Scalar Proof: No Solution For Linear Combination

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Introduction

Hey guys! Today, we're diving into a cool problem from linear algebra. We need to prove that there are no scalars c1{c_1}, c2{c_2}, and c3{c_3} that satisfy the equation:

c1(1,0,1,0)+c2(1,0,βˆ’2,1)+c3(2,0,1,2)=(1,βˆ’2,2,3).{c_1(1,0,1,0) + c_2(1,0,-2,1) + c_3(2,0,1,2) = (1, -2, 2, 3).}

In simpler terms, we want to show that it’s impossible to find numbers c1{c_1}, c2{c_2}, and c3{c_3} that, when multiplied by the given vectors and added together, result in the vector (1, -2, 2, 3). This involves setting up a system of linear equations and demonstrating that the system is inconsistent, meaning it has no solution. Let’s break it down step by step to make it super clear. This problem highlights the importance of linear combinations and the conditions under which a vector can be expressed as a combination of other vectors. Understanding these concepts is crucial for grasping more advanced topics in linear algebra, such as vector spaces, linear independence, and basis vectors. So, let's get started and show that no such scalars exist!

Setting Up the Equations

First, let's set up the system of linear equations from the given vector equation. We can rewrite the equation as:

(c1+c2+2c3,0,c1βˆ’2c2+c3,c2+2c3)=(1,βˆ’2,2,3).{(c_1 + c_2 + 2c_3, 0, c_1 - 2c_2 + c_3, c_2 + 2c_3) = (1, -2, 2, 3).}

This gives us the following system of equations:

  1. c1+c2+2c3=1{c_1 + c_2 + 2c_3 = 1}
  2. 0=βˆ’2{0 = -2}
  3. c1βˆ’2c2+c3=2{c_1 - 2c_2 + c_3 = 2}
  4. c2+2c3=3{c_2 + 2c_3 = 3}

Notice anything strange? The second equation, 0=βˆ’2{0 = -2}, is clearly a contradiction. This immediately tells us that the system of equations is inconsistent, meaning there are no values for c1{c_1}, c2{c_2}, and c3{c_3} that can satisfy all four equations simultaneously. This contradiction arises because the target vector (1, -2, 2, 3) cannot be expressed as a linear combination of the given vectors (1,0,1,0), (1,0,-2,1), and (2,0,1,2). The structure of these vectors, particularly the presence of a zero in the second component of each, prevents any combination of them from resulting in a vector with a non-zero second component. The implication is that the target vector lies outside the span of the given vectors.

Showing Inconsistency

The fact that we have a contradiction right away makes our job much easier. The second equation, 0=βˆ’2{0 = -2}, is always false, no matter what values we assign to c1{c_1}, c2{c_2}, and c3{c_3}. This means the system of equations has no solution. Therefore, there are no scalars c1{c_1}, c2{c_2}, and c3{c_3} that satisfy the original vector equation.

To further illustrate, let’s ignore the second equation for a moment and try to solve the remaining equations. We have:

  1. c1+c2+2c3=1{c_1 + c_2 + 2c_3 = 1}
  2. c1βˆ’2c2+c3=2{c_1 - 2c_2 + c_3 = 2}
  3. c2+2c3=3{c_2 + 2c_3 = 3}

From equation (3), we can express c2{c_2} in terms of c3{c_3}:

c2=3βˆ’2c3.{c_2 = 3 - 2c_3.}

Substitute this into equation (1):

c1+(3βˆ’2c3)+2c3=1{c_1 + (3 - 2c_3) + 2c_3 = 1}

c1+3=1{c_1 + 3 = 1}

c1=βˆ’2.{c_1 = -2.}

Now substitute c1=βˆ’2{c_1 = -2} and c2=3βˆ’2c3{c_2 = 3 - 2c_3} into equation (2):

βˆ’2βˆ’2(3βˆ’2c3)+c3=2{-2 - 2(3 - 2c_3) + c_3 = 2}

βˆ’2βˆ’6+4c3+c3=2{-2 - 6 + 4c_3 + c_3 = 2}

5c3=10{5c_3 = 10}

c3=2.{c_3 = 2.}

Finally, find c2{c_2}:

c2=3βˆ’2(2)=3βˆ’4=βˆ’1.{c_2 = 3 - 2(2) = 3 - 4 = -1.}

So, we have c1=βˆ’2{c_1 = -2}, c2=βˆ’1{c_2 = -1}, and c3=2{c_3 = 2}. However, remember that we initially ignored the second equation, which stated 0=βˆ’2{0 = -2}. Since we have a contradiction, our attempt to find a solution is invalid. Thus, there are indeed no scalars that satisfy the original equation.

Conclusion

In summary, guys, we’ve shown that there are no scalars c1{c_1}, c2{c_2}, and c3{c_3} such that:

c1(1,0,1,0)+c2(1,0,βˆ’2,1)+c3(2,0,1,2)=(1,βˆ’2,2,3).{c_1(1,0,1,0) + c_2(1,0,-2,1) + c_3(2,0,1,2) = (1, -2, 2, 3).}

The presence of the contradiction 0=βˆ’2{0 = -2} when setting up the system of equations proves that the vector (1, -2, 2, 3) cannot be written as a linear combination of the vectors (1,0,1,0), (1,0,-2,1), and (2,0,1,2). This means that the target vector lies outside the span of the given vectors. Understanding these concepts is essential for solving linear algebra problems and grasping more complex topics. Keep practicing, and you’ll get the hang of it in no time!

Additional Insights

To deepen our understanding, let's consider the implications of this result. The fact that we cannot find scalars to satisfy the given equation tells us something important about the vector space spanned by the vectors (1,0,1,0), (1,0,-2,1), and (2,0,1,2). Specifically, it means that the vector (1, -2, 2, 3) does not lie within this span. The span of a set of vectors is the set of all possible linear combinations of those vectors.

In this case, since all three given vectors have a 0 in their second component, any linear combination of them will also have a 0 in the second component. Therefore, any vector in the span of these vectors must have a 0 in its second component. However, the vector (1, -2, 2, 3) has a -2 in its second component, which means it cannot be in the span of the given vectors. This is a key reason why no scalars c1{c_1}, c2{c_2}, and c3{c_3} can satisfy the equation.

Moreover, this problem illustrates the concept of linear independence. If the vectors (1,0,1,0), (1,0,-2,1), and (2,0,1,2) were linearly dependent, it might be possible to express (1, -2, 2, 3) as a linear combination of them. However, even if they were linearly dependent, the constraint of having a 0 in the second component would still prevent us from achieving the desired result.

This type of problem is fundamental in linear algebra because it helps us understand the structure of vector spaces and the relationships between vectors within those spaces. By showing that certain vectors cannot be expressed as linear combinations of others, we gain insight into the dimensionality and properties of the vector spaces involved. Keep exploring such problems to build a strong foundation in linear algebra.

Practical Applications

The concepts we've discussed here aren't just abstract mathematical ideas; they have practical applications in various fields. For example, in computer graphics, linear combinations are used to transform and manipulate images and 3D models. Understanding the span of a set of vectors is crucial for determining the range of transformations that can be applied.

In engineering, linear algebra is used to analyze systems of equations that model physical phenomena. The consistency of these systems determines whether a solution exists, which can have significant implications for the design and performance of engineering systems. For instance, in structural analysis, engineers use linear algebra to determine whether a structure is stable under certain loads. If the system of equations is inconsistent, it may indicate that the structure is not stable and needs to be redesigned.

In data science, linear algebra is used for dimensionality reduction techniques like Principal Component Analysis (PCA). PCA involves finding a set of orthogonal vectors that capture the most variance in a dataset. These vectors form a basis for a lower-dimensional subspace that approximates the original data. Understanding the span of these vectors is essential for interpreting the results of PCA and making informed decisions about data analysis.

So, while the problem we tackled today might seem theoretical, it's rooted in concepts that are widely used in various practical applications. By mastering these concepts, you'll be well-equipped to tackle real-world problems in science, engineering, and technology.