Math Problem Statement

  1. For each set of vectors in R 3 given below, describe geometrically the set of all linear combinations ( line, plane or all of R 3 ). (a) {(1, −1, 1),(1, 2, 0),(4, 5, 1)} (b) {(1, 2, −3),(−2, −4, 6),(−3, −6, 9)} (c) {(2, −1, 0),(−1, 2, −1),(0, −1, 2)}

Solution

To describe the geometric set of all linear combinations of the given vectors in R3\mathbb{R}^3, we need to determine the span of each set of vectors. The span of a set of vectors can describe a point, a line, a plane, or all of R3\mathbb{R}^3, depending on the linear independence of the vectors.

Let's analyze each set:

(a) {(1,1,1),(1,2,0),(4,5,1)}\{(1, -1, 1), (1, 2, 0), (4, 5, 1)\}

To determine the geometric description, we first check if these vectors are linearly independent by finding the determinant of the matrix formed by these vectors as columns.

Matrix A=(114125101)\text{Matrix } A = \begin{pmatrix} 1 & 1 & 4 \\ -1 & 2 & 5 \\ 1 & 0 & 1 \end{pmatrix}

Calculate the determinant:

det(A)=1(2150)1(1151)+4(1021)\text{det}(A) = 1 \cdot (2 \cdot 1 - 5 \cdot 0) - 1 \cdot (-1 \cdot 1 - 5 \cdot 1) + 4 \cdot (-1 \cdot 0 - 2 \cdot 1)

=121(6)+4(2)= 1 \cdot 2 - 1 \cdot (-6) + 4 \cdot (-2)

=2+68=0= 2 + 6 - 8 = 0

Since the determinant is zero, the vectors are linearly dependent. Therefore, the set of all linear combinations of these vectors forms a plane in R3\mathbb{R}^3.

(b) {(1,2,3),(2,4,6),(3,6,9)}\{(1, 2, -3), (-2, -4, 6), (-3, -6, 9)\}

Examine the linear dependence of these vectors. Notice that:

(2,4,6)=2(1,2,3)(-2, -4, 6) = -2 \cdot (1, 2, -3) (3,6,9)=3(1,2,3)(-3, -6, 9) = -3 \cdot (1, 2, -3)

All vectors are scalar multiples of each other, meaning they are linearly dependent. The span of these vectors is a line in R3\mathbb{R}^3, as they are all collinear.

(c) {(2,1,0),(1,2,1),(0,1,2)}\{(2, -1, 0), (-1, 2, -1), (0, -1, 2)\}

Again, let's check for linear independence by calculating the determinant of the matrix formed by these vectors:

Matrix B=(210121012)\text{Matrix } B = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix}

Calculate the determinant:

det(B)=2(22(1)(1))(1)(12(1)0)+0((1)(1)20)\text{det}(B) = 2 \cdot (2 \cdot 2 - (-1) \cdot (-1)) - (-1) \cdot (-1 \cdot 2 - (-1) \cdot 0) + 0 \cdot ((-1) \cdot (-1) - 2 \cdot 0)

=2(41)(1)(2)= 2 \cdot (4 - 1) - (-1) \cdot (-2)

=232=62=4= 2 \cdot 3 - 2 = 6 - 2 = 4

Since the determinant is non-zero, these vectors are linearly independent. Therefore, the span of these vectors is all of R3\mathbb{R}^3.

Summary

  • (a) The set of all linear combinations forms a plane in R3\mathbb{R}^3.
  • (b) The set of all linear combinations forms a line in R3\mathbb{R}^3.
  • (c) The set of all linear combinations spans all of R3\mathbb{R}^3.

Would you like more details on any part of this explanation, or do you have any questions?


Here are 5 related questions to expand on this topic:

  1. How do you determine if vectors in Rn\mathbb{R}^n are linearly dependent or independent?
  2. What does the span of a set of vectors represent geometrically?
  3. Can you explain the significance of the determinant in relation to vector independence?
  4. What are the possible geometric interpretations of the span of vectors in R2\mathbb{R}^2?
  5. How does the concept of linear independence extend to higher dimensions beyond R3\mathbb{R}^3?

Tip: When calculating determinants for larger matrices, breaking down the determinant into smaller parts using cofactor expansion can simplify the process.

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Math Problem Analysis

Mathematical Concepts

Linear algebra
Vectors
Span

Formulas

Determinant calculation for 3x3 matrices

Theorems

Linear dependence and independence in vector spaces

Suitable Grade Level

Advanced undergraduate level