Kuwait University. Math 111. Date: Dec 15, 2010 ... (b) Let X, Y â Rn. Find X · Y given that X + Y = 1 and X â Y =
Kuwait University Dept of Mathematics
Math 111 2nd Midterm Exam
Date: Dec 15, 2010 Duration: 90 minutes
Answer all of the following questions. Calculators, pagers and mobile telephones are NOT allowed. 1. (3 pts.) Find parametric equations for the line through the point R(1, −3, 2) which is perpendicular to the plane 2x − y + 5z = 7.
1 2 −2 3 0 −1 3 1 2. (4 pts.) Let A = . Find det(A) by 0 2 −6 1 2 4 −3 0 (a) reducing A to triangular form. (b) using a cofactor expansion of det(A).
3. (4 pts.) (a) Let A be an n × n nonsingular matrix. Show that if A is symmetric, then adj A is symmetric. (b) Let X, Y ∈ Rn . Find X · Y given that kX + Y k = 1 and kX − Y k = 5.
4. (4 pts.) Which of the following are subspaces of R3 ? Prove your assertion. (a) W1 = {(a, b, c) : (a + b)c = 0} (b) W2 is the set of all vectors in R3 which are perpendicular to X = (1, −1, 2).
5. (4 pts.) Let P be the plane containing the points A(1, −1, 2), B(2, 0, 3) and C(−1, 2, 2). (a) Find an equation for P. (b) Verify that D(2, −5, 1) lies on P. (c) Find the area of the triangle with vertices A, B, C.
6. (6 pts.) Answer each question with “True” or “False” and justify your answer. (a) For a 5 × 5 matrix A, the sign of the term a13 a21 a34 a45 a52 in det(A) is “+”. b) If A is a nonsingular 3 × 3 matrix with adj A = −A, then det(A) = −1. c) There exist X, Y ∈ R4 such that kXk = kY k = 2 and X · Y = 6. (d) In R3 , if X = 4Y , then X × Y = (0, 0, 0).
Solutions
Math 111, Fall 2010, Second Exam
1. The parallel vector of the line is (2, −1, 5) (since it is perpendicular to the plane). Thus a set of parametric equations for the line is x = 2t + 1, y = −t − 3, z = 5t + 2 where t ∈ R. 1 2 −2 3 0 −1 3 1 2. (a) 0 2 −6 1 2 4 −3 0 3.
=
1 2 −2 3 0 −1 3 1 0 2 −6 1 0 0 1 −6
=
1 2 −2 3 0 −1 3 1 0 0 0 3 0 0 1 −6
= −
1 2 −2 3 0 −1 3 1 = 0 0 1 −6 0 0 0 3
(b) We use the cofactor expansion of det(A) along the first column. That is 1 2 −2 3 −1 2 −2 3 3 1 0 −1 3 1 2 −6 1 − 2 −1 = 27 − 24 = 3. 3 1 = 1 0 2 −6 1 4 −3 0 2 −6 1 2 4 −3 0 T T T T 3. (a) A adj A = |A|I n =⇒ A adj A = |A|I n =⇒ adj A AT = |A|I n =⇒ adj A A = |A|I n T On the other hand adj A A = |A|I n so we have adj A A = adj A A. Now multiplying both T sides by A−1 from right we get adj A = adj A. (b) We have ||X + Y ||2 = (X + Y ) · (X + Y ) = 1 and ||X − Y ||2 = (X − Y ) · (X − Y ) = 25, thus −24 = ||X + Y ||2 − ||X − Y ||2 = (X · X + 2X · Y + Y · Y ) − (X · X − 2X · Y + Y · Y ) = 4X · Y, that is X · Y = −6. 4. (a) We have (1, 1, 0), (1, −1, 1) ∈ W1 but (1, 1, 0) + (1, −1, 1) = (2, 0, 1) 6∈ W1 . Thus W1 is not a subspace of R3 since it does not satisfy the property (α). (b) We verify that W2 is a subspace of R3 . • O · X = 0 =⇒ O ∈ W2 =⇒ W2 6= ;; • Y, Z ∈ W2 =⇒ (Y + Z) · X = Y · X + Z · X = 0 + 0 = 0 =⇒ Y + Z ∈ W2 ; • c ∈ R and Y ∈ W2 =⇒ (cY ) · X = c (Y · X ) = c0 = 0 =⇒ cY ∈ W2 . ALTERNATE SOLUTION: W2 is a subspace of R3 since it is the solution space of the homogeneous linear equation x − y + 2z = 0. → − → − 5. (a) We may take n = AB × AC = (1, 1, 1) × (−2, 3, 0) = (−3, −2, 5) as a normal to P . Then an equation for P is −3(x − 1) − 2( y + 1) + 5(z − 2) = 0 or 3x + 2 y − 5z + 9 = 0. (b) 3(2) + 2(−5) − 5(1) + 9 = −9 + 9 = 0. p → − → − (c) The area of the triangle is 12 ||AB × AC|| = 12 ||(−3, −2, 5)|| = 38/2. 6. (a) True: The permutation 31452 has 4 inversions, so it is even. (b) True: A adj A = |A|I3 ⇒ −A2 = |A|I3 ⇒ (−1)3 |A2 | = |A|3 |I3 | ⇒ −|A|2 = |A|3 ⇒ |A| = −1. (c) False: We have X · Y = ||X || ||Y || cos θ ≤ ||X || ||Y || = 4, thus X · Y cannot be 6. (d) True: X × Y = (4Y ) × Y = 4 (Y × Y ) = 4 (0, 0, 0) = (0, 0, 0).
