1) Show the following vectors are linearly independent:

a) (π, 0) and (0, 1)

b) (1, 1, 0), (1, 1, 1) and (0, 1, -1)

* Solutions*:

We write out: a (π, 0) + b (0, 1) = (0,0)

i.e. to show linear independence we require the
sum of the linear combination to be zero, and hence a = 0, b = 0.

Then:

πa + b(0) = 0

a(0) + b = 0

from which we see: b = 0 and also a = 0

Hence, linear independence applies.

b) (1, 1, 0), (1, 1, 1) and (0, 1, -1)

We are to generalize to 3 dimensions here so:

a(1,1,0) + b(1,1,1) + c(0, 1, -1) = (0, 0 , 0)

Whence we find:

1) a + b = 0

2) a + b + c = 0

3) 0 + b - c = 0

Subtract (1) from (2): c = 0

From (3): b = c = 0

From (1): a = 0

Hence, linear independence applies

2) Express X as a linear combination of the given vectors A, B and find the coordinates of X with respect to A, B:

a) X = (1, 0), A = (1, 1), B = (0, 1)

b) X = (1,1), A = (2, 1), B = (-1, 0)

__Solutions:__

Write: a (1,1) + b(0, 1) = (1, 0)

Which leads to:

a + b(0) = 1

a + b = 0

Thus: a = 1 and b = -a = -1

So the coordinates are: (1, -1)

b) X = (1,1), A = (2, 1), B = (-1, 0)

Write the linear combination:

a (2,1) + b(-1, 0) = (1, 1)

This leads to:

2a - b = 1

a + 0 = 1

Therefore: a = 1 and b = 2a - 1 = 2(1) -1 = 1

The coordinates are (1,1)

3) Show the following vectors are linearly independent over **C** or **R**:

a) (1, 1, 1) and (0, 1, -2)

b) (-1, 1, 0) and (0, 1, 2)

c) (0, 1, 1), (0, 2, 1) and (1, 5, 3)

__Solutions:__

To show linear independence we require the sum of the linear combination to be zero, and hence a = 0, b = 0, c=0 in each case

a) Write out: a (1,1, 1) + b(0, 1, -2) = (0, 0, 0)

Then: a1 + b 0 = 0, a1 + b1 = 0, a1 + b2 = 0

Or in normal form:

a = 0

a + b = 0

a - 2b = 0

From which we find a = 0 and also b = 0. Hence, linear independence applies.

b) Write out: a (-1,1, 0) + b(0, 1, 2) = (0, 0, 0)

Then: - a1 + b 0 = 0, a1 + b1 = 0, a0 + b2 = 0

Or in normal form:

- a + 0 = 0

a + b = 0

0 - 2b = 0

From which we find a = 0 and also b = 0. Hence, linear independence applies.

c)Write out:

a (0, 1, 1) + b(0, 2, 1) + c(1, 5 , 3 ) = (0, 0, 0)

Then: a0 + b0 + c1 = 0, a1 + b2 + c5 =0, a1 + b1 + c3 =0

Or in normal form:

0 + 0 + c = 0

a + 2b + 5c = 0

a + b + 3c = 0

From which we can see: c= 0, a = 0, b = 0

Hence, linear independence applies.

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