1. Consider the following system of linear equations x − 2y + 3z = 2 2x − 2y − z = 1 2x − 4y + 6z = α (a) (1 mark) Write the equation in the form of an augmented matrix (A | b). Solution: 1 −2 3 | 2 (A | b) = 2 −2 −1 | 1 2 −4 6 | α (b) (3 marks) Using Gaussian elimination, find an echelon form of (A | b). Solution: 1 0 −4 | −1 1 −2 3 | 2 −3 −3 → 0 2 −7 | (A | b) → 0 2 −7 | 0 0 0 | α−4 0 0 0 | α−4 (c) (1 mark) For which values of α does the system of equations have solution(s)? Solution: The system is consistent iff α = 4. (d) (1 mark) Assuming solutions exist, and by considering the rank of A, how many parameters are in the general solution. Solution: There is one parameter in the general solution. (e) (2 marks) Express the general solution in vector form. Solution: The general solution is x 8 −1 y = λ 7 + −3/2 z 2 0 for some λ ∈ R. 2. Let A, B ∈ Rn×n . Suppose that A2 = 1 and B T = B. Simplify the following expressions −1 −1 (a) (2 marks) AB −1 A Solution: −1 −1 AB −1 A = BA−2 = B T T (b) (2 marks) (BA) AB −1 Solution: T T T (BA) AB −1 = AT B T B −1 AT = I 3. (a) (3 marks) Consider the vectors v1 , v2 , v3 where 0 1 1 v1 = 2 v2 = 1 v3 = −1 1 −1 2 1 Show that {v1 , v2 , v3 } is a linearly independent set. Solution: We show that the only solution to λ1 v1 + λ2 v2 + λ3 v3 = 0 is λ1 = λ2 = λ3 = 0. We can write the system of equations in matrix form, and use Gaussian elimination to solve: 1 1 0 | 0 1 1 0 | 0 1 1 0 2 1 −1 | 0 → 0 −1 −1 | 0 → 0 −1 −1 1 −1 2 | 0 0 −2 2 | 0 0 0 4 This shows that the only solution is λ1 = λ2 = λ3 = 0. (b) (1 mark) What is span {v1 , v2 , v3 } Solution: R3 4. Let A = 2 1 1 3 3 −1 (a) (2 marks) Find the inverse of A. Solution: A −1 = 1 5 −1 2 (b) (2 marks) Use part a) to solve 2x + y = b1 x + 3y = b2 Solution: x y = = 1 5 1 5 3 −1 −1 2 3b1 − b2 2b2 − b1 b1 b2 5. State whether the following subsets W of R3 are subspaces. In each case, if W is a subspace of R3 , then find a basis for W . If not, then find two vectors u, v ∈ W such that u + v 6∈ W . (a) (3 mark) W = (x, y, z) ∈ R3 | xyz = 0 Solution: The subset W ⊂ R3 is not a subspace, since (1, 1, 0), (0, 1, 1) ∈ W , but their sum (1, 2, 1) is not in W . (b) (3 mark) W = im (B) = Bx | x ∈ R4 where 1 2 1 2 B = 2 4 3 0 1 2 1 2 Solution: The subset W is a subspace of R3 . We perform Gaussian elimination to obtain 1 2 1 2 B → 0 0 1 −4 0 0 0 0 so a basis for col (A) consists of the first and third columns of B. 2 | 0 | 0 | 0 For parts a) and b) above, I gave 1 mark for stating whether W is a subspace. For part a) I gave 2 additional marks for a correct proof that W is not a subspace. For part b) I gave 1 additional mark for each basis element of W . 6. Consider the matrix A 1 = 2 1 2 1 1 4 3 1 (a) (1 mark) Is A invertible? Solution: Yes. (b) (3 marks) If A is invertible, find its inverse. If not, find a linear relation between the columns of A. Solution: 1 2 4 | 1 0 0 (A | I3 ) → 0 −3 −5 | −2 1 0 0 −1 −3 | −1 0 1 1 2 4 | 1 0 0 → 0 −1 −3 | −1 0 1 0 −3 −5 | −2 1 0 1 2 4 | 1 0 0 → 0 −4 −12 | −4 0 4 0 0 4 | 1 1 −3 1 2 4 | 1 0 0 → 0 −4 −12 | −4 0 4 0 0 4 | 1 1 −3 2 4 0 | 0 −2 6 → 0 −4 0 | −1 3 −5 0 0 4 | 1 1 −3 2 0 0 | −1 1 1 → 0 −4 0 | −1 3 −5 0 0 4 | 1 1 −3 so A−1 = −2 1 1 4 1 3 2 −3 1 2 5 −3

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