr. M.Fabbri
QUIZ#3 Solutions
Date: Tues Feb 5, 2008
1. i) False. Adding the first two columns as the 4th and 5th columns we get the matrix: a-1 1 0 a-1 1 1 -2 a + 2 1 -2, 3 0 a 3 0 Now using the method of diagonals to compute the determinant, we get det A = [(-2a(a-1)+3(a+2)+0]-(0-0+a) = -2a2 +2a+3a+6-a = -2a2 +4a+6 = -2(a+1)(a-3). ii) True. A is not invertible if a = -1 or a = 3. iii) True. If a = 2 then A is invertible, and so system has a solution for all values b1 , b2 the b1 x y = A-1 b2 . and b3 . In fact, the solution is b3 z Therefore, C) is the answer. 2. i) False. Adding the first two columns as the 4th and 5th columns we get the matrix: 1 0 1 1 0 0 -1 0 0 -1, 1 a 0 1 a Again, using the method of diagonals to compute the determinant, we get det A = (0 + 0 + 0) - (-1 - 0 + 0) = 1. ii) True. Since the determinant is non-zero for all a, the matrix A is invertible for all a. iii) False. If you reduce the matrix A to I, then the matrix I is reduced to 0 a 1 A-1 = 0 -1 0 1 -a -1 and hence (A)32 = -a. Therefore, B) is the answer. 3. i) False. The reduced row echelon form of the augmented matrix is 1 0 3 -2 0 1 -7 5 ii) False. z is a free variable, and so there are infinitely many solutions. iii) False. The set of all solutions are the ordered triples of the form (-3z - 2, 7z + 5, z) where z can be any real number. If -3z - 2 = 1 then z = -1 and so the solution with = x 1 is (1, -2, -1). Therefore, E) is the answer.
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4. i) False. A is in reduced row echelon form, but det A = 0 (there's a row of zeros), and so A is not invertible. ii) False. All 4 conditions are satisfied, so B is in reduced row echelon form. iii) True. The powers of a diagonal matrix are easily calculated. Just take the powers of 9 1 0 0 0 , and we conclude that the elements on the main diagonal. Therefore C 9 = 0 39 0 0 (-2)9 9 9 (C )33 = (-2) = -512. Therefore, D) is the answer. 0 1 2 3 5. If you reduce the given matrix G = 4 0 0 to the identity matrix, I, then I will in turn 1 0 1 3 2 4 0 3 0 1 -1 be reduced to G = - 2 0 3 . To see this, we compute: 3 0 -3 4 2 3 0 1 2 1 0 0 0 0 0 1 0 4 R R 0 1 2 1 0 0 - - -2 - A = 3 0 0 0 1 0 - - 1- - 4 1 1 1 0 2 3 0 0 1 0 1 3 0 0 1 2 3 0 0 0 4 0 4R 0 1 2 1 0 0 - - 1- -- 6R3 0 3 2 0 0 6 1 0 0 0 4 0 3 -3R +R 1 0 0 - - - - 3- - -1 2 - 0 1 2 - R1 3 0 0 -4 -3 0 6 4 1 0 0 0 3 0 1 - R3 0 1 2 1 0 0 - - 4- - - - - 3 0 0 1 3 0 -2 4 4 1 0 0 0 3 0 -2R +R - - - - 2- - - 3 - 0 1 0 -1 0 3 - 2 3 0 0 1 0 -3 4 2 We use this matrix to move backwards. Therefore, the population at stage i - 1 is: 120 0 4/3 0 120 240 3 180 = 30 . pi-1 = G-1 pi = G-1 180 = -1/2 0 30 3/4 0 -3/2 30 45 Finally, the population at stage i - 2 is: 240 0 4/3 0 40 3 15 pi-2 = G-1 pi-1 = G-1 30 = -1/2 0 45 3/4 0 -3/2 225/2 Therefore, A) is the answer.