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Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{ll} 1 & -1 \end{array}\right] \quad \vec{v}=\left[\begin{array}{l} 3 \\ 3 \end{array}\right] $$

Short Answer

Expert verified
The product \(\vec{u} \vec{v}\) is 0.

Step by step solution

01

Identify Vector Dimensions

We begin by noting the dimensions of the vectors. \(\vec{u}\) is a row vector with dimensions \(1 \times 2\), and \(\vec{v}\) is a column vector with dimensions \(2 \times 1\). This allows us to multiply them since the inner dimensions match.
02

Perform Matrix Multiplication

Perform the matrix multiplication by calculating the dot product of \(\vec{u}\) and \(\vec{v}\). This involves summing the products of corresponding elements:\[\vec{u} \vec{v} = \begin{bmatrix} 1 & -1 \end{bmatrix} \begin{bmatrix} 3 \ 3 \end{bmatrix} = 1 \cdot 3 + (-1) \cdot 3\]Calculate this to find the result \(\vec{u} \vec{v}\).
03

Calculate the Dot Product

Calculate the products and sum them up:\[1 \cdot 3 = 3\]\[-1 \cdot 3 = -3\]Summing these gives:\[3 + (-3) = 0\]Thus, the result of the product \(\vec{u} \vec{v}\) is 0.

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Key Concepts

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

Vector Dimensions
When dealing with vectors, understanding their dimensions is crucial. Vectors are essentially arrays of numbers that can appear in either row or column form. The dimensions of a vector refer to the number of elements it contains. This understanding is vital when performing operations like matrix multiplication.

In our specific example, vector \(\vec{u}\) is a row vector with dimensions \(1 \times 2\). This means it has 1 row and 2 columns, hence two elements.
  • The first element is 1.
  • The second element is -1.
Similarly, vector \(\vec{v}\) is a column vector with dimensions \(2 \times 1\), meaning it has 2 rows and 1 column.
  • The first element is 3.
  • The second element is 3.
For an operation to be valid, such as multiplying a row vector by a column vector, the number of columns in the row vector must equal the number of rows in the column vector. In this case, both have a common dimension of 2.
Dot Product
The dot product is a fundamental operation for multiplying vectors. It gives us a measure of how much one vector goes in the direction of another. When we calculate the dot product of a row vector and a column vector, we obtain a single scalar value.

To compute the dot product of \(\vec{u}\) and \(\vec{v}\), we multiply the corresponding elements from each vector and then add these products together. In our case, we have:
  • \(1 \cdot 3 = 3\)
  • \(-1 \cdot 3 = -3\)
Adding these results gives us:
  • \(3 + (-3) = 0\)
Thus, the dot product of \(\vec{u}\) and \(\vec{v}\) is 0. This result also indicates that the vectors are orthogonal, meaning they are at right angles to each other in a geometric sense.
Row Vector
A row vector is simply a 1-dimensional array of numbers displayed horizontally. It's the opposite of a column vector. Row vectors are useful in various mathematical computations, including matrix multiplication.

For instance, in our example, \(\vec{u}\) is defined as a row vector:
  • \(\begin{bmatrix} 1 & -1 \end{bmatrix}\)
This notation indicates that it spans across one row and contains two elements.

Row vectors play a significant role in linear algebra processes. One of their main properties is that they can be multiplied with column vectors of matching inner dimensions, enabling the computation of their dot product.
Column Vector
Contrary to row vectors, column vectors are organized vertically as a single column of numbers. They are often used in operations involving linear transformations and solving systems of equations.

In our case, \(\vec{v}\) is a column vector:
  • \(\begin{bmatrix} 3 \3 \end{bmatrix}\)
It is structured with two rows, and thus contains two elements.

Column vectors are a building block in matrix mathematics. Their arrangement makes them compatible for use in certain operations, like our example where a column vector multiplies a row vector to produce a scalar dot product. This perpendicular nature of column vectors to row vectors facilitates various vector space manipulations.

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Most popular questions from this chapter

Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ \begin{array}{l} A=\left[\begin{array}{ll} 3 & 3 \\ 6 & 4 \end{array}\right] \\ B=\left[\begin{array}{ll} 15 & -39 \\ 16 & -66 \end{array}\right] \end{array} $$

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 2 \\ -6 \end{array}\right], \vec{u}=\left[\begin{array}{c} 0 \\ -1 \end{array}\right], \vec{v}=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ \begin{array}{l} A=\left[\begin{array}{ccccc} 3 & 0 & -2 & -4 & 5 \\ 2 & 3 & 2 & 0 & 2 \\ -5 & 0 & 4 & 0 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} -1 \\ -5 \\ 4 \end{array}\right] \end{array} $$

A diagonal matrix \(D\) and a matrix \(A\) are given. Find the products \(D A\) and \(A D,\) where possible. $$ \begin{array}{l} D=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right] \\ A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{ll} -4 & -1 \\ -3 & -2 \end{array}\right], \vec{b}=\left[\begin{array}{l} 1 \\ 4 \end{array}\right] $$

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