Question3. Determine whether the sets defined by the following vectors are subspaces of $\mathbf{R}^{3}$ . (a) $(a, b, 2 a+3 b)$

Studdy Solution

STEP 1

1. We are given a vector in the form $(a, b, 2a + 3b)$ .
2. We need to determine if the set of all such vectors forms a subspace of $\mathbf{R}^3$ .
3. A subspace must satisfy three conditions: it must contain the zero vector, be closed under vector addition, and be closed under scalar multiplication.

STEP 2

1. Check if the zero vector is in the set.
2. Check if the set is closed under vector addition.
3. Check if the set is closed under scalar multiplication.

STEP 3

Check if the zero vector $(0, 0, 0)$ is in the set.
Substitute $a = 0$ and $b = 0$ into the vector $(a, b, 2a + 3b)$ :
$(0, 0, 2 \times 0 + 3 \times 0) = (0, 0, 0)$
Since the zero vector is in the set, this condition is satisfied.

STEP 4

Check if the set is closed under vector addition.
Take two arbitrary vectors $(a_1, b_1, 2a_1 + 3b_1)$ and $(a_2, b_2, 2a_2 + 3b_2)$ from the set.
Add them:
$(a_1 + a_2, b_1 + b_2, (2a_1 + 3b_1) + (2a_2 + 3b_2))$
$= (a_1 + a_2, b_1 + b_2, 2(a_1 + a_2) + 3(b_1 + b_2))$
The result is in the same form as the original vector, so the set is closed under addition.

STEP 5

Check if the set is closed under scalar multiplication.
Take an arbitrary vector $(a, b, 2a + 3b)$ and a scalar $c$ .
Multiply the vector by the scalar:
$c \cdot (a, b, 2a + 3b) = (ca, cb, c(2a + 3b))$
$= (ca, cb, 2(ca) + 3(cb))$
The result is in the same form as the original vector, so the set is closed under scalar multiplication.
Since the set satisfies all three conditions, it is a subspace of $\mathbf{R}^3$ .

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