Question

Given \(u=\left[\begin{array}{l}5 \\ 1\end{array}\right]\) and \(v=\left[\begin{array}{l}0 \\ 3\end{array}\right],\) find the following graphically: \((a) 2 v\) (c) \(u-v\) \((e) 2 u+3 v\) (b) \(u-v\) \((d) v-u\) (f) \(4 u-2 v\)

Short answer

\( 2v = \begin{pmatrix} 0 \\ 6 \end{pmatrix} \), \( u-v = \begin{pmatrix} 5 \\ -2 \end{pmatrix} \), \( v-u = \begin{pmatrix} -5 \\ 2 \end{pmatrix} \), \( 2u+3v = \begin{pmatrix} 10 \\ 11 \end{pmatrix} \), \( 4u-2v = \begin{pmatrix} 20 \\ -2 \end{pmatrix} \).

Step-by-step solution

- Calculate 2v

Given that \( v = \begin{pmatrix} 0 \ 3 \end{pmatrix} \), we need to multiply each component of \( v \) by 2. So, \( 2v = 2 \times \begin{pmatrix} 0 \ 3 \end{pmatrix} = \begin{pmatrix} 0 \ 6 \end{pmatrix} \). Thus, \( 2v = \begin{pmatrix} 0 \ 6 \end{pmatrix} \).

- Calculate u-v

Given \( u = \begin{pmatrix} 5 \ 1 \end{pmatrix} \) and \( v = \begin{pmatrix} 0 \ 3 \end{pmatrix} \), subtract each component of \( v \) from \( u \): \( u-v = \begin{pmatrix} 5-0 \ 1-3 \end{pmatrix} = \begin{pmatrix} 5 \ -2 \end{pmatrix} \). Thus, \( u-v = \begin{pmatrix} 5 \ -2 \end{pmatrix} \) for both (c) and (b).

- Calculate v-u

Now, we subtract each component of \( u \) from \( v \): \( v-u = \begin{pmatrix} 0-5 \ 3-1 \end{pmatrix} = \begin{pmatrix} -5 \ 2 \end{pmatrix} \). Thus, \( v-u = \begin{pmatrix} -5 \ 2 \end{pmatrix} \).

- Calculate 2u + 3v

First, compute \( 2u \) and \( 3v \). \( 2u = 2 \times \begin{pmatrix} 5 \ 1 \end{pmatrix} = \begin{pmatrix} 10 \ 2 \end{pmatrix} \) and \( 3v = 3 \times \begin{pmatrix} 0 \ 3 \end{pmatrix} = \begin{pmatrix} 0 \ 9 \end{pmatrix} \). Now, add \( 2u \) and \( 3v \): \( 2u + 3v = \begin{pmatrix} 10 \ 2 \end{pmatrix} + \begin{pmatrix} 0 \ 9 \end{pmatrix} = \begin{pmatrix} 10 \ 11 \end{pmatrix} \).

- Calculate 4u - 2v

Calculate \( 4u \) and \( -2v \). \( 4u = 4 \times \begin{pmatrix} 5 \ 1 \end{pmatrix} = \begin{pmatrix} 20 \ 4 \end{pmatrix} \) and \( -2v = -2 \times \begin{pmatrix} 0 \ 3 \end{pmatrix} = \begin{pmatrix} 0 \ -6 \end{pmatrix} \). Now, perform the subtraction: \( 4u - 2v = \begin{pmatrix} 20 \ 4 \end{pmatrix} - \begin{pmatrix} 0 \ 6 \end{pmatrix} = \begin{pmatrix} 20 \ -2 \end{pmatrix} \).

Key concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. It is fundamental in various fields of science and engineering.

Key structures in linear algebra include vectors, which are quantities that have both magnitude and direction. They are often represented as an array of numbers or variables. Matrices, another cornerstone of this field, are rectangular arrays of numbers or functions that can represent linear transformations or systems of equations.

Linear algebra provides tools to solve systems of linear equations, which can be visualized as planes intersecting in multi-dimensional space. Understanding linear algebra is crucial for applications in computer graphics, machine learning, and optimization problems.

Vector Addition

Vector addition is a basic operation in vector algebra that combines two or more vectors to form a new vector. This operation involves adding corresponding components from the vectors. For example, if you have two vectors, \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \ b_2 \end{pmatrix} \), their sum \( \mathbf{a} + \mathbf{b} \) would be determined by adding each pair of corresponding components: \[ \mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \ a_2 + b_2 \end{pmatrix} \].

Vector addition is commutative, meaning the order doesn't matter: \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \). It is also associative: \((\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c}) \).

Geometrically, adding vectors corresponds to placing them tip-to-tail, and drawing the resultant vector from the starting point to the end point of this configuration.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a single number), scaling the magnitude of the vector without altering its direction.

For a vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) and a scalar \( k \), the product \( k \mathbf{v} \) is given by: \[ k \mathbf{v} = \begin{pmatrix} kx \ ky \end{pmatrix} \].

This operation stretches or shrinks a vector. If \( k \) is positive, the vector maintains its direction; if negative, the vector points in the opposite direction.

• If \( |k| > 1 \), the vector increases in magnitude.
• If \( 0 < |k| < 1 \), the vector decreases in magnitude.

Understanding scalar multiplication is essential in scaling transformations and adjusting vector sizes without changing their directional properties.

Vector Subtraction

Vector subtraction is another fundamental operation where one vector is subtracted from another. This forms a new vector that represents the change between the two original vectors.

For vectors \( \mathbf{u} = \begin{pmatrix} u_1 \ u_2 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} v_1 \ v_2 \end{pmatrix} \), the subtraction \( \mathbf{u} - \mathbf{v} \) is calculated by subtracting the corresponding components: \[ \mathbf{u} - \mathbf{v} = \begin{pmatrix} u_1 - v_1 \ u_2 - v_2 \end{pmatrix} \].

This operation can also be thought of as adding the negative of a vector: \( \mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) \). In graphical terms, vector subtraction can be visualized by making the vectors' tails coincide and drawing the resulting vector from the head of the subtracted vector to the head of the other vector.