(c)

1. Find the reduced echelon form of the matrix 1 1 5 1 8 5. 1 1 1 (a) 3 1 3 0 1 3 1 (b) 0 0 1 (c) 3 0 0 1 0 (d) 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 (e) 1 0 5 0 0 1 3 0 0 0 0 Solution. 1 1 1 1 1 1 1 1 1 1 1 1 8 5 1 8 5 0 3 9 3 0 1 3 1 1 1 1 1 1 5 3 0 0 0 3 0 1 3 1 where denotes row equivalence. 0 0 0 0. Which of the following equations involving 3 3-matrices A, B, C and I 3 (the identity matrix) could be false for some such matrices A, B, C? (a) (A + B) = A + AB + B (b) (A + B)C = AC + BC (c) (AB)C = A(BC) (d) A + B = B + A (e) (I 3 + A)(I 3 A) = I 3 A Solution. (I 3 +A)(I 3 A) = I 3 (I 3 A)+A(I 3 A) = I 3 I 3 I 3 A+AI 3 AA = I 3 A+A A = I 3 A so (e) is true. (A + B) = (A + B)(A + B) = A(A + B) + B(A + B) = AA + AB + BA + BB = A + AB + BA + B could be different from A + AB + B if AB BA, which can happen for some A, B. So (a) could be false. The other formulae (b), (c), (d) are all standard matrix laws which are true for all 3 3 matrices. 3. Determine by inspection which of the following sets of vectors is linearly independent. (a) (d) 1, 1 3 0 (b) 1 0, 0 1, 1 0 0 0 3, 4 6, 1, 4 8 4 (c) 3 6, 5, 4 3, 4 3 9 1 6 7 (e) 3 6, 0 0, 1 9 0 0 Solution. First vector in (a) is non-zero and second is not a scalar multiple of it; so they are linearly independent. (b) Not linearly independent; v = v 1. (c) Four vectors in R 3 must be linearly dependent. (d) Not linearly independent; v 3 = v 1 + v (e) Not linearly independent; contains the zero vector.

4. Let T : R R be the linear map given by counterclocwise rotation of the plane about the origin by an angle of π (in radians). Let A be the standard matrix of T. Which of the 4 following matrices is equal to A? [ ] [ ] [ ] [ ] [ ] 0 1 (a) (b) (c) (d) (e) 1 1 [ ] [ cos θ sin θ Solution. Let θ = π. One has A = ] [ ] = and A 4 sin θ cos θ = by matrix multiplication. Alternatively, note that A is the matrix of the composite linear transformation given by rotating counterclocwise twice [ around the origin] by [ angle θ] i.e. A cos θ sin θ is the standard matrix of a rotation by θ = π which is =. sin θ cos θ 5. Consider the linear system 3 [ 6 9 x = 1 h where h and are real numbers. Which 4 7 one of the following statements is true about the solution? (a) The system is inconsistent if h 3. (b) The system is inconsistent if. (c) The system is not consistent for any value of h and. (d) The system is consistent for all values of h and. (e) For some values of h and, the system has more than one solution. Solution. 3 1 6 9 h 3 1 0 0 h + 3 3 1 0 7 3 4 7 0 0 h + 3 0 0 h + 3 (7 3)/ 0 1. The system is inconsistent precisely when h + 3 0 i.e. h 3. If 0 0 h + 3 h = 3, it has the unique solution x = (7 3)/, y =. 6. The dimension of the null space of a 7 8 matrix B is 5. How many rows of zeros does the row reduced echelon form of B contain? (a) 4 (b) (c) 3 (d) 5 (e) 1 Solution. ran(b) = 8 nullity(b) = 8 5 = 3. The reduced echelon form of B has 3 pivot rows and 7 rows altogether, so there are 7 3 = 4 rows of zeros. 7. Let B denote the basis of R 3 given by B = 1 1, 1 0, 1 1 1 1 and let v denote the vector v = 0 1. The coordinate vector [v] B of v with respect to B is [v] B = a b. Which of 0 c the following is the value of a?

