Problem 27 - Advanced Engineering Mathematics

Question

Without expanding, show that \(\left|\begin{array}{ccc}1 & 1 & 1 \\\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ b c & a c & a b\end{array}\right|=0\)

Step-by-Step Solution

Answer

The rows of the matrix are linearly dependent, making the determinant zero.

Step 1: Understand the Problem

We are given a 3x3 determinant and need to show that its value is zero without expanding it. The determinant can be expressed as:\[\left| \begin{array}{ccc}1 & 1 & 1 \\frac{1}{a} & \frac{1}{b} & \frac{1}{c} \bc & ac & ab \end{array} \right| = 0\]This problem requires using properties of determinants.

Step 2: Apply Determinant Properties

One useful property of determinants is that the determinant of a matrix is zero if its rows (or columns) are linearly dependent. We will check for linear dependence among the rows.

Step 3: Analyze Row 1 and Row 2

Consider the first and second rows of the determinant:\[\begin{array}{ccc}1 & 1 & 1 \\frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\end{array}\]The elements in row 2 are the reciprocals of the variables corresponding to row 1's ones. We look for a combination of row operations that can show dependability.

Step 4: Analyze Row 2 and Row 3

Now consider the second and third rows:\[\begin{array}{ccc}\frac{1}{a} & \frac{1}{b} & \frac{1}{c} \bc & ac & ab \\end{array}\]If we multiply the second row by \(abc\), we get:\[\begin{array}{ccc}1 & 1 & 1 \abc \cdot \frac{1}{a} = bc & abc \cdot \frac{1}{b} = ac & abc \cdot \frac{1}{c} = ab \end{array} \]This shows that the second row is a scalar multiple of the third row.

Step 5: Conclude Linear Dependence

Since the second row is a scalar multiple of the third row, the rows of the determinant are linearly dependent. According to the properties of determinants, if any row is a scalar multiple of another row, the determinant is zero.

Key Concepts

Linear Dependence

Linear dependence is a concept that helps us understand relations between rows or columns in a matrix. When we say that rows or columns are linearly dependent, it means that at least one row or column can be expressed as a combination of the others. In simpler terms, some rows or columns might be repetitive in nature in terms of the information they convey.

This idea is crucial when dealing with determinants. If any two rows or columns in a matrix are linearly dependent, the determinant of that matrix is zero. This can be a helpful shortcut, as it allows one to determine the value of a determinant without performing complex expansions. In the given problem, recognizing the linear dependence between the rows helps quickly show the determinant is zero.

Understanding linear dependence allows for efficient problem-solving in linear algebra without unnecessarily labor-intensive calculations.

Matrix Properties

Matrices are essential structures in linear algebra, characterized by their unique properties. Here are some fundamental properties that are particularly useful when dealing with determinants:

• **Zero Determinant**: If any row or column in a matrix is a scalar multiple of another, the matrix's determinant is zero. This highlights the importance of checking dependence among rows or columns.
• **Row and Column Operations**: Performing elementary operations such as swapping rows, scaling rows, or adding multiples of one row to another can significantly simplify a matrix.
• **Determinant Consistency**: Elementary row operations do not affect the determinant's value except for row swaps, which change its sign.

By understanding these properties, solving determinant-related problems becomes much less daunting. These properties allow you to manipulate matrices and verify results without diving into tedious arithmetic details.

Matrix Row Operations

Matrix row operations encompass a set of techniques used to manipulate and simplify matrices. These operations can help reveal specific properties, like linear dependence, often without changing the matrix's determinant value.

• **Row Addition**: Adds a multiple of one row to another. This operation is helpful in creating zeros in the matrix, which can simplify computations.
• **Row Scaling**: Multiplies the elements of a single row by a non-zero scalar. When this operation helps one row become a multiple of another, it can indicate linear dependence, making it very useful in determinant evaluations.
• **Row Swapping**: Swaps the positions of two rows. Although this operation changes the sign of the determinant, it is often the first step in simplification.

Mastering these row operations is like having a toolbox for matrices. They allow for efficient determination of key properties like linear dependence, ultimately simplifying the evaluation process of determinants.