would you be able to solve all six questions in the attachment file? The Vector
ID: 3081066 • Letter: W
Question
would you be able to solve all six questions in the attachment file?
The Vector Space Axioms u + v V u + v = v + u (u + v) + w = u + (v + w) There is a vector 0 such that u + 0 = u For any u there exists a vector -u such that u + -u = 0 cu V c(u + v) = cu + cv (c + d)u = cu + du (cd)u = c(du) lu = u Each question corresponds to the question of the same number on the test.. Given vectors u = (1,-2,1), v = (0,-1,0) and w = (2,1,1), Calculate the components of u - 4v + 2w Find a vector x such that u + x = 3w - v In each case, decide whether the set given is a vector space or not. If NOT, give a vector space axiom that is not satisfied in the set. If the set IS a vector space, describe the zero vector of the space. The set of all functions f(x) with domain [0,1] such that f is twice different and f"(x) = f'(x)-3xf(x). You do not need to solve this differential equation. The set of all power series such that the power series has a nonzero radius of convergence. Addition and scalar multiplication term-by-term. Prove or disprove each statement: The set V of polynomials a0 + a1x + a2x2 where at least one of the ai's is zero is a subspace of P2. The set of continuous functions f such that f(x) - f(x + 1) = for all x is a subspace of C(-infinity, +infinity), the set of all continuous functions defined on the whole real line. Given the set of matrices: in the vector space M2,2 of two by two matrices: Does S span M2,2 Given a complete reason. The set V of solutions to the system of linear equations Ax = 0 for any given 3 times 3 matrix A is a a subspace of R3. Remember that subspaces of R3 are (i) the zero vector only (ii) lines through the origin (iii) planes through the origin or (iv) all of R3 If det(A) 0, describe the solution space V as one of (i) through (iv) and give a reason. Given A below, find if possible a complete description of the vectors in V V and decide which type [(i) through (iv)] V is Find the adjoint matrix for the matrix A in question 5.Explanation / Answer
6) A= [1 -2 0]
[0 1 1]
[3 -4 2]
cofactor(aij) = (-1)^(i+j) *Minor(aij)
cofactor(1) = (-1)^1+1 *|1 1|
|-4 2|
=(2-(-4)) = 6
cofactor(-2) = (-1)^(1+2) *|0 1|
|3 2|
= -(0-3) = 3
In this way find remaining cofactors and form a cofactor matrix
Adjoint matrix is transpose of cofactor matrix
AdjA = [6 4 -2]
[3 2 -1]
[-3 -2 1]
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