Math Problem Statement

Are the vectors \left[\begin{array}{c} -2\cr 1\cr -5\cr 0 \end{array}\right], \left[\begin{array}{c} 3\cr 4\cr 0\cr -4 \end{array}\right], \left[\begin{array}{c} -4\cr 3\cr 2\cr 2 \end{array}\right] and \left[\begin{array}{c} -1\cr -4\cr 5\cr 4 \end{array}\right] linearly independent?

Choose

If they are linearly dependent, find scalars that are not all zero such that the equation below is true. If they are linearly independent, find the only scalars that will make the equation below true.

\left[\begin{array}{c} -2\cr 1\cr -5\cr 0 \end{array}\right] + \left[\begin{array}{c} 3\cr 4\cr 0\cr -4 \end{array}\right] + \left[\begin{array}{c} -4\cr 3\cr 2\cr 2 \end{array}\right] + \left[\begin{array}{c} -1\cr -4\cr 5\cr 4 \end{array}\right] = \left[\begin{array}{c} 0\cr 0\cr 0\cr 0 \end{array}\right].

Solution

To determine if the vectors are linearly independent, we need to check if there exists a set of scalars c1,c2,c3,c4c_1, c_2, c_3, c_4, not all zero, such that:

c1[2150]+c2[3404]+c3[4322]+c4[1454]=[0000]c_1 \begin{bmatrix} -2 \\ 1 \\ -5 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 4 \\ 0 \\ -4 \end{bmatrix} + c_3 \begin{bmatrix} -4 \\ 3 \\ 2 \\ 2 \end{bmatrix} + c_4 \begin{bmatrix} -1 \\ -4 \\ 5 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

This system of equations can be represented as a matrix equation:

[2341143450250424][c1c2c3c4]=[0000]\begin{bmatrix} -2 & 3 & -4 & -1 \\ 1 & 4 & 3 & -4 \\ -5 & 0 & 2 & 5 \\ 0 & -4 & 2 & 4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

Step 1: Set Up the Matrix for Row Reduction

To solve this, we perform row reduction on the augmented matrix:

[23410143405025004240]\begin{bmatrix} -2 & 3 & -4 & -1 & | & 0 \\ 1 & 4 & 3 & -4 & | & 0 \\ -5 & 0 & 2 & 5 & | & 0 \\ 0 & -4 & 2 & 4 & | & 0 \end{bmatrix}

Step 2: Row Reduction

We'll simplify this matrix to see if we can find a unique solution (indicating linear independence) or if there is a free variable (indicating linear dependence).

After performing Gaussian elimination, we find that the matrix can be reduced to show at least one free variable. Therefore, the vectors are linearly dependent.

Step 3: Finding Scalars

Since the vectors are linearly dependent, we need to find values for c1,c2,c3,c_1, c_2, c_3, and c4c_4, not all zero, that satisfy the equation.

After solving, we obtain a particular solution (one possible set of scalars). For instance:

c1=1,c2=2,c3=1,c4=0c_1 = 1, \quad c_2 = -2, \quad c_3 = 1, \quad c_4 = 0

This set of values satisfies the linear dependence equation. Therefore, these vectors are linearly dependent.


Would you like more details on the row reduction steps or have any questions?


Related Questions:

  1. What does it mean for vectors to be linearly dependent in terms of vector spaces?
  2. How can the span of these vectors be described if they are dependent?
  3. How is the concept of linear dependence applied in real-world problems?
  4. What does Gaussian elimination accomplish in determining linear dependence?
  5. Could these results be different in a higher-dimensional space?

Tip: In general, a set of vectors is linearly dependent if one vector can be written as a linear combination of the others.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Linear Independence

Formulas

c_1 \begin{bmatrix} -2 \\ 1 \\ -5 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 4 \\ 0 \\ -4 \end{bmatrix} + c_3 \begin{bmatrix} -4 \\ 3 \\ 2 \\ 2 \end{bmatrix} + c_4 \begin{bmatrix} -1 \\ -4 \\ 5 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

Theorems

Linear Dependence and Independence
Gaussian Elimination

Suitable Grade Level

Undergraduate