Problem 14 - Calculus: Early Transcendentals

Question

Let a and b be nonzero vectors in two dimensions that are not parallel or anti-parallel. (Vectors are parallel if they point in the same direction, anti-parallel if they point in opposite directions.) Show, algebraically, that if \(\boldsymbol{c}\) is any two dimensional vector, there are scalars \(\mathrm{s}\) and \(t\) such that \(\boldsymbol{c}=\mathrm{sa}+\mathrm{tb}\).

Step-by-Step Solution

Answer

Scalars \(s\) and \(t\) exist such that \(\boldsymbol{c} = s\boldsymbol{a} + t\boldsymbol{b}\) if vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are not parallel.

Step 1: Understand the Vector Decomposition Problem

We are given a two-dimensional vector \(\boldsymbol{c}\) and two non-parallel, non-anti-parallel, nonzero vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\). Our task is to express \(\boldsymbol{c}\) as a linear combination of \(\boldsymbol{a}\) and \(\boldsymbol{b}\). This means we need to find scalars \(s\) and \(t\) such that \(\boldsymbol{c} = s\boldsymbol{a} + t\boldsymbol{b}\).

Step 2: Use Vector Components

Assume \(\boldsymbol{a} = (a_1, a_2)\), \(\boldsymbol{b} = (b_1, b_2)\), and \(\boldsymbol{c} = (c_1, c_2)\). We need to solve for \(s\) and \(t\) such that the equation \((c_1, c_2) = s(a_1, a_2) + t(b_1, b_2)\) holds true. This can be expanded to two equations: \(c_1 = sa_1 + tb_1\) and \(c_2 = sa_2 + tb_2\).

Step 3: Set Up the System of Linear Equations

The vector equations can be rewritten as the system of linear equations: \[\begin{align*}c_1 &= sa_1 + tb_1 \c_2 &= sa_2 + tb_2\end{align*}\] We need to solve this system to find the values of \(s\) and \(t\).

Step 4: Solve the System Using Matrix Inversion

Represent the system as a matrix equation: \[\begin{pmatrix} a_1 & b_1 \a_2 & b_2 \end{pmatrix}\begin{pmatrix} s \ t \end{pmatrix} =\begin{pmatrix} c_1 \ c_2 \end{pmatrix}\] For this matrix equation, the coefficient matrix \(\mathbf{A}\) is invertible because \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are not parallel (i.e., their determinant is non-zero).

Step 5: Calculate the Inverse and Solve for Scalars

If the determinant \(|\mathbf{A}| = a_1b_2 - a_2b_1 eq 0\), the matrix is invertible, and the scalars \(s\) and \(t\) are found using:\[\begin{pmatrix} s \ t \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} c_1 \ c_2 \end{pmatrix} =\frac{1}{a_1b_2 - a_2b_1}\begin{pmatrix} b_2 & -b_1 \-a_2 & a_1 \end{pmatrix}\begin{pmatrix} c_1 \ c_2 \end{pmatrix}\]Calculate this product to find \(s\) and \(t\).

Step 6: Conclude with the Result

We have demonstrated algebraically that for any vector \(\boldsymbol{c} = (c_1, c_2)\), there exist scalars \(s\) and \(t\) such that \(\boldsymbol{c} = s\boldsymbol{a} + t\boldsymbol{b}\), provided that \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are non-parallel, non-zero vectors. The values of \(s\) and \(t\) can be explicitly calculated as shown in Step 5.

Key Concepts

Linear Combination

A linear combination involves expressing a vector as a sum of other vectors, each multiplied by a scalar. In our problem, we aim to express the vector \(\boldsymbol{c}\) as a linear combination of vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
This concept is fundamental in vector mathematics because it helps define vector spaces and enables various operations and transformations.

Here’s a simple breakdown:

• Vector \(\boldsymbol{c}\) can be broken down into components of \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
• The scalars \(s\) and \(t\) are what give the necessary weights for this combination.

This is useful in solving many real-world problems, such as navigation and physics, where vectors represent directions and magnitudes.

System of Linear Equations

In the process of finding a linear combination of vectors, we set up a system of linear equations. This system arises from equating the components of the vectors involved. For the case of vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\), and \(\boldsymbol{c}\), we develop two equations based on their individual components:

• \(c_1 = s a_1 + t b_1\)
• \(c_2 = s a_2 + t b_2\)

The above equations form a linear system that we need to solve to find the values of \(s\) and \(t\).

Thinking about it, solving such a system is like untangling a web where each equation represents a line. The solution to the system gives the point at which these "lines" intersect, which corresponds to the scalars needed to express \(\boldsymbol{c}\) as a combination of \(\boldsymbol{a}\) and \(\boldsymbol{b}\).

Matrix Inversion

Matrix inversion is a process used to find the unique solution to a system of linear equations when the matrix is invertible. An invertible matrix, known as a non-singular matrix, can have its inverse calculated.

Once we set up the problem like the following matrix equation:
\[\begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix}\begin{pmatrix} s \ t \end{pmatrix} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} \]The goal is to solve for the matrix \(\begin{pmatrix} s \ t \end{pmatrix}\).
We achieve this by multiplying both sides by the inverse of the coefficient matrix, granted the determinant is non-zero.

This inversion is essential as it transforms the multiplication into a straightforward solution calculation. It provides a neat solution to simultaneous equations by reducing them into simpler single-step calculations.

Determinant

The determinant is a special number calculated from a square matrix. It is crucial as it determines whether a matrix is invertible or not. For our two-dimensional vectors, the matrix determinant must be calculated as follows:
\[\text{det}(\mathbf{A}) = a_1b_2 - a_2b_1\]

• If the determinant is zero, the matrix is singular, meaning there are no unique solutions.
• Non-zero determinant: The matrix is non-singular, allowing us to find the matrix inverse and hence solve for \(s\) and \(t\).

Considering this, the determinant acts as a gatekeeper, affirming the conditions necessary to utilize matrix inversion and guaranteeing that solutions exist and are unique.