Introduction to Linear Algebra (Math 220, Section 2) – Fall 2013 Brief Solutions to Practice Midterm Exam 1. (12) Consider the following system of linear equations. 3x1 + 4x2 + 0.3x3 = −3 x2 + 6x3 = 5 −2x1 − 5x2 + 7x3 = 0 (a) Write the system as a matrix equation. −3 3 4 0.3 x1 0 1 6 x2 = 5 . 0 −2 −5 7 x3
(b) Write the system as a vector equation.
3 4 0.3 −3 0 + x2 1 + x3 6 = 5 . x1 −2 −5 7 0 (c) Write the augmented matrix for the system.
3 4 0.3 | −3 0 1 6 | 5 −2 −5 7 | 0
1 1 −3 1 6 0 1 −2 1 , and b = 5. 2. (16) Let A = −1 −1 3 0 −3 (a) Solve the system Ax = b, and write the solution in parametric vector form.
1 1 −3 1 | 6 1 − − − − − → 0 1 −2 1 | 5 R3 + R1 0 −1 −1 3 0 | −3 0 1 0 −1 −−−−−→ R2 − R3 0 1 −2 0 0 0
1 −3 1 | 6 1 0 −1 − − − − − → 1 −2 1 | 5 R1 − R2 0 1 −2 0 0 1 | 3 0 0 0 x1 = 1 + x3 0 | 1 x = 2 + 2x3 0 | 2 =⇒ 2 x3 = x3 1 | 3 x4 = 3 1 1 2 2 The parametric vector form of the solutions is x = 0 + s 1 , s ∈ R. 3 0 (b) Using the result from Part (a), write the solution to the homogeneous system Ax the parametric vector form. 1 2 The parametric vector form of the solutions to Ax = 0 is x = s 1 , s ∈ R. 0
0 | 1 1 | 5 1 | 3
= 0 in
Math220 2 (Fall 2013) – Solutions to Practice Midterm
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3. (10) Let
1 2 0.5 u = 3 , v = 5 , and w = 2 . −1 7 −5 It can shown that 3u − v = 2w. Use this fact (and no row operations) to find a non-trivial solution to the homogeneous system Ax = 0, where 2 1 0.5 3 2 . A = 5 7 −1 −5 Notice that −v 2w= 0, and hence a non-trivial solution to the homogeneous system is +3u − −1 x1 given by x = x2 = 3. −2 x3 4. (12) Construct a 3 × 3 matrix A with every entry non-zero such that the following vector b is not in the span of the columns of A. Justify your answer. 8 b = −3 1 We need to create a matrix A such that the system Ax = b is inconsistent. One straightforward way to do this task is to make an A with all three columns being b itself, and then change all entries in one of the rows to a different number. For example, consider 8 8 8 8 8 8 | 8 −−−−−−−−−→ 8 8 8 | 8 A = −3 −3 −3 . Then [A | b] = −3 −3 −3 | −3 R2 + (3/2)R3 0 0 0 | −3/2 , 2 2 2 2 2 2 | 1 2 2 2 | 1 showing that the system is inconsistent. 5. (12) Let 3 6 5 5 v1 = 1 , v2 = 2 , v3 = −2 , and v4 = 0 . 4 1 1 2 (a) Does {v1 , v2 , v3 , v4 } span R3 ? Why or why not? 3 6 5 5 1 2 −2 0 1 2 −2 0 −−−−−→ −−−−−−−−−−−−−−−→ 1 2 −2 0 − 5 5 R2 − 3R1 , R3 − 4R1 0 −7 9 2 . R1 R2 3 6 4 1 1 2 4 1 1 2 0 0 11 5 There is a pivot in every row, and hence {v1 , v2 , v3 , v4 } spans R3 (b) Does {v1 , v2 } span R3 ? Why or why not? We need a pivot in every row, i.e., we need three pivots. With only two vectors, we can have only two pivots at the best. Hence {v1 , v2 } does not span R3 .
Math220 2 (Fall 2013) – Solutions to Practice Midterm
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6. (11) Find the standard matrix of the linear transformation T : R2 → R2 that first does a horizontal shear transformation mapping e2 to e2 + 2e1 (leaving e1 unchanged), and then reflects points through the vertical axis.
−1 −2 T (e1 ) = −e1 , and T (e2 ) = e2 − 2e1 . Hence T (x) = Ax, where A = . 0 1 7. (15) Consider the following system. x1 + 3x2 = k x1 − hx2 = 2 Determine all the values of the parameters h and k for which each of the following statements are true. 1 3 | k −−−−−→ 1 3 | k R2 − R1 . 1 −h | 2 0 −h − 3 | 2 − k (a) The system has no solution. We need the second row to be [0 0 | ] for the system to be inconsistent. Hence we need −h − 3 = 0 and 2 − k 6= 0. Thus, the system has no solution when h = −3, and k is any real number other than 2. (b) The system has a unique solution. We need two pivots (and hence o free variables). Hence −h − 3 6= 0, and k can be any real number. (c) The system has many solutions. We need a free variable. Hence, −h − 3 = 0, 2 − k = 0, i.e., h = −3, k = 2. 8. (12) Decide whether each of the following statements is True or False. Justify your answer.
(a) A 3 × 3 matrix can have more than three echelon forms. TRUE. We can have four echelon forms. 0 0 0 0 0 0 , and 0 0 , 0 0 , 0 0 . 0 0 0 0 0 0 0 0 (b) Let v1 and v2 be two vectors in R2 that are not collinear (i.e., they do not lie along the same line), and let A = v1 v2 . Then the system Ax = b cannot have infinitely many solutions for any b. TRUE. The columns are linearly independent, and hence there is a pivot in every column, so there are no free variables. (c) If a linear transformation is onto, then it cannot be one-to-one. FALSE. It can be both 1-to-1 and onto at the same time, e.g., when A is a square matrix with a pivot in each diagonal entry – and hence has a pivot in each row and each column. (d) If A is an m × n matrix, the range of the matrix transformation x 7→ Ax is Rm . FALSE. The co-domain is Rm . The range can be a subset of Rm .