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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.
P1= t² + 2t + 1
P2= 2t + 5t + 4
P3 = t² + 3t + 6
V1= 1, 1, -2, -2
V2= 2, -3, 0, 2
V3= -2, 0, 2, 2
V4= 3, -3 -2, 2
Express 5,9-4 as a linear combination of the vectors
V1= ( 2, 4, -5)
V2= (3, 7, -8)
V3= (7, 6, 1)
(0, 9)? Justify your answer.
It is possible for a set S of three vectors from R^3 to be linearly dependent, even though every subset of two vectors from LaTeX: S\:Sis linearly independent
with partial pivoting:
2x1 −x2 +x3 = 4
3x1 +2x2 −4x3 = 1
x1 +4x2 −2x3 = 2
#340153 on Dec 2023