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Last name: name: 1 Quiz 3 (Notes, books, and calculators are not authorized) Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with no justification will not be graded. a 4 1 Question 1: Consider the system X= . 2 a b (a) For which values of a does the system have a unique solution? The system has a unique solution if and only if the determinant is non zero: a 4 2 2 a = a − 8 6= 0 √ √ √ This is equivalent to a 6= ±2 2. The admissible values are a ∈ R\{−2 2, +2 2}. (b) Find those values of (a, b) for which the system has no solution. √ √ Observe first that it is necessary that a = 2 2 or a = −2 2 for the above system not to have a solution (otherwise the determinant is nonzero). In particular, a cannot be zero. Then, the augmented matrix is reduced in echelon form as follows: a 4 1 a 4 1 a 4 1 ∼ M∼ ∼ 2 a b 0 12 a2 − 4 12 ab − 1 a 12 a2 21 ab √ The system has no solution iff 12 a2 − 4 = 0 and 21 ab − 1 6= 0. This means a = ±2 2 and b 6= ± √12 . √ In conclusion the above system has no solution if and only if either a = 2 2 and b 6= √12 , or √ a = −2 2 and b 6= − √12 . −x3 + 2x1 + 3x2 = 1 x2 + 2x1 = 1 Question 2: Consider the following linear system 2x2 = 2 (a) Write the augmented matrix. The system can be rewritten into the following form: 2x1 + 3x2 − x3 = 1 2x1 + x2 + 0x3 = 1 0x1 + 2x2 + 0x3 = 2 and the augmented matrix is 2 2 0 3 1 2 −1 0 0 1 1 2 (b) Determine the pivot and free variables. We reduce the augmented 2 3 2 1 0 2 matrix in echelon form. −1 1 2 3 −1 0 1 ∼ 0 −2 1 0 2 0 2 0 1 2 0 ∼ 0 2 0 3 −2 0 −1 1 1 1 0 2 There is no free variable. There are three pivots, which are 2, −2 and 1 in the first, second and third column, respectively. 2 Quiz 3, September 18, 2011 2 Question 3: (a) Compute the echelon form of the augmented matrix M = 4 8 2 M = 4 8 2 4 8 −1 1 −1 6 4 2 10 13 ∼ 0 26 23 0 −1 3 3 2 0 0 6 −2 2 4 2 2 5 ∼ 0 0 7 0 0 −1 3 0 2 1 6 ∼ 0 1 0 2 0 1 0 −1 1 −1 2 4 8 6 −2 4 6 4 10 13 26 23 4 5 2 (b) Compute the reduced echelon form of M . 2 2 M ∼ 0 0 0 0 −1 3 0 2 4 5 ∼ 0 2 0 6 −2 4 2 0 0 −1 3 0 0 0 1 2 0 0 0 0 1 3 2 1 2 The reduced echelon form of M is 1 1 0 0 M ∼ 0 0 1 0 0 0 0 1 1 Question 4: Let E = −3 0 R2 and R3 are row vectors. 0 1 0 (a) 3 2 2 1 2 R1 0 0 and let A = R2 be an arbitrary 3 × 3 matrix, where R1 , R3 1 Compute EA E is the matrix on the elementary row operation that consists of replacing the second row of a matrix by itself minus three times the first row: R1 EA = −3R1 + R2 R3 (b) Give the inverse of E E −1 1 = 3 0 0 1 0 0 0 1 because 1 3 0 0 1 0 0 1 0 EA = 3 1 0 0 1 0 0 R1 R1 0 −3R1 + R2 = 3R1 − 3R1 + R2 = A. 1 R3 R3 Using A = I in the above equality means that 1 0 3 1 0 0 0 0 E = I 1