Problem 52 - Algebra and Trigonometry

Question

Solve for \(x\) $$ \left|\begin{array}{lll}{x} & {1} & {1} \\ {1} & {1} & {x} \\ {x} & {1} & {x}\end{array}\right|=0 $$

Step-by-Step Solution

Answer

The solution for \(x\) is 1.

Step 1: Set Up the Determinant

The problem involves solving for \(x\) in a 3x3 determinant problem. The determinant is expressed as: \[ \begin{vmatrix} x & 1 & 1 \ 1 & 1 & x \ x & 1 & x \end{vmatrix} \]We need to expand this determinant to find the equation.

Step 2: Expand the Determinant

Using the method of cofactor expansion along the first row, we write the determinant as:\[x \cdot \big( \begin{vmatrix} 1 & x \ 1 & x \end{vmatrix} \big) - 1 \cdot \big( \begin{vmatrix} 1 & x \ x & x \end{vmatrix} \big) + 1 \cdot \big( \begin{vmatrix} 1 & 1 \ x & 1 \end{vmatrix} \big)\]

Step 3: Calculate the 2x2 Determinants

Each of the 2x2 determinants are calculated as follows:- For the first determinant: \( \begin{vmatrix} 1 & x \ 1 & x \end{vmatrix} = 1 \cdot x - 1 \cdot x = 0 \)- For the second determinant: \( \begin{vmatrix} 1 & x \ x & x \end{vmatrix} = 1 \cdot x - x \cdot x = x - x^2 \)- For the third determinant: \( \begin{vmatrix} 1 & 1 \ x & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot x = 1 - x \)

Step 4: Substitute and Simplify

Substituting back the values of the 2x2 determinants into the expansion, we get:\[ x \cdot 0 - 1(x - x^2) + 1(1 - x) = 0 \]Simplify:\[-(-x^2 + x) + (1 - x) = 0 \]\[ x^2 - x + 1 - x = 0 \]\[ x^2 - 2x + 1 = 0 \]

Step 5: Solve the Quadratic Equation

The equation \( x^2 - 2x + 1 = 0 \) is a quadratic equation that can be factored as:\[ (x - 1)(x - 1