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The solution to problem 1 of exam #1 for the matrix methods course, which involves finding the row echelon form, lu decomposition, and general solution of a linear system. It also includes the definition of the range of a matrix and its basis, as well as solutions to problems 2 and 3, which cover even symmetric polynomials, vector subspaces, and properties of skew-symmetric matrices.
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Solution: APPM 3310: Matrix Methods — Exam #1 — June 17, 2011
Problem 1: (25 pts) Consider the linear system
x 1 + 2x 3 + 3x 4 = 4 −x 1 + x 2 + x 3 − 5 x 4 = − 4 2 x 1 + 2x 2 + 10x 3 + 3x 4 = 9
The system above has the compact matrix-vector form Ax = b.
(a) (4 pts) Find the row echelon form (REF) of the matrix A. What is the rank of the matrix? (b) Find the LU decomposition of the matrix A.(Write down both L and U.) (c) Find the general solution of the system Ax = b given above. (d) Write down the general definition of the range of an m × n matrix. Now, find a basis for range(A), what is the dimension of range(A)?
Solution: (a) We have
[ A | ~b ] =
1 0 2 3 b 1 − 1 1 1 − 5 b 2 2 2 10 3 b 3
1 0 2 3 b 1 0 1 3 − 2 b 1 + b 2 0 2 6 − 3 b 3 − 2 b 1
1 0 2 3 b 1 0 1 3 − 2 b 1 + b 2 0 0 0 1 b 3 − 4 b 1 − 2 b 2
So the REF of A is
(^) and so the rank of A is 3.
(b) Note E 1 =
(^) and so,
(^) and U =
(c) The augmented matrix
[ A | ~b ] =
and x 3 is free, so the general solution is
~x =
1 − 2 t 2 − 3 t t 1
t,^ ∀t^ ∈^ R.
(d) In general, if A is m × n, then range(A) = {b ∈ Rm
∣∣Ax = b, for some x ∈ Rn}.
Now by the Fundamental Theorem of Linear Algebra, if A is m × n with rank r then dim(range(A)) = r = 3, now note that since A is 3 × 4 we have that range(A) ⊂ R^3 and so
range(A) = R^3 = span
a basis for range(A) Problem 2: (35 pts) Let EP(4)^ denote the set of all even symmetric polynomials of degree ≤ 4, i.e.,
EP(4)^ ≡
p(x) = a 0 + a 1 x^2 + a 2 x^4
∣∣ ∀ a 0 , a 1 , a 2 ∈ R
(a) Prove that EP(4)^ is a subspace of P(4)^ (the space of all polynomials of degree ≤ 4). (b) What is the dimension of the subspace EP(4)? Justify your answer. (c) Given the vectors p 1 (x) = − 2 − 2 x^4 , p 2 (x) = 3 + 2x^2 + 5x^4 , and, p 3 (x) = 1 + x^2 + 3x^4 , is p 3 in the span of p 1 and p 2? Justify your answer. (d) Is the set {p 1 (x), p 2 (x), p 3 (x)} a basis of EP(4)? Justify your answer. (e) Let λ be a real number, for what values of λ will the vectors p 1 (x) = − 2 − 2 x^4 , p 2 (x) = 3 + 2x^2 + 5x^4 and p 3 (x) = 1 + x^2 + λx^4 be linearly independent? Justify your answer.
Solution: (a) Note that if p, q ∈ EP(4)^ then ∀c ∈ R,
p(x)+c·q(x) = (a 0 +a 1 x^2 +a 2 x^4 )+c·(b 0 +b 1 x^2 +b 2 x^4 ) = (a 0 +cb 0 )+(a 1 +cb 1 )x^2 +(a 2 +cb 2 )x^4 = d 0 +d 1 x^2 +d 2 x^4 ∈ EP(4)
and so, by closure, EP(4)^ is a subspace.
(b) Note that EP(4)^ = span
1 , x^2 , x^4
is a basis and so dim(EP(4)) = 3.
(c) Note that EP(4)^ = span
1 , x^2 , x^4
, and we wish to find c 1 and c 2 such that
c 1
(^) + c 2
(^) → inconsistent system
so p 3 ∈/ span {p 1 , p 2 }.
(d) Note that
det
(^) = det
and so since dim(EP(4)) = 3 and p 1 , p 2 , and p 3 are linearly independent we have a basis by a dimension argument. (Note that p 3 ∈/ span {p 1 , p 2 } does not imply that p 1 , p 2 , and p 3 are linearly independent since p 1 and p 2 could be linearly dependent.) (e) Here, (^)
− 2 5 λ
0 2 λ − 1
0 0 λ − 2
and so p 1 , p 2 , and p 3 will be linearly independent if λ 6 = 2.
Problem 3: (40 pts) Short answer. Answer the questions below.
(a) Let A be an invertible, n × n matrix. Show that (cA)−^1 =
c
A−^1 , ∀c ∈ R. Explain your answer.