(a) 1 3 (b) 1 3 (c) 1 6 (d) 1 6 (e) 0 Solution. We have to solve the linear system with augmented matrix the first matrix in: 1 1 1 1 1 3 1 1 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 3 1 3 0 3 1 0 0 0 1/3 0 0. The coordinate vector is 1/3 0 so a = 1. 3 0 0 1 1/3 0 0 1 1/3 1/3 8. Let A be an n n square matrix. Suppose that for some b in R n, the linear system Ax = b is inconsistent. Which of the following statements must be true? (a) The linear system Ax = c has more than one solution for some c in R n. (b) A has a pivot in every column. (c) The linear map T : R n R n given by T (x) = Ax is one-to-one. (d) There is an n n-matrix B with AB = I n. (e) The linear system Ax = 0 has only the trivial solution. Solution. Since Ax = b is inconsistent for some b, the linear map T : R n R n given by T (x) = Ax is not onto. If T is not onto, the equivalent conditions for matrix invertibility show that A is not invertible, and T is not one-to-one either, so Ax = c has at least two solutions for some c in R n. The conditions for invertibility also show that the other listed conditions (b),(c),(d),(e) are all equivalent to invertibility of A, so cannot hold. [ [ ] [ 3 1 x h 9. Which of the following is the solution of the matrix equation =? 17 5] y ] x 5/ 1/ h x 3/ 1/ h (a) = (b) = 17/ 3/] 17/ 5/] x 3/ 1/ h x 3/ 17/ h (c) = (d) = 17/ 5/] 1/ 5/] x 5/ 17/ h (e) = 1/ 3/] [ ] [ [ 3 1 x h Solution. The equation is Av = b where A =, v = and b =. Note 17 5 ] [ ] 5 1 det A = 3 5 ( 1) ( 17) = so A is invertible and A 1 = 1. The solution is 17 3 x 5/ 1/ h v = A 1 b i.e. =. 17/ 3/] 10. Compute the inverse of the matrix A = 1 1 3 3 6. 1 1

Solution. Row-reduce 1 1 3 0 3 6 0 1 1 3 0 0 1 3 3 1 1 0 0 0 1 1 0 0 0 1 3 3 0 0 1 3. The inverse is 0 1 3. 0 0 1 0 1 1 1 11. Express the solution set of x 1 4x + 5x 3 + x 4 = 3 x 1 x + x 3 + x 4 = 1 x 1 x + 3x 3 = in parametric vector form. 4 5 1 3 1 1 1 1 1 Solution. Row reduce: 1 1 1 4 5 1 3 0 0 1 1 1 1 3 0 1 3 0 0 0 1 1 1 1 0 3 1 0 0 1 1 1. The equation is equivalent to 0 0 0 0 0 x 1 x + 3x 4 = 1 x 3 x 4 = 1 where the free variables x and x 4 can tae arbitrary values (the last row of the matrix gives the equation 0 = 0, which we omit because it is always true). The bound variables are x 1, x 3 (corresponding to pivot columns) and the free variables are x, x 4. Rewriting with free variables on the right, x 1 = 1 + x 3x 4 x = x x 3 = 1 + x 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 tae arbitrary values). In parametric form x 1 1 3 x x 3 = 1 + x 1 0 + x 4 1 0 0 1 or writing x = r, x 4 = s, x 4 x 1 x x 3 x 4 1 3 = 1 + r 1 0 + s 1. 0 0 1

1. The row-reduced echelon form of the 3 6 matrix A = 0 4 1 5 6 0 1 1 7 5 is given 0 by B = 0 1 0 4 0 0 0 0 1 3 0. (You may assume this; you do not have to chec it.) 0 0 0 0 0 1 (a) Determine a basis for the null space null(a). (b) Determine a basis for the column space col(a). Solution. (a) The pivot columns of A and B are, 4, 6, so x 1, x 3 and x 5 are free variables. Writing the homogeneous equations from B with free variables on the right gives x 1 = x 1 x = x 3 4x 5 x 3 = x 3 x 4 = 3x 5 x 5 = x 5 x 6 = 0 We include the equation x i = x i, for i = 1, 3 or 5, to indicate that the free variable x i can tae arbitrary values. The system has 3 basic solutions given by setting one free variable equal to 1 and the others equal to 0. Setting x 1 = 1 and x 3 = x 5 = 0 gives the solution v 1 = [ 0 0 0 0 ] T. Setting x3 = 1 and x 1 = x 5 = 0 gives the solution v = [ 0 0 0 ] T. Setting x5 = 1 and x 1 = x 3 = 0 gives the solution v 3 = [ ] T 0 4 0 3. Then { v1, v, v 3 } is a basis for null(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 (rd, 4th, 6th) of B form a basis for col(b) so the pivot columns (rd, 4th, 6th) 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 1 ] T, w = [ 1 1 ] T and w 3 = [ 6 5 0 ] T.