Let A1=[1, 3, 4] A2=[2,3,7] And B=[-1,-2,-4]Is B A Linear Combination Of A And A2? A. Yes, B Is A Linear

Let A1=[1, 3, 4] A2=[2,3,7] And B=[-1,-2,-4]Is B A Linear Combination Of A And A2? A. Yes, B Is A Linear

Understanding whether a vector can be expressed as a linear combination of other vectors is a fundamental concept in linear algebra. This idea is crucial in various applications, including solving systems of equations, understanding vector spaces, and more. In this article, we will analyze whether the vector B = [-1, -2, -4] can be written as a linear combination of the vectors A1 = [1, 3, 4] and A2 = [2, 3, 7]. We will explore the methodology step by step, providing a comprehensive explanation suitable for learners at different levels.

What Is a Linear Combination?

Before delving into the specific problem, it is essential to understand what it means for a vector to be a linear combination of other vectors.

Definition of Linear Combination

A vector v in a vector space can be expressed as a linear combination of vectors u₁, u₂, ..., uₙ if there exist scalars (coefficients) c₁, c₂, ..., cₙ such that:

\[ \mathbf{v} = c1 \mathbf{u}1 + c2 \mathbf{u}2 + \cdots + cn \mathbf{u}n \]

In our context, the question becomes: Can we find scalars α and β such that:

\[ \mathbf{B} = \alpha \mathbf{A}1 + \beta \mathbf{A}2 \]

where A₁ = [1, 3, 4], A₂ = [2, 3, 7], and B = [-1, -2, -4]?

Formulating the Problem

Given the vectors:

• \( \mathbf{A}_1 = [1, 3, 4] \)


• \( \mathbf{A}_2 = [2, 3, 7] \)


• \( \mathbf{B} = [-1, -2, -4] \)

We need to find scalars \( \alpha \) and \( \beta \) such that:

\[ \alpha \times [1, 3, 4] + \beta \times [2, 3, 7] = [-1, -2, -4] \]

which translates into a system of equations:

\[
\begin{cases}
\alpha \times 1 + \beta \times 2 = -1 \quad &(1) \\
\alpha \times 3 + \beta \times 3 = -2 \quad &(2) \\
\alpha \times 4 + \beta \times 7 = -4 \quad &(3)
\end{cases}
\]

Our goal is to solve this system for \( \alpha \) and \( \beta \).

Solving the System of Equations

Let's analyze the equations step by step.

Equation (1):

\[ \alpha + 2 \beta = -1 \] which gives: \[ \alpha = -1 - 2 \beta \quad (4) \]

Equation (2):

\[ 3 \alpha + 3 \beta = -2 \] Divide both sides by 3: \[ \alpha + \beta = -\frac{2}{3} \] Now, substitute \( \alpha \) from (4): \[ (-1 - 2 \beta) + \beta = -\frac{2}{3} \] Simplify: \[ -1 - 2 \beta + \beta = -\frac{2}{3} \] \[ -1 - \beta = -\frac{2}{3} \] Solve for \( \beta \): \[ -\beta = -\frac{2}{3} + 1 \] \[ -\beta = \frac{1}{3} \] \[ \beta = -\frac{1}{3} \]

Now, substitute \( \beta = -\frac{1}{3} \) into (4):
\[ \alpha = -1 - 2 \times \left(-\frac{1}{3}\right) = -1 + \frac{2}{3} = -\frac{3}{3} + \frac{2}{3} = -\frac{1}{3} \]

Verify with Equation (3):

\[ 4 \alpha + 7 \beta = -4 \] Substitute \( \alpha = -\frac{1}{3} \), \( \beta = -\frac{1}{3} \): \[ 4 \times \left(-\frac{1}{3}\right) + 7 \times \left(-\frac{1}{3}\right) = -\frac{4}{3} - \frac{7}{3} = -\frac{11}{3} \]

Compare with RHS:
\[ -4 = -\frac{12}{3} \]

Since \( -\frac{11}{3} \neq -\frac{12}{3} \), the equations are inconsistent, indicating that B cannot be expressed as an exact linear combination of A₁ and A₂.

But wait! Our initial steps suggest the coefficients satisfy the first two equations but not the third. Does this mean B is not a linear combination? Not necessarily.

Re-evaluating the system

The inconsistency shows that B cannot be expressed as a linear combination of A₁ and A₂ exactly if we only consider these two vectors. However, in some contexts, especially in vector spaces with more dimensions, the question may be whether B lies within the span of A₁ and A₂—that is, whether B can be expressed as a linear combination of A₁ and A₂ approximately or as part of a larger basis.

But based on our calculations, the system does not have an exact solution, so B is not a linear combination of A₁ and A₂ in the strict sense.

---

Wait, the initial statement says: "A. Yes, B is a linear" — perhaps indicating that B can be expressed as a linear combination, possibly with different coefficients or considering an extended basis.

Let's explore if perhaps the initial problem is asking whether B is a linear combination of A₁ and A₂ or if the statement is just a partial answer.

---

Understanding the Context: Is B a Linear Combination of A₁ and A₂?

Based on the calculations, B is not an exact linear combination of A₁ and A₂. But perhaps the problem is asking whether B can be approximated or expressed as a linear combination, or if there's a different interpretation.

When is a vector considered a linear combination?

• Exact: When the coefficients satisfy all equations simultaneously.


• Approximate: When the equations are nearly satisfied, maybe in a least-squares sense.


• Within the span: If B lies in the subspace spanned by A₁ and A₂.

Given the inconsistency in our system, B does not lie in the span of A₁ and A₂, unless additional vectors are considered.

---

Alternative Approach: Using Matrix Methods

Let's formalize this problem using matrix algebra to clarify.

Constructing the Matrix

Create a matrix M with A₁ and A₂ as columns:

\[
M = \begin{bmatrix}
1 & 2 \\
3 & 3 \\
4 & 7 \\
\end{bmatrix}
\]

And the vector:

\[
\mathbf{b} = \begin{bmatrix}
-1 \\
-2 \\
-4 \\
\end{bmatrix}
\]

We want to solve:

\[
M \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \mathbf{b}
\]

which is a least squares problem if an exact solution does not exist.

Computing the Least Squares Solution

Using the normal equations:

\[
M^T M \mathbf{x} = M^T \mathbf{b}
\]

Calculate \( M^T M \):

\[
M^T M = \begin{bmatrix}
1 & 3 & 4 \\
2 & 3 & 7 \\
\end{bmatrix}
\begin{bmatrix}
1 & 2 \\
3 & 3 \\
4 & 7 \\
\end{bmatrix} = \begin{bmatrix}
1^2 + 3^2 + 4^2 & 1 \times 2 + 3 \times 3 + 4 \times 7 \\
2 \times 1 + 3 \times 3 + 7 \times 4 & 2^2 + 3^2 + 7^2 \\
\end{bmatrix}
\]

Compute each element:

• \( (1,1) \): \( 1 + 9 + 16 = 26 \)


• \(