Kuwait University, Department of Mathematics Math 111, Linear Algebra 11 January, 2012, 2-4pm. Final Exam. Answer all questions. Calculators and mobile phones are NOT allowed. ( 1. (3+2 pts) Let A =
3 −2
6 2
)
( and B =
3 −3 1 k
) .
(a) Express the columns of AB as linear combinations of the columns of A. (b) Find all values of k for which AB = BA. 2. (3+2 pts) (a) Find parametric equations for the line of intersection of the two planes x + y + z = 3 and x − 2y + z = 0. (b) Let U = (a, b, c) and V = (x, y, z) be two vectors in ℜ3 . Show that (ax + by + cz)2 ≤ (a2 + b2 + c2 )(x2 + y 2 + z 2 ).
3. (3+2 pts) (a) Show that the vector U = (1, 2, 1, 1) does not belong to span{(1, 2, 3, 0), (2, 1, 6, 0)}. (b) Let X = (10, 0, c), Y = (1, 0, 5) and Z = (0, −1, 0)} be three vectors in ℜ3 . Determine all values of c for which the vector X is orthogonal to the plane containing the vectors Y and Z. 4. (3+2 pts) (a) Does there exist a ∈ ℜ such that S = {(1, 1, 1), (2, 2, 2), (3, a, 4)} is a basis for ℜ3 . (b) Let A be a 3 × 3 matrix such that the system AX = 0 has the following two vectors X1 = (1, 0, 1)T and X2 = (0, 1, 1)T as solutions. Find one more nontrivial solution to the system AX = 0. 5. (3+1+2+1 pts) Let A =
1 0 0 0
0 3 0 0
4 9 0 0
1 3 2 3
.
(a) Find a basis for the null space of A. (b) Find a basis for the column space of A containing only columns of A. (c) Determine whether the vector X = (0, 3, 2, 3)T belongs to the column space of A. (d) Determine the rank and nullity of A. 6. (7 pts) Let A =
2 0 0 1
0 2 0 0
0 0 1 0
0 0 0 1
.
If possible, find a non-singular matrix P and a diagonal matrix D such that P −1 AP = D. 7. (6 pts) Answer each of the following as True or False (justify your answer). (a) Let A be an n × n matrix. If 0 is an eigenvalue of A then rank(A) = n. (b) If A and B are two n × n matrices with |A| = 1 and |B| = −1, then the columns of AB span ℜn . (c) If U and V are two orthogonal vectors in ℜn , then ||U + V ||2 − ||U − V ||2 = 0. (d) W = {(x, y, z)|2x + 3y + z = 0, 3x + 2y − z = 0 and x − y + 7z = 0} is a subspace of ℜ3 .
µ 1. (3+2 pts) Let A =
3 −2
6 2
¶
µ and B =
3 1
−3 k
¶ .
(a) Express the columns of AB as linear combinations of the columns of A. µ ¶ µ ¶ µ ¶ 3 6 15 col1 (AB) = A col1 (B) = 3 +1 = . −2 2 −4 µ ¶ µ ¶ µ ¶ 3 6 −9 + 6k col2 (AB) = A col2 (B) = −3 +k = . −2 2 6 + 2k (b) Find all values of k for which AB = BA. µ ¶ µ 15 −9 + 6k 15 AB = and BA = −4 6 + 2k 3 − 2k i.e., when k = 27 .
12 6 + 2k
¶ . AB = BA when 3 − 2k = −4 and −9 + 6k = 12
2. (3+2 pts) (a) Find parametric equations for the line of intersection of the two planes x + y + z = 3 and x − 2y + z = 0. µ ¶ µ ¶ µ ¶ µ ¶ 1 1 1 3 1 1 1 3 1 1 1 3 1 0 1 2 → → → : Parametric equations for 1 −2 1 0 0 −3 0 −3 0 1 0 1 0 1 0 1 2−t x 1 , where t ∈ <. the line L: y = z t (b) Let U = (a, b, c) and V = (x, y, z) be two vectors in <3 . Show that (ax + by + cz)2 ≤ (a2 + b2 + c2 )(x2 + y 2 + z 2 ). ||U ||2 = a2 + b2 + c2 , ||V ||2 = x2 + y 2 + y 2 , and (U · V )2 = (ax + by + cz)2 . Using Cauchy-Schwartz inequality we get: (ax + by + cz)2 ≤ (a2 + b2 + c2 )(x2 + y 2 + z 2 ). 3. (3+2 pts) (a) Show 1 2 3 0
that the vector U = (1, 2, 1, 1) does not belong to span{(1, 2, 3, 0), (2, 1, 6, 0)}. 2 1 1 2 has no solution. 6 1 0 1
(b) Let X = (10, 0, c), Y = (1, 0, 5) and Z = (0, −1, 0)} be three vectors in <3 . Determine all values of c for which the vector X is orthogonal to the plane containing the vectors Y and Z. A normal to the plane is: N = Y × Z = (5, 0, −1). Now X = αN i.e., (10, 0, c) = α(5, 0, −1) when c = −2. 4. (3+2 pts) (a) Does there exist a ∈ < such that S = {(1, 1, 1), (2, 2, 2), (3, a, 4)} is a basis for <3 . It is clear that (2, 2, 2) = 2(1, 1, 1). Therefore, no value for a ∈ < makes S a basis for <3 . (b) Let A be a 3 × 3 matrix such that the system AX = 0 has the following two vectors X1 = (1, 0, 1)T and X2 = (0, 1, 1)T as solutions. Find one more nontrivial solution to the system AX = 0. One more nontrivial solution: X1 + X2 = (1, 1, 2)T . 1 0 4 1 1 0 4 0 0 3 9 3 0 1 3 0 5. (3+1+2+1 pts) Let A = 0 0 0 2 . RREF: 0 0 0 1 0 0 0 3 0 0 0 0 (a) Find a basis for the null space of A. −4 −3 T X = t 1 , where t ∈ <. Thus a basis for the null space of A:{(−4, −3, 1, 0) } 0 (b) Find a basis for the column space of A containing only columns of A. A basis for the column space of A: {(1, 0, 0, 0)T , (0, 3, 0, 0)T , (1, 3, 2, 3)T )} (c) Determine whether the vector 1 0 1 0 1 0 0 3 3 3 0 1 0 0 2 2 → 0 0 0 0 3 3 0 0
X = (0, 3, 2, 3)T belongs to the column space of A. 0 −1 0 0 : Thus, (0, 3, 2, 3)T = −(1, 0, 0, 0) + (1, 3, 2, 3)T . 1 1 0 0
(d) Determine the rank and nullity of A. rank(A) = 3, nullity(A) = 1.
2 0 6. (7 pts) Let A = 0 1
0 2 0 0
0 0 1 0
0 0 . 0 1
If possible, find a non-singular matrix P and a diagonal matrix D such that P −1 AP = D. ¯ ¯ ¯ λ−2 0 0 0 ¯¯ ¯ ¯ 0 λ−2 0 0 ¯¯ |λI − A| = ¯¯ = (λ − 2)2 (λ − 1)2 . The eigenvalues of A are λ = 2, 2, 1, 1. 0 0 λ−1 0 ¯¯ ¯ ¯ −1 0 0 λ−1 ¯ For λ = 2:
0 0 (λI − A)X = 0 : 0 −1 where s, t ∈ <. Two linearly For λ = 1:
0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 independent
1 0 0 −1 0 t 0 1 0 0 1 0 0 s 1 0 → 0 0 0 0 0 . Thus X = 0 = s 0 + t 0 , 0 0 0 0 0 t 0 1 eigenvectors are associated with λ = 2, namely: (0, 1, 0, 0)T and (1, 0, 0, 1)T .
−1 0 0 0 0 1 0 0 0 0 0 0 −1 0 0 0 → 0 1 0 0 0 . Thus X = 0 = s (λI − A)X = 0 : 0 0 0 0 0 0 s 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 t where s, t ∈ <. Two linearly independent eigenvectors are associated with λ = 1, namely: (0, 0, 1, 0)T 0 1 0 0 2 0 0 0 1 0 0 0 0 2 0 0 P = 0 0 1 0 and D = 0 0 1 0 . 0 1 0 1 0 0 0 1
0 0 0 0 + t , 0 1 1 0 and (0, 0, 0, 1)T .
7. (6 pts) Answer each of the following as True or False (justify your answer). (a) Let A ne an n × n matrix. If 0 is an eigenvalue of A then rank(A) = n. FALSE (0 is an eigenvalue of A iff A is singular iff rank(A) < n). (b) If A and B are two n × n matrices with |A| = 1 and |B| = −1, then the columns of AB span
