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show that ( 1, 1, 1), ( 0, 1, 1) and ( 0, 1, -1) generate or spanned R^3

Let V be the subspace of R3 spanned by

{(1, 1, 0), (1, 1, 1)} and T : V —> V be defined by T(x1, x2, x3) = (0, x1, x2).

Find the kernel of T.


If V is a finite dimensional vector space and v not equal to 0 is a vector in V, show that there is a linear functional f E V* such that f(v) not equal to 0


Solve by Gaussian elimination method the
following system of equations :
x+y+z+t=5
x-y+z+t=1
x + z + t = 3

Find the radius and the centre of the circular section of the sphere I r I = 4 cut off by the plane r . (2i -j + 4k) = 3.


Let T : R3 -> R3 be the linear operator defined by T(x1, x2, x3) = (x1, x3, -2x2- x3).

Let f(x) = - x³+ 2.

Find the operator f(T)


Show that the vectors (3, 0, -3), (-1, 1, 2), (2, 1, 1) and (4, 2, -2) are linearly dependent in R3.


Suppose al = (1, 0, 1), a2 = (0, 1, -2) and  a3 = (-1, -1, 0) are vectors in R3 and  f : R3 -> R is a linear functional such that  f(al) = 1, f(a2) = -1 and f(a3) = 3.

If  a = (p,q,r) belongs to R3, find f(a). 


Let B = (a1, a2, a3) be an ordered basis of  R3 with a1 = (1, 0, -1), a2 = (1, 1, 1),  a3 = (1, 0, 0).

Write the vector v = (a, b, c) as  a linear combination of the basis vectors  from B.


Express V= 2t² + 5t + -5 as a linear combination of the polynomials
P1= t² + 2t + 1
P2= 2t + 5t + 4
P3 = t² + 3t + 6
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