Problemas álgebra lineal
Páginas: 23 (5663 palabras)
Publicado: 7 de enero de 2012
Linear Algebra. Degrees in Engineering. Solved problems.
Chapters 1 & 2.
Questions
Q1.- Is the following statement true or false? Justify your answer by citing appropriate facts or theorems if true or, if false, explain why or give a counterexample: If a set is linearly dependent, then the set contains more vectors that there are entries in the vectors. Answer: It is false. The set 2 1 0 , 0 S= 0 0
is dependent (the second vector is two times the first one) and has only two R3 vectors. Q2.- If A, B and C are n × n invertible matrices, does the equation C −1 (A + X)B −1 = In have a solution? (In is the n × n identity matrix). If so, find it. Answer: Multiplying by C on the left and by B on the right we get A + X = CB, from where we find X = CB − A. Q3.- Is thefollowing statement true or false? Justify your answer by citing appropriate facts or theorems if true or, if false, explain why or give a counterexample: If a set contains fewer vectors that there are entries in the vectors, then the set is linearly independent. Answer: It is false. The set 2 1 S = 0 , 0 0 0
has two R3 vectors and is dependent (the second vector is twotimes the first one). Q4.- Suppose A and B are n × n matrices, B is invertible, and AB is invertible. Show that A is invertible. Answer: Let us call C = AB. By hypothesis, B is invertible, and we can multiply by both sides of the previous equation by B −1 . We thus obtain CB −1 = A. Therefore, A is the product of two invertible matrices and hence it is itself invertible. Q5.- Let B, C and D bematrices. If BC = BD, then C = D. Answer: False. If B is not invertible, it can not be taken out of the equation. Q6.- If A is a 3 × 3 matrix and the equation Ax = (1 0 0)T has a unique solution, then A is invertible. Answer: True. For the system to have a unique solution, A must have 3 pivots and, being 3 × 3 it is henceforth invertible. 1
Problems
P1.- Let T : R3 −→ R2 , given by
x1 T x2 = x3
x1 − 5x2 + 4x3 x2 − 6x3
Show that T is a linear transformation by finding a matrix that implements it and discuss whether T is onto, one-to-one, both or none of them. Answer: To find the matrix one has to compute T (e1 ), T (e2 ) and T (e3 ), i.e., the transformation of the columns of the identity, which are straightforwardly obtained as T (e1 ) = 1 0 , T (e2 ) = −5 1 , T (e3 ) = 4−6 ,
and hence the matrix is given by A= 1 −5 4 0 1 −6
The matrix has obviously two pivots, so the system Ax = b will always be consistent, and the transformation is onto (its range is all of R2 ). It cannot be one-to-one because two many vectors are mapped to the same one in R2 , as the system will always have a free variable. P2.- Let A be given below. Compute A−1 . By row operationsone finds A−1 8 3 1 4 1 = 10 7/2 3/2 1/2
1 0 −2 1 4 A = −3 2 −3 4
P3.- Let T : R3 −→ R3 , with T (x) = Ax, and 1 −4 2 A= 0 3 5 8 −4 −2 Find Range(T ) and discuss whether T is onto, one-to-one, both or none of them. Answer: To find Range(T ) is to find the set of vectors b ∈ R3 such that the equation Ax = b has a solution. The augmented matrix of the system is 1 −4 2b1 1 −4 2 b1 0 0 3 5 b2 3 5 b2 −2 8 −4 b3 0 0 0 (b3 + 2b1 ) Therefore, consistency of the system requires b3 + 2b1 = 0 ⇒ b1 = −b3 /2, b2 free, and 0 −1 Range(T ) = Span 0 , 1 2 0 2
T is neither onto (its range is not all of R3 ) nor one-to-one (the system has an infinite number of solutions and therefore an infinite number of vectors that share everyimage). P4.- Let A be given below. Discuss without any calculation why it should be invertible. Compute A−1 . 5 0 0 A = −3 1 0 8 5 −1 The matrix is invertible because its columns form clearly an independent set. By row operations one finds 1/5 0 0 A−1 = 3/5 1 0 23/5 5 −1 P5.- Let A be given below. Compute A−1 . 1 2 3 A = 1 1 2 . 0 1 2
Answer: The matrix is invertible...
Chapters 1 & 2.
Questions
Q1.- Is the following statement true or false? Justify your answer by citing appropriate facts or theorems if true or, if false, explain why or give a counterexample: If a set is linearly dependent, then the set contains more vectors that there are entries in the vectors. Answer: It is false. The set 2 1 0 , 0 S= 0 0
is dependent (the second vector is two times the first one) and has only two R3 vectors. Q2.- If A, B and C are n × n invertible matrices, does the equation C −1 (A + X)B −1 = In have a solution? (In is the n × n identity matrix). If so, find it. Answer: Multiplying by C on the left and by B on the right we get A + X = CB, from where we find X = CB − A. Q3.- Is thefollowing statement true or false? Justify your answer by citing appropriate facts or theorems if true or, if false, explain why or give a counterexample: If a set contains fewer vectors that there are entries in the vectors, then the set is linearly independent. Answer: It is false. The set 2 1 S = 0 , 0 0 0
has two R3 vectors and is dependent (the second vector is twotimes the first one). Q4.- Suppose A and B are n × n matrices, B is invertible, and AB is invertible. Show that A is invertible. Answer: Let us call C = AB. By hypothesis, B is invertible, and we can multiply by both sides of the previous equation by B −1 . We thus obtain CB −1 = A. Therefore, A is the product of two invertible matrices and hence it is itself invertible. Q5.- Let B, C and D bematrices. If BC = BD, then C = D. Answer: False. If B is not invertible, it can not be taken out of the equation. Q6.- If A is a 3 × 3 matrix and the equation Ax = (1 0 0)T has a unique solution, then A is invertible. Answer: True. For the system to have a unique solution, A must have 3 pivots and, being 3 × 3 it is henceforth invertible. 1
Problems
P1.- Let T : R3 −→ R2 , given by
x1 T x2 = x3
x1 − 5x2 + 4x3 x2 − 6x3
Show that T is a linear transformation by finding a matrix that implements it and discuss whether T is onto, one-to-one, both or none of them. Answer: To find the matrix one has to compute T (e1 ), T (e2 ) and T (e3 ), i.e., the transformation of the columns of the identity, which are straightforwardly obtained as T (e1 ) = 1 0 , T (e2 ) = −5 1 , T (e3 ) = 4−6 ,
and hence the matrix is given by A= 1 −5 4 0 1 −6
The matrix has obviously two pivots, so the system Ax = b will always be consistent, and the transformation is onto (its range is all of R2 ). It cannot be one-to-one because two many vectors are mapped to the same one in R2 , as the system will always have a free variable. P2.- Let A be given below. Compute A−1 . By row operationsone finds A−1 8 3 1 4 1 = 10 7/2 3/2 1/2
1 0 −2 1 4 A = −3 2 −3 4
P3.- Let T : R3 −→ R3 , with T (x) = Ax, and 1 −4 2 A= 0 3 5 8 −4 −2 Find Range(T ) and discuss whether T is onto, one-to-one, both or none of them. Answer: To find Range(T ) is to find the set of vectors b ∈ R3 such that the equation Ax = b has a solution. The augmented matrix of the system is 1 −4 2b1 1 −4 2 b1 0 0 3 5 b2 3 5 b2 −2 8 −4 b3 0 0 0 (b3 + 2b1 ) Therefore, consistency of the system requires b3 + 2b1 = 0 ⇒ b1 = −b3 /2, b2 free, and 0 −1 Range(T ) = Span 0 , 1 2 0 2
T is neither onto (its range is not all of R3 ) nor one-to-one (the system has an infinite number of solutions and therefore an infinite number of vectors that share everyimage). P4.- Let A be given below. Discuss without any calculation why it should be invertible. Compute A−1 . 5 0 0 A = −3 1 0 8 5 −1 The matrix is invertible because its columns form clearly an independent set. By row operations one finds 1/5 0 0 A−1 = 3/5 1 0 23/5 5 −1 P5.- Let A be given below. Compute A−1 . 1 2 3 A = 1 1 2 . 0 1 2
Answer: The matrix is invertible...
Leer documento completo
Regístrate para leer el documento completo.