PLEASE SOLVE QUESTION 2 TO 4 IN DETAILED Let A be the matrix given by A=(3 1 3 1

Question

PLEASE SOLVE QUESTION 2 TO 4 IN DETAILED

Let A be the matrix given by A=(3 1 3 1 2 1 0 10) Find Cartesian equations for Col(A) and Null(A). Are the columns of A linearly independent? Give full reasons for your answer (based on the definition of linear independence). Let V be the set of all real polynomials p of degree at most 2 satisfying p(0) = p(1), i.e. poly- ­nomials with the same values at x = 0 and x = 1. Use the subspace theorem to decide whether the above set V is a real vector space with the usual operations. Find the explicit form of all polynomial p epsilon V, that is, determine all a, b and c epsilon R such that p(x)=a+bx=cx^2 satisfies p(0) = p(l). Determine the dimension of the space V. The set R^3 of all column vectors of length three, with real entries, is a vector space. Is the subset B={(x y z) epsilon R^3 vertical line x^2-y^2+z^2=0} a subspace of R^3? Justify your answer. Show that the set of all twice differentiable functions f: R right arrow R satisfying the differential equation x^3f"(x)-tan(x)f'(x)-sin(x)f(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.

Explanation / Answer

|A|=0 hence Ax=0 has infinite solutions

Ax=0 gives 3x+y =0 and x+2y+z =0

Simplify to get 6x-z=0 Or z =6x and y =-3x

Hence (x,y,z) = (1, -3, 6) is the null space

ii) No not linearly independent as determinant =0

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p(0) = p(1)

If p1 and p2 are two polynomials in this set

p1+p2(0) = (p1+p2)1 hence closure is true.

Identity is 0 polynomial and inverse for p is -p

Also polynomial addition is transitive

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0 p(x) = 0

1 p(x)= p(x)

cd(x) = c(d(x))

Also distributive

Hence forms a vector space

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p(x) = a+bx+cx2

p(0) = p(1) gives

a = a+b+c or b+c =0

Hence p(x) = a+bx-bx2

c) Dim space V = 2

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3) if (x,y,z) and (x1,y1,z1) belong to B

then (x+x1)^2+.... need not be 0 hence not a vector space

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4) If f and g satisfy that DE, then f+g also satisfies hence closed under addition

identity is 0 function and inverse is -f

1*f = f and 0*f = 0

Also distributive over addition and scalar multiplication

Hence a vector space.

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