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MATH 21 Study Guide - Midterm Guide: Solution Set, Coordinate Vector, Zero Element

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20 Jul 2018

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What if implies: (1) pivot columns (2) check that there is a pivot in each column, determines the one solution. Express solution set in the form of +(cid:1845)(cid:1853) (cid:4666)(cid:1871)(cid:4667) is the solution and (cid:1845) is the basis of null space of given matrix. Free variables are (cid:1876)(cid:2871),(cid:1876)(cid:2873) (cid:1876)(cid:2871)=(cid:1871), (cid:1876)(cid:2873)=(cid:1872) ( (cid:1876)(cid:2869)(cid:1876)(cid:2870)(cid:1876)(cid:2871)(cid:1876)(cid:2872)(cid:1876)(cid:2873)) =( (cid:886) (cid:884)(cid:1845)(cid:1872)+(cid:1872) (cid:887)+(cid:886)(cid:1845)+(cid:885)(cid:1872) (cid:1845) (cid:885)+(cid:884)(cid:1872) (cid:1872) ( (cid:886)(cid:887)(cid:882) (cid:885)(cid:888)) +(cid:1845)( (cid:884)(cid:886)(cid:883)(cid:882)(cid:882)) +(cid:1872)( (cid:883)(cid:885)(cid:882)(cid:884)(cid:883)) Let (cid:1827)=(cid:885) (cid:886) matrix ( (cid:884)(cid:886)(cid:883)(cid:882)(cid:882)) and ( (cid:883)(cid:885)(cid:882)(cid:884)(cid:883)) are the basis of the null space (span) (cid:1876)+(cid:1877)= distinct 4 vectors (cid:1827)(cid:1876)=(cid:1827)(cid:1877)=((cid:883)(cid:884)(cid:885)) Choose the value of (cid:1876) that makes last column the pivot column (zero row) Determine the coordinate vector (cid:3438)(cid:886)(cid:884)(cid:885)(cid:885)) with respect to (cid:1828) Given 3 vectors that are basis of subspace =(cid:3438)(cid:1876)(cid:2869)(cid:1876)(cid:2870)(cid:1876)(cid:2871)(cid:1876)(cid:2872))=(cid:1876)(cid:2869)+(cid:1876)(cid:2870) (cid:1876)(cid:2871) (cid:1876)(cid:2872)=(cid:882) Use gaussian elimination (cid:883) (cid:883) (cid:884)|(cid:886)(cid:884)(cid:885)(cid:885)) (cid:3438)(cid:883) (cid:882) (cid:882) (cid:882) (cid:882) (cid:882)| (cid:886)(cid:883)(cid:885)(cid:882)) (cid:3438)(cid:883) (cid:884) (cid:884) (cid:882) (cid:883) (cid:882) (cid:883) (cid:885) (cid:883) (cid:882) (cid:882) (cid:883) (cid:883) (cid:886) (cid:883)

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Related Questions

1. Consider the augmented matrix: (linear algebra)

a) Convert the matrix to echelon form

b) Obtain the reduced echelon form from the previous step

c) write a solution set for the corresponding system of equations

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

This is Linear Algebra, please help!