The problem asks which of the given sets $S_1$, $S_2$, and $S_3$ form a basis for $R^3$. $S_1 = \{(1, 0, 1), (0, -1, 1)\}$, $S_2 = \{(-1, -2, 1), (0, 2, -3), (1, -1, 0)\}$, and $S_3 = \{(1, 1, 2), (0, 1, 1), (1, 0, 1)\}$. A basis for $R^3$ must consist of 3 linearly independent vectors.

1. Problem Description

The problem asks which of the given sets $S_1$, $S_2$, and $S_3$ form a basis for $R^3$. $S_1 = \{(1, 0, 1), (0, -1, 1)\}$, $S_2 = \{(-1, -2, 1), (0, 2, -3), (1, -1, 0)\}$, and $S_3 = \{(1, 1, 2), (0, 1, 1), (1, 0, 1)\}$. A basis for $R^3$ must consist of 3 linearly independent vectors.

2. Solution Steps

To determine if a set of vectors forms a basis for $R^3$, we need to check if the vectors are linearly independent and span $R^3$. Since $R^3$ is a 3-dimensional vector space, any set of three linearly independent vectors will form a basis. We can check for linear independence by forming a matrix with the vectors as columns and computing the determinant. If the determinant is non-zero, the vectors are linearly independent.

For $S_1$, there are only two vectors. Since $R^3$ is a 3 dimensional space, a basis for $R^3$ requires 3 vectors. Thus $S_1$ cannot be a basis.

For $S_2$, we form a matrix with the vectors as columns:

$A_2 = \begin{bmatrix} -1 & 0 & 1 \\ -2 & 2 & -1 \\ 1 & -3 & 0 \end{bmatrix}$

We compute the determinant:

$det(A_2) = -1(2(0) - (-1)(-3)) - 0 + 1((-2)(-3) - 2(1)) = -1(0-3) + (6-2) = 3 + 4 = 7$

Since the determinant is non-zero, the vectors in $S_2$ are linearly independent and thus form a basis for $R^3$.

For $S_3$, we form a matrix with the vectors as columns:

$A_3 = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 2 & 1 & 1 \end{bmatrix}$

We compute the determinant:

$det(A_3) = 1(1(1) - 0(1)) - 0 + 1(1(1) - 1(2)) = 1(1-0) + (1-2) = 1 - 1 = 0$

Since the determinant is zero, the vectors in $S_3$ are linearly dependent and do not form a basis for $R^3$.

3. Final Answer

The set $S_2$ forms a basis of $R^3$.