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Chapter 1: Problem 63

Let \(\mathbf{a}=\left\langle a_{1}, a_{2}\right\rangle, \mathbf{b}=\left\langle b_{1}, b_{2}\right\rangle\), and \(\mathbf{c}=\left\langle c_{1}, c_{2}\right\rangle\) be three nonzero vectors. If \(a_{1} b_{2}-a_{2} b_{1} \neq 0\), then show there are two scalars, \(\alpha\) and \(\beta\), such that \(\mathbf{c}=\alpha \mathbf{a}+\beta \mathbf{b}\).

Short Answer

Expert verified
Scalars \(\alpha\) and \(\beta\) exist because the vectors are linearly independent.

Step by step solution

01

Understand the condition

The condition given is \(a_{1} b_{2} - a_{2} b_{1} eq 0\). This is the determinant of the 2x2 matrix formed by \(\mathbf{a}\) and \(\mathbf{b}\), which implies that the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are linearly independent. Thus, any vector in the plane can be expressed as a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\).
02

Express \(\mathbf{c}\) as a linear combination

We want to show that \(\mathbf{c} = \alpha \mathbf{a} + \beta \mathbf{b}\). This translates to the system of equations: \(c_{1} = \alpha a_{1} + \beta b_{1}\) and \(c_{2} = \alpha a_{2} + \beta b_{2}\).
03

Write the system in matrix form

We can write the system as a matrix equation: \[ \begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix} \begin{pmatrix} \alpha \ \beta \end{pmatrix} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} \].
04

Use the inverse matrix to solve for \(\alpha\) and \(\beta\)

Since the determinant \(a_{1} b_{2} - a_{2} b_{1} eq 0\), the matrix \(\begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix}\) is invertible. Multiply both sides by the inverse matrix to solve for \(\begin{pmatrix} \alpha \ \beta \end{pmatrix}\): \(\begin{pmatrix} \alpha \ \beta \end{pmatrix} = \begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix}^{-1} \begin{pmatrix} c_1 \ c_2 \end{pmatrix}\).
05

Conclusion

Since the solution for \(\alpha\) and \(\beta\) exists because the determinant is non-zero, we have shown that \(\mathbf{c}\) can be expressed as a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Combination
A linear combination involves expressing one vector as a sum of multiples of other vectors. Basically, if you want to construct a vector like \(\mathbf{c}\), using other vectors \(\mathbf{a}\) and \(\mathbf{b}\), you can do so by introducing scalars, typically denoted \(\alpha\) and \(\beta\). These scalars determine how much of each vector \(\mathbf{a}\) and \(\mathbf{b}\) to use to arrive at \(\mathbf{c}\).
For example, if \(\mathbf{c} = \alpha \mathbf{a} + \beta \mathbf{b}\), it means you are combining the effects of vectors \(\mathbf{a}\) and \(\mathbf{b}\) by multiplying them with the scalars \(\alpha\) and \(\beta\), respectively. This is particularly useful when you need to express a vector in terms of other vectors, which can help simplify calculations or provides a deeper understanding of the vector space.
  • This method helps in solving linear systems as it breaks down complex problems into manageable parts.
  • Every vector in the span of \(\{\mathbf{a}, \mathbf{b}\}\) can be written as a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\).
Matrix Determinant
The determinant of a matrix provides important information about the matrix itself, especially in the context of solving equations. For a 2x2 matrix, the determinant can be easily computed and tells us if the matrix is invertible. In this particular case, if the determinant is non-zero, the matrix is invertible, indicating the vectors are linearly independent.
The given expression \(a_{1}b_{2} - a_{2}b_{1} \) is actually the formula for the determinant of the matrix formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\). A non-zero determinant thus confirms that the vectors \(\mathbf{a}\) and \(\mathbf{b}\) do not lie on the same line (or are not scaled versions of each other), and can be independently combined to form any vector \(\mathbf{c}\) in the plane they define.
  • If the determinant is zero, the vectors are linearly dependent, indicating redundancy.
  • The determinant is crucial for determining solutions to systems of equations, as it signifies if unique solutions exist.
Vector Space Basis
In linear algebra, a basis of a vector space is a set of vectors that is both - linearly independent, and - spans the vector space. This means that any vector in the vector space can be expressed as a linear combination of the basis vectors.
In this exercise with vectors \(\mathbf{a}\) and \(\mathbf{b}\), if they are linearly independent (which is guaranteed due to a non-zero determinant), they act as a basis for the vector space defined by their span. This means any vector, such as \(\mathbf{c}\), within that space can be uniquely constructed using \(\mathbf{a}\) and \(\mathbf{b}\) by defining appropriate scalars \(\alpha\) and \(\beta\).
  • An ideal basis simplifies representing complex vector operations efficiently.
  • Using a basis, one can transform complex vector spaces into simpler, more manageable forms like Cartesian coordinates.

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Most popular questions from this chapter

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates.\(x^{2}+y^{2}+z^{2}=144\)

Write the standard form of the equation of the ellipsoid centered at the origin that passes through points \(A(2,0,0), B(0,0,1)\), and \(C\left(\frac{1}{2}, \sqrt{11}, \frac{1}{2}\right)\)

Consider \(\mathbf{r}(t)=\langle\sin t, \cos t, 2 t\rangle\) the position vector of a particle at time \(t \in[0,3]\), where the components of \(\mathbf{r}\) are expressed in centimeters and time is measured in seconds. Let \(\overrightarrow{O P}\) be the position vector of the particle after 1 sec a. Determine the velocity vector \(\mathbf{v}(1)\) of the particle after 1 sec. b. Find the scalar equation of the plane that is perpendicular to \(\mathbf{v}(1)\) and passes through point \(P\). This plane is called the normal plane to the path of the particle at point \(P\). c. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point \(P\).

For the following exercises, find the equation of the plane with the given properties.The plane that passes through point \((4,7,-1)\) and has normal vector \(\mathbf{n}=\langle 3,4,2\rangle\)

John allocates \(d\) dollars to consume monthly three goods of prices \(a, b\), and \(c .\) In this context, the budget equation is defined as \(a x+b y+c z=d\), where \(x \geq 0, y \geq 0\), and \(z \geq 0\) represent the number of items bought from each of the goods. The budget set is given by \(\\{(x, y, z) \mid a x+b y+c z \leq d, x \geq 0, y \geq 0, z \geq 0\\}\), and the budget plane is the part of the plane of equation \(a x+b y+c z=d\) for which \(x \geq 0, y \geq 0\), and \(z \geq 0 .\) Consider \(a=\$ 8, b=\$ 5\), \(c=\$ 10\), and \(d=\$ 500 .\) a. Use a CAS to graph the budget set and budget plane. b. For \(z=25\), find the new budget equation and graph the budget set in the same system of coordinates.
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