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MATH 304 Midterm: MATH 304 Binghamton Math304 Fall2018 PrEx1

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turquoisegnu128

31 Jan 2019

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Binghamton University

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Mathematical Science

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Notations: rm n is the vector space of all m n real matrices, and rm = rm. 1 . (1) (15 points) let la : r4 r4 be the linear function la(x) = ax associated with the matrix. What does this tell you about lk for k 5? (3) (15 points) answer each question separately (a) if a rm n , b rn p are matrices such that the associated functions. 2 such that l = la, and check that l(x) = la(x) = ax for all x r2. Practice exam 1 solutions: (a) (6 points) to nd ker(l) we must solve a linear system by row reducing. 0 so x1 = 2r + 3s x2 = r 2s x3 = r r x4 = s r. 0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (b) (6 points) y = [yi] range(l) i the following system is consistent: 3y1 2y2 + y3 y1 + y2 y4 is consistent i .

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Related Questions

K(x,y,z) = (x, x + y,x + y + z)

L(x,y,z) = (2x - y, x + 2y)

S(x,y,z) = (z,y,x)

T(x,y) = (2x + y,x + y,x - y,x - 2y)

G: R^3 -> M(2,2) defined by G(x,y,z) = (y z , (-x + y) -x)

H: M(2,2) -> P2 defined by H(a b, c d) = (a + b) + (a - 2d)x+ dx^2

Find the indicated linear transformation if it is defined. ifnot defined explain why it is not:

1) K + S

2) GS

3)2S^3 - S^2 + 3S - 4I

Find the matrix that represent each of the linear transformationK,L,S,and T for the following

1)TL

2)K + S

Find the matrix that represent each of linear transformation intwo ways, first using the result normal way and using therepresention of the matrix.

1)

T:R^3 --> R^4 given by T(X) = Ax, where

A = [1 1 0

2 0 3

-1 1 1

1 1 2]

and v1 = (-3,3,2) v2 = (0,0,0)

2) T: R^4 --> R given by T(x1,x2,x3,x4) = x1 + 2x2 + 3x3 +4x4 and v1 = (1,2,1,-2) v2 = (2,3,0,-2)

3)T: M(2,3) --> P2

P2 = (a11 + a13)x^2 + (a21 - a22)x + a23

v1 = [-1 3 1

2 2 0]

v2 = [2 1 -2

1 1 0]

& determine wheter w1 or w2 is the iamge of thecorresponding linear transformation

w1 = x^2 + 2x + 1,w2 = x - 2.

true false

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ceruleangoose957

I have the answer for the following question. But some of theanswers are not correct.

Please do the following question by yourself and check which arethe wrong answers,please post your answers and compare to it.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S