Problem 47 - Advanced Engineering Mathematics

Question

Supply a first column so that the matrix is orthogonal: $$ \left(\begin{array}{c:cc} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ & 0 & \frac{1}{\sqrt{3}} \\ & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \end{array}\right) \text {. } $$

Step-by-Step Solution

Answer

First column can be \( \begin{bmatrix} \frac{1}{\sqrt{6}} \\ -\sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{6}} \end{bmatrix} \) or \( \begin{bmatrix} -\frac{1}{\sqrt{6}} \\ \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{6}} \end{bmatrix} \).

Step 1: Understand the Concept of Orthogonal Matrix

An orthogonal matrix is a square matrix whose columns are orthogonal unit vectors. This means each column vector has a magnitude of 1 (is normalized) and is orthogonal (perpendicular) to the other column vectors.

Step 2: Calculate Required Magnitude for First Column

The first column must be a unit vector. To satisfy the orthogonality, let's denote it as \( \begin{bmatrix} a \ b \ c \end{bmatrix} \). The condition is \( a^2 + b^2 + c^2 = 1 \).

Step 3: Set Orthogonality Conditions with Existing Columns

The new column must be orthogonal to the given columns. For the column \( \begin{bmatrix} a \ b \ c \end{bmatrix} \) to be orthogonal to \( \begin{bmatrix} -\frac{1}{\sqrt{2}} \ 0 \ \frac{1}{\sqrt{2}} \end{bmatrix} \), the dot product must be zero: \( a \cdot (-\frac{1}{\sqrt{2}}) + b \cdot 0 + c \cdot \frac{1}{\sqrt{2}} = 0 \ \, \implies \, \ -\frac{a}{\sqrt{2}} + \frac{c}{\sqrt{2}} = 0 \rightarrow a = c. \)

Step 4: Solve Further Orthogonality

Apply orthogonality condition with the second column \( \begin{bmatrix} \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \end{bmatrix} \):\( a \cdot \frac{1}{\sqrt{3}} + b \cdot \frac{1}{\sqrt{3}} + c \cdot \frac{1}{\sqrt{3}} = 0 \rightarrow \frac{a + b + c}{\sqrt{3}} = 0 \rightarrow a + b + c = 0. \)

Step 5: Solve the System of Equations

From \(a = c\) and \(a + b + c = 0\), substitute \(c = a\) in the equation: \(a + b + a = 0 \rightarrow 2a + b = 0 \rightarrow b = -2a. \)

Step 6: Specify the Magnitude Program

From \(b = -2a\), and \(a^2 + b^2 + c^2 = 1\), substitute \(b = -2a\) and \(c = a \): \(a^2 + (-2a)^2 + (a)^2 = 1 \rightarrow 6a^2 = 1 \rightarrow a = \pm \frac{1}{\sqrt{6}}. \) Then, \( c = a = \pm \frac{1}{\sqrt{6}} \) and \( b = -2a = \mp \frac{2}{\sqrt{6}} = \mp \sqrt{\frac{2}{3}}. \)

Step 7: Formulate the First Column

Hence, the first column can be either\[ \begin{bmatrix} \frac{1}{\sqrt{6}} \ -\sqrt{\frac{2}{3}} \ \frac{1}{\sqrt{6}} \end{bmatrix} \] or \[ \begin{bmatrix} -\frac{1}{\sqrt{6}} \ \sqrt{\frac{2}{3}} \ -\frac{1}{\sqrt{6}} \end{bmatrix} \]. Both satisfy the orthogonal conditions and unit normalization.

Key Concepts

Matrix Orthogonality

Matrix orthogonality is an essential concept in linear algebra. An orthogonal matrix is a square matrix where each column is both a unit vector and orthogonal to other columns. This means that the columns of the matrix are perpendicular to each other, ensuring no overlap in their direction. Additionally, each column must have a magnitude of one to qualify as a unit vector.
Orthogonal matrices hold special properties that make them invaluable in computations. For instance, the product of an orthogonal matrix and its transpose results in the identity matrix. This property is particularly useful in fields such as 3D graphics and signal processing.

Vector Dot Product

In the world of vectors, the dot product is a key operation that helps in determining orthogonality among vectors. The dot product is calculated by multiplying corresponding entries of two vectors and summing the results. Mathematically, for vectors \( \mathbf{u} = \begin{bmatrix} u_1 & u_2 & u_3 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \), the dot product is \( u_1v_1 + u_2v_2 + u_3v_3 \).
If the dot product is zero, the vectors are orthogonal to each other. This is because the zero result indicates no overlap in directions between the two vectors, meeting the perpendicularity condition.

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. When two vectors are orthogonal, their dot product is zero, signifying no shared direction. Having orthogonal vectors means they form the basis for the space, maintaining independence from one another.
This property is beneficial in many applications, such as simplifying complex systems in engineering or representing rotations and projections in computer graphics. Orthogonal vectors are crucial as they maintain numerical stability in solutions and transformations.

Unit Vector

A unit vector is a vector with a magnitude of one. It is used to signify direction without impacting scale. In calculations involving orthogonal matrices, each column vector must be a unit vector to ensure correct normalization.
The formula to transform any vector \( \mathbf{v} \) into a unit vector \( \mathbf{u} \) is \( \mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} \), where \( \| \mathbf{v} \| \) is the magnitude of \( \mathbf{v} \). Unit vectors play an informative role in determining the directionality of transformations and rotations, particularly in physics and engineering.

Linear Algebra

Linear algebra is the branch of mathematics focusing on vector spaces and linear mappings between these spaces. It includes systems of linear equations, matrices, determinants, and vector spaces. This field forms the backbone of several applications in science and engineering, especially in data analysis, optimization, and computer graphics.
Understanding linear algebra enables mathematicians and scientists to break down complex systems into simpler, solvable components. Core concepts such as orthogonal matrices and unit vectors are indispensable tools within this area, allowing for efficient computation and solution of multi-dimensional problems.