1. (b), (c), and (e) are true; the others are false. Note that projections always have rank 1, and that the composition of two linear transformations is always linear (and its matrix is the product of the matrices). 2. The initials NTSA on your paper mean that you did not use the standard algorithm -- there was no deduction for that. The standard algorithm goes this way: Multiply the first row by 1/2 to get: 1 2 3 4 1 2 4 6 3 6 10 14 Subtract the first row from the second to get: 1 2 3 4 0 0 1 2 3 6 10 14 Subtract 3 times the first row from the third to get: 1 2 3 4 0 0 1 2 0 0 1 2 Subract three times the second row from the first to get: 1 2 0 -2 0 0 1 2 0 0 1 2 Subtract the second row from the third to get: 1 2 0 -2 0 0 1 2 0 0 0 0 (b) There are two leading 1's (two nonzero rows), so the rank is 2. (c) The leading variables are x and z while y is free, and we may solve to get x = -2 -2y, z = 2. The solution set is therefore -2 - y { y : y in R } 2 Our R^3 is 3 by 1 column vectors. There was no penalty for the use of the notation (-2-y, y, 2), however. Also, note that in set notation it must be specified how y is allowed to vary, and that the condition on y is separated from the vector by a colon : or a vertical line | . Do not use a comma or semi-colon. Also, do not put an extra column vector x y z inside the set notation. It is completely correct if you replace "y" by another variable such as "s" or "t": in set notation, it is a "dummy variable". (d) The system referred to in (c) is the system given by A (there is no reference to the reduced row echelon form) --- your answer should use the columns of A, not of rref(A) (nor rref(A) with its 0 row deleted). The system you write down from rref(A) has the same solutions as the original system referred to, but it is a different system. (One point deduction for using rref(A).) The scalars x, y, z should be written to the left of the column vectors: there was no penalty for putting them on the right, however. 2 4 6 8 x 1 + y 2 + z 4 = 6 3 6 10 14 3. (a) (1 + 2^2 + 2^2)^(1/2) = 9^(1/2) = 3 1/3 (b) Divide the vector by 3 to get u = 2/3 2/3 (c) The projection of $v$ will be (v.u)u. The dot product is 6(1/3) + 0(2/3) + 3(2/3) = 2+2 = 4, and so the projection is 4/3 4u = 8/3 8/3 Typically, anyone who made the mistake of thinking the dot product is a vector lost all credit on parts (c), (d), (e). There was a substantial penalty for using the original vector 1 2 2 instead of u, which is 1/3 2/3 2/3 In the formula, (v.u)u some people used the correct value of u in one spot and the wrong value in another. This also incurred a substantial loss of credit that tended to carry through (c), (d), (e). (d) Here, (v.u) is x/3 + 2y/3 + 2z/3 and multiplying by u gives x/9 + 2y/9 + 2z/9 2x/9 + 4y/9 + 4z/9 2x/9 + 4y/9 + 4z/9 (e) Thus, the matrix is 1/9 2/9 2/9 2/9 4/9 4/9 2/9 4/9 4/9 The answer should not involve variables. Little or no credit was given for matrices coming from thoroughly incorrect answers to part (d). Some people thought that the intention here was to generalize to the case of a line in the direction of an arbitrary unit vector u = a b c where a^2 + b^2 + c^2 = 1. In that generality, the correct matrix is a^2 ab ac ab b^2 bc ac bc c^2 for which full credit was given. (If one does not assume that the length is one, one should divide all entries by the square of the length, i.e. by a^2 + b^2 + c^2.) If you are not sure what is meant in a question, please ask. 4. (a) (1) 30 degree rotation counterclockwise (2) dilation by a factor of 3 (3) shear parallel to the e_1 - axis (b) The image of the unit square is a parallelogram with edges 2 1 and 1 3 and T(e_1/2 + e_2/2) = T(e_1)/2 + T(e_2)/2 will correspond to the center of the parallelogram, which is the midpoint of the line joining T(e_1) to T(e_2). Quite a few people did not answer the question fully: they did not show the parallelogram which is the image of the unit square. 5. (a) We row reduce the matrix 1 0 2 1 0 0 0 2 0 0 1 0 1 0 k 0 0 1 Subtract the first row from the last. Divide the second row by 2. This gives: 1 0 2 1 0 0 0 1 0 0 1/2 0 0 0 k-2 -1 0 1 It is now clear that the matrix will be invertible (rank 3) if and only if k is NOT 2. Assuming this, divide the third row by k-2: 1 0 2 1 0 0 0 1 0 0 1/2 0 0 0 1 -1/(k-2) 0 1/(k-2) Finally, subtract twice the third row from the first: 1 0 0 1 + 2/(k-2) 0 -2/(k-2) 0 1 0 0 1/2 0 0 1 0 -1/(k-2) 0 1/(k-2) Thus, the inverse is 1 + 2/(k-2) 0 -2/(k-2) 0 1/2 0 -1/(k-2) 0 1/(k-2) The entry on the upper left is equal to k/(k-2). Either form of the answer received full credit. (b) 1(1) + 2(1) + 3(0) 1(2) + 2(0) + 3(1) 1(1) - 1(1) + 2(0) 1(2) - 1(0) + 2(1) or 3 5 0 4