Determining Linear Independence of Vectors Using Gaussian Elimination and Cross Product

Linear independence is a fundamental concept in linear algebra that helps us understand the structure of vector spaces. A set of vectors is said to be linearly independent if no vector in the set can be expressed as a linear combination of the others. In this article, we will explore how to determine if the set of vectors and are linearly independent using Gaussian elimination and the cross product method.

Introduction to Linear Independence

Formally, the vectors and are linearly independent if the only solution to the equation c_1 c_2 is c_1 0 and c_2 0. This means that the only way to obtain the zero vector using a linear combination of these vectors is by having both coefficients as zero.

Method 1: Using Gaussian Elimination

Step 1: Set Up the Equation

We can express the equation as:

c_1 1 2 3 - c_2 2 -2 0  0 0 0

This leads to the following system of equations:

c_1 - 2c_2 0, for the first component 2c_1 - 2c_2 0, for the second component 3c_1 0c_2 0, for the third component

Step 2: Solve the System of Equations

From the third equation, we have:

3c_1 0 implies c_1 0

Substituting c_1 0 into the first equation gives:

0 - 2c_2 0 implies c_2 0

The second equation becomes:

2(0) - 2c_2 0 implies 0 0

Step 3: Conclusion

Since the only solution to the system is c_1 0 and c_2 0, the vectors and are linearly independent.

Method 2: Using the Cross Product

Another way to show linear independence for two vectors in is to check if they are not scalar multiples of each other. We can calculate the cross product of the two vectors:

mathbf{v_1} times mathbf{v_2}  | mathbf{i} mathbf{j} mathbf{k}  1   2   3  2  -2  0 |

This determinant can be calculated as follows:

mathbf{v_1} times mathbf{v_2} mathbf{i}2 cdot 0 - 3 cdot -2 - mathbf{j}1 cdot 0 - 3 cdot 2 mathbf{k}1 cdot -2 - 2 cdot 2

Which simplifies to:

mathbf{i}0 6 - mathbf{j}0 - 6 mathbf{k}-2 - 4

6mathbf{i} 6mathbf{j} - 6mathbf{k} 6 6 -6

Since the cross product is not the zero vector , the vectors are linearly independent.

Alternative Method: Simplified Linear Combination

Let the linear combination be:

[a 1 2 3] [b 2 -2 0] [2a -2b 3a] [0 0 0]

This leads to the following system of equations:

2a - 2b 0 — 1 3a 0 — 3

From equation 3a 0, we get: a 0.

Substituting a 0 into equation 2a - 2b 0 gives:

0 - 2b 0 implies b 0

Hence, the vectors and are independent vectors.

Understanding these methods is crucial for linear algebra and has applications in various fields such as physics, computer graphics, and engineering.