Determine whether the set S5647 is linearly independent Does

Solution

(1) To check whether the set S = { (-5, 6), (4, 7) } is linearly independent.

Let A and B be scalars.

And let us set, A(-5, 6) + B(4, 7) = (0, 0)

That is, (-5A, 6A) + (4B, 7B) = (0, 0)

That is, (-5A + 4B, 6A + 7B) = (0, 0)

So, we get the equations as, -5A + 4B = 0 --------- (equation 1)

and 6A + 7B = 0 --------- (equation 2)

To solve them, let us multiply equation 1 by 6, so we get, -30A + 24B = 0

multiply equation 2 by 5, so we get, 30A + 35B = 0

Adding, ----------------------------

59 B = 0

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59 B = 0 implies, B = 0.

Pluggin in B = 0 in equation 1, we get -5A = 0 which implies A = 0

Hence, for any two scalars A and B, A(-5, 6) + B(4, 7) = (0, 0) implies A = 0 and B = 0

Hence the set S = {(-5, 6), (4, 7)} is linearly independent.

X----------X----------X

(2) To check whether the set S = { (3, 6, 4), (0, 3, 4), (0, 0, 7) } form a basis in R3

Let us check whether the set S is lineraly independent.

Let A, B a nd C be scalars.

Let us set, A(3, 6, 4) + B(0, 3, 4) + C(0, 0, 7) = (0, 0, 0)

(3A + 0B + 0C, 6A + 3B + 0C, 4A + 4B + 7C) = (0, 0, 0)

So we get, 3A = 0 ----------- (Equation 1)

6A + 3B = 0 ----------- (Equation 2)

4A + 4B + 7C = 0 ------------ (Equation 3)

Equation 1 implies, A = 0.

Plugging in A as 0 in equation 2, we get B = 0.

Plugging in A as 0 and B as 0 in equation 3, we get C = 0.

Hence, the set S is lineraly independent.

And any set of three linearly independent vectors in R3 spans R3.

Hence, the given set, S, form a basis in R3.