Understanding Basis and Linear Independence in R³

To determine if a set of vectors can be enlarged to a basis for $\text{R}^{3}$, we first need to understand the concept of a basis and linear independence. A basis for a vector space like $\text{R}^{3}$ is a set of vectors that are:

• Linearly Independent: No vector in the set can be written as a linear combination of the others.
• Span the space: Any vector in the space can be written as a linear combination of the vectors in the set.

For $\text{R}^{3}$, any basis must contain exactly three linearly independent vectors. If a given set of vectors is linearly independent and its size is less than the dimension of the space (i.e., less than 3 for $\text{R}^{3}$), then it can be enlarged to a basis by adding more linearly independent vectors. However, if the given set of vectors is linearly dependent, it cannot be enlarged to a basis, because a basis must always consist of linearly independent vectors.

Analyzing Subset S₁ for Basis Enlargement

We are given the first subset $\text{S}_{1} = \{[1, -1, 2], [3, 2, -1]\}$. To check if $\text{S}_{1}$ can be enlarged to a basis for $\text{R}^{3}$, we must first determine if its vectors are linearly independent. For two vectors, they are linearly dependent if and only if one is a scalar multiple of the other. Let's check if there is a scalar $c$ such that $[3, 2, -1] = c \cdot [1, -1, 2]$.

• From the first component: $3 = c \cdot 1 \implies c = 3$
• From the second component: $2 = c \cdot (-1) \implies c = -2$
• From the third component: $-1 = c \cdot 2 \implies c = -\frac{1}{2}$

Since we obtain different values for $c$ for each component ($3, -2, -\frac{1}{2}$), there is no single scalar $c$ that satisfies the equation. This means that the vector $[3, 2, -1]$ is not a scalar multiple of $[1, -1, 2]$.

Therefore, the two vectors in $\text{S}_{1}$ are linearly independent. Since $\text{S}_{1}$ contains two linearly independent vectors in $\text{R}^{3}$ (and $2 < 3$), it means that these vectors form a linearly independent set that does not yet span $\text{R}^{3}$. We can add one more vector that is linearly independent of $[1, -1, 2]$ and $[3, 2, -1]$ to form a basis for $\text{R}^{3}$. Thus, $\text{S}_{1}$ can be enlarged to a basis for $\text{R}^{3}$.

Analyzing Subset S₂ for Basis Enlargement

Next, we consider the second subset $\text{S}_{2} = \{[2, 7, -3], [-6, -21, 9]\}$. We apply the same test for linear independence. Let's check if there is a scalar $k$ such that $[-6, -21, 9] = k \cdot [2, 7, -3]$.

• From the first component: $-6 = k \cdot 2 \implies k = -3$
• From the second component: $-21 = k \cdot 7 \implies k = -3$
• From the third component: $9 = k \cdot (-3) \implies k = -3$

In this case, we find a consistent scalar value $k = -3$ across all components. This means that $[-6, -21, 9] = -3 \cdot [2, 7, -3]$.

Therefore, the two vectors in $\text{S}_{2}$ are linearly dependent because one vector is a scalar multiple of the other. A set of vectors that is linearly dependent cannot be part of a basis for a vector space. If we add more vectors to a linearly dependent set, the resulting set will still be linearly dependent (unless we remove the dependent vectors, which is not what "enlarge" implies). Hence, $\text{S}_{2}$ cannot be enlarged to a basis for $\text{R}^{3}$.

Conclusion for Both Subsets

Summarizing our findings:

• The subset $\text{S}_{1}$ consists of linearly independent vectors, which means it has the potential to be part of a basis, and since it has fewer than 3 vectors, it can be enlarged to a basis for $\text{R}^{3}$.
• The subset $\text{S}_{2}$ consists of linearly dependent vectors, which means it cannot form a basis, nor can it be enlarged to a basis for $\text{R}^{3}$.

Based on this analysis, $\text{S}_{1}$ can be enlarged but $\text{S}_{2}$ cannot be enlarged to a basis for $\text{R}^{3}$.