Final answer:
a) {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for ( mathbbR^3 ). b) {(1, 2, 3), (2, 4, 6), (3, 6, 9)} is not a basis but a spanning set. c) {(1, 1, 0), (0, 1, 1), (1, 0, 1)} is a spanning set but not linearly independent. d) {(1, 0, 0), (1, 1, 1), (0, 1, 0)} is linearly independent but not a basis.
Step-by-step explanation:
a. The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for (mathbbR^3) because it forms the standard unit vectors along the x-axis, y-axis, and z-axis respectively. These vectors are linearly independent and span the entire space of (mathbbR^3).
b. The set {(1, 2, 3), (2, 4, 6), (3, 6, 9)} is not a basis for (mathbbR^3) because all three vectors are parallel and lie on the same line. Therefore, they are linearly dependent and do not span the entire space.
c. The set {(1, 1, 0), (0, 1, 1), (1, 0, 1)} is a spanning set for (mathbbR^3) because any vector in (mathbbR^3) can be written as a linear combination of these three vectors. However, this set is not linearly independent as the third vector can be expressed as the sum of the first two vectors.
d. The set {(1, 0, 0), (1, 1, 1), (0, 1, 0)} is a linearly independent set but is not a basis for (mathbbR^3) because it does not span the entire space. These vectors form a plane in (mathbbR^3).