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In Exercises 1 and \(2,\) compute \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-2 \mathbf{v}\) \(\mathbf{u}=\left[\begin{array}{r}{-1} \\ {2}\end{array}\right], \mathbf{v}=\left[\begin{array}{c}{-3} \\ {-1}\end{array}\right]\)

Short Answer

Expert verified
\(\mathbf{u}+\mathbf{v} = \begin{bmatrix} -4 \\ 1 \end{bmatrix}\) and \(\mathbf{u}-2\mathbf{v} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}\)."}

Step by step solution

01

Identify the Vectors

We are provided with two vectors: \(\mathbf{u} = \begin{bmatrix} -1 \ 2 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -3 \ -1 \end{bmatrix}\). The task is to compute \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-2\mathbf{v}\).
02

Compute \(\mathbf{u} + \mathbf{v}\)

To add two vectors, we add their corresponding components. This gives us: \(\mathbf{u} + \mathbf{v} = \begin{bmatrix} -1 \ 2 \end{bmatrix} + \begin{bmatrix} -3 \ -1 \end{bmatrix} = \begin{bmatrix} -1 + (-3) \ 2 + (-1) \end{bmatrix} = \begin{bmatrix} -4 \ 1 \end{bmatrix}\).
03

Compute \(2\mathbf{v}\)

To multiply a vector by a scalar, we multiply each component of the vector by the scalar. Thus, \(2\mathbf{v} = 2 \times \begin{bmatrix} -3 \ -1 \end{bmatrix} = \begin{bmatrix} 2\times(-3) \ 2\times(-1) \end{bmatrix} = \begin{bmatrix} -6 \ -2 \end{bmatrix}\).
04

Compute \(\mathbf{u} - 2\mathbf{v}\)

Subtract \(2\mathbf{v}\) from \(\mathbf{u}\) by subtracting their corresponding components: \(\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} -1 \ 2 \end{bmatrix} - \begin{bmatrix} -6 \ -2 \end{bmatrix} = \begin{bmatrix} -1 - (-6) \ 2 - (-2) \end{bmatrix} = \begin{bmatrix} 5 \ 4 \end{bmatrix}\).

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Key Concepts

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

Vector Addition
Vector addition is a fundamental operation in linear algebra that combines two vectors to produce a new vector. To add vectors, you simply add their corresponding components. This means each element in the first vector is added to the corresponding element of the second vector. For example, if you have two vectors \[ \mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix} \] and \[ \mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix} \], the sum of these two vectors, \(\mathbf{a} + \mathbf{b}\), is calculated as:\[ \mathbf{a} + \mathbf{b} = \begin{bmatrix} a_1 + b_1 \ a_2 + b_2 \end{bmatrix} \]In this operation, the size (or dimension) of both vectors must be the same. Addition is associative and commutative, which means the order of addition does not affect the result. This makes vector addition a versatile and powerful tool.
Vector Subtraction
Vector subtraction allows us to find the vector that points from one vector to another. It is just like vector addition but instead involves subtracting the corresponding components. For vectors \[ \mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix} \] and \[ \mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix} \], the subtraction \( \mathbf{a} - \mathbf{b} \) is defined as:\[ \mathbf{a} - \mathbf{b} = \begin{bmatrix} a_1 - b_1 \ a_2 - b_2 \end{bmatrix} \]Subtraction, like addition, requires the vectors to be of the same dimension. It is not commutative, meaning \( \mathbf{a} - \mathbf{b} \) is not the same as \( \mathbf{b} - \mathbf{a} \). This operation is useful in calculating differences between vectors, often applying to physical quantities like displacement.
Scalar Multiplication
In scalar multiplication, each component of a vector is multiplied by a number, known as a scalar. This operation is significant in changing the magnitude of a vector without affecting its direction (if the scalar is positive). Applying a scalar to a vector \[ \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix} \] with a scalar value \( c \), the resulting vector \( c \mathbf{v} \) will be:\[ c \mathbf{v} = \begin{bmatrix} c \cdot v_1 \ c \cdot v_2 \end{bmatrix} \]If the scalar is negative, the direction of the vector is reversed. Scalar multiplication scales the vector according to the scalar, and it's critical in linear transformations, vector projections, and as you'll often see in solving linear equations.
Linear Algebra
Linear Algebra is a branch of mathematics focusing on vectors, vector spaces (or linear spaces), linear mappings, and systems of linear equations. It provides the language and framework for much of mathematics and is widely used in fields like physics, computer science, and engineering. Key concepts include:
  • Vectors and Vector Spaces: Fundamental entities representing magnitude and direction, directly applicable to numerous areas.
  • Matrices: Rectangular arrays of numbers or expressions used to represent linear transformations and solve systems of linear equations.
  • Determinants and Eigenvalues: These concepts dig deeper into the properties of matrices, helping in simplifying matrix operations and understanding transformations.
Linear Algebra helps solve real-world problems by expressing complex systems in manageable mathematical structures. Whether it's robotics, graphics, or optimizing real-life processes, mastering linear algebra principles opens the door to efficient problem-solving techniques.

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Most popular questions from this chapter

Balance the chemical equations in Exercises \(5-10\) using the vector equation approach discussed in this section. The following reaction between potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\) and manganese sulfate in water produces manganese dioxide, potassium sulfate, and sulfuric acid: \(\mathrm{KMnO}_{4}+\mathrm{MnSO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\) [For each compound, construct a vector that lists the numbers of atoms of potassium (K), manganese, oxygen, sulfur, and hydrogen.]

Exercises \(17-20\) refer to the matrices \(A\) and \(B\) below. Make appropriate calculations that justify your answers and mention an appropriate theorem. $$ A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \\ {-1} & {-1} & {-1} & {1} \\\ {0} & {-4} & {2} & {-8} \\ {2} & {0} & {3} & {-1}\end{array}\right] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \\ {0} & {1} & {1} & {-5} \\\ {1} & {2} & {-3} & {7} \\ {-2} & {-8} & {2} & {-1}\end{array}\right] $$ Can each vector in \(\mathbb{R}^{4}\) be written as a linear combination of the columns of the matrix \(A\) above? Do the columns of \(A\) span \(\mathbb{R}^{4} ?\)

If \(\mathbf{b} \neq \mathbf{0},\) can the solution set of \(A \mathbf{x}=\mathbf{b}\) be a plane through the origin? Explain.

Construct a \(2 \times 2\) matrix \(A\) such that the solution set of the equation \(A \mathbf{x}=\mathbf{0}\) is the line in \(\mathbb{R}^{2}\) through \((4,1)\) and the origin. Then, find a vector \(\mathbf{b}\) in \(\mathbb{R}^{2}\) such that the solution set of \(A \mathbf{x}=\mathbf{b}\) is \(n o t\) a line in \(\mathbb{R}^{2}\) parallel to the solution set of \(A \mathbf{x}=\mathbf{0} .\) Why does this not contradict Theorem 6\(?\)

[M] Budget® Rent A Car in Wichita, Kansas, has a fleet of about 500 cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix below. Suppose that on Monday there are 295 cars at the airport (or rented from there), 55 cars at the east side office, and 150 cars at the west side office. What will be the approximate distribution of cars on Wednesday? \(\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {.00} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right]\)

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