1. Determine by inspection which of the following sets of vectors is linearly independent. 3 3.

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1 1. Determine by inspection which of the following sets of vectors is linearly independent. (a) (d) 1, 3 4, 1 { [ [,, 1 1] 3]} (b) 1, 4 5, (c) 3 6 (e) 1, 3, Solution. The answer is (a): v 1 is non-zero, v is not a scalar multiple of v 1 and v 3 is not a linear combination of v 1, v (since the last entry in any such linear combination is zero). So they are linearly independent. (b) Linearly dependent; contains the zero vector. (c) Linearly dependent; v 3 = v 1 + v. (d) Three vectors in R must be linearly dependent. (e) Linearly dependent, as the set contains a zero vector.. A linear transformation T : R R 4 satisfies ([ () 1 T = 1 ]), T = ([ Find T. 3]) (a) 7 3 (b) not defined (c) (d) Solution. The answer is (a). The standard matrix is A = = (e) () Hence, T 3 1 = 3. Let v 1 = 1, v = 1, v 3 = 1 1. The span of the set {v 1, v, v 3 } is: 5 (a) R 3 (b) a plane in R 3 (c) a line in R (d) not defined (e) The set {v 1, v, v 3 }.

2 Solution. The answer is (a): v 1, v, v 3 are the columns of the matrix A = Row 5 reduce: A There is a pivot in every row, hence the columns of A span the entire R Assume that the augmented matrix of a linear system was reduced to the echelon form Then the linear system: (a) has infinitely many solutions (b) has a unique solution (c) is inconsistent (d) has exactly two solutions (e) none of the other answers. Solution. The answer is (a): the second column is non-pivotal and corresponds to a free variable, hence no uniqueness. The last column is non-pivotal, hence the system is consistent. 5. For which values of h is the system consistent? x 1 + x = 1 x x 3 = 1 x 1 + x 3 = h (a) h = 4 (b) Consistent for any h. (c) Consistent for no value of h. (d) h = (e) h =. Solution. The answer is (a): the augmented matrix is h The last column is non-pivotal if and 4 h + h + 4 only if h + 4 = Let A =, B = 1 4 and C = AB. Which of the following is equal to the (1, )-entry c 1 of C? (a) 5 (b) (c) 3 (d) 4 (e) 1

3 Solution. The answer is (a): the (1, ) entry is the sum of products of corresponding entries from the first row of A and the second column of B, so c 1 = = Let B be an n n matrix. Suppose that the homogeneous equation Bx = has a non-trivial solution. Determine which of the following statements must be true. (a) The system Bx = c is inconsistent for some c in R n. (b) The linear transformation x Bx is one-to-one. (c) The matrix B has n pivot positions. (d) The columns of B span R n. (e) There is an n n matrix C such that BC = Id n. Solution. The answer is (a). A square matrix B is invertible if and only if Bx = has only the trivial solution, so the given condition means B is not invertible. Invertibility of B is also equivalent to the system Bx = c being consistent for all c in R n, so (a) is also equivalent to non-invertibility of B. The remaining conditions (b) (d) are all equivalent to invertibility of B. 8. Let A = A 1? 3. Which of the following is equal to the entry in the top left corner of 4 5 (a) 5 (b) 5 (c) 1 (d) 5 (e) 5 Solution. The answer is (a); the determinant of A is = so A 1 = 1 and the top left entry is A 4 8 matrix A has rank 3. What is the dimensions of the null space of A? (a) 5 (b) 1 (c) 3 (d) 7 (e) Solution. The answer is (a). For rank(a)+dim NulA = n where n is the number of columns of A. In this case, n = 8 so dim Nul = 8 3 = Is the matrix invertible? If so, find the inverse A 1.

4 Solution. Row reduce the augmented matrix: [A I] = = [I A 1 ] We find that A is invertible and A 1 = The row-reduced echelon form of the 3 5 matrix A = is given by 3 5 B = (You may assume this; you do not have to check it.) 1 (a) Determine a basis for the null space Nul A. (b) Determine a basis for the column space Col A. Solution. (a) The pivot columns of A and B are 1, 3, 5, so x and x 4 are free variables. Writing the homogeneous equations from B with free variables on the right gives x 1 = 3x 4 x = x x 3 = x 4 x 4 = x 4 x 5 = We include the equation x i = x i, for i = or 4, to indicate that the free variable x i can take arbitrary values. The system has basic solutions given by setting one free variable equal to 1 and the others equal to. Setting x = 1 and x 4 = gives the solution v 1 = [ 1 ] T. Setting x = and x 4 = 1 gives the solution v = [ 3 1 ] T. Then { v 1, v } is a basis for Nul A. (Alternatively, proceeding as in the following question shows the general solution of the homogeneous system is [ x 1 x x 3 x 4 x 5 ] T = x v 1 +x 4 v from which it follows that { v 1, v } is a basis for Nul A). (b) Row operations don t change the solution space of the homogeneous equation or the linear dependences of columns of a matrix. The pivot columns (1st, 3rd 5th) of B form a basis for col B so the pivot columns (1st, 3rd 5th) of A form a basis for Col A. A basis of col A is given by { w 1, w, w 3 } where w 1 = [ 1 3 ] T, w = [ 3 1 ] T and w3 = [ 4 3 ] T. 1. Express the solution set of in parametric vector form. x 1 6x 4x 4 = 4 x 1 3x + x 3 + x 4 = 1 3x 1 9x + x 3 + x 4 = 4

5 6 4 4 Solution. Row reduce: The bound variables are x 1, x 3 (corresponding to pivot columns) and the free variables are x, x 4. The equation is equivalent to x 1 3x x 4 = x 3 + 4x 4 = 1 where the free variables x and x 4 can take arbitrary values (the last row of the matrix gives the equation =, which we omit because it is always true). Rewriting with free variables on the right, x 1 = + 3x + x 4 x = x x 3 = 1 4x 4 x 4 = x 4 (we include the equation x i = x i, for i = or 4, to indicate that the free variable x i can take arbitrary values). In parametric form x 1 3 x x 3 = 1 + x 1 + x or writing x = r, x 4 = s, x 4 x 1 x x 3 x 4 3 = 1 + r 1 + s 4. 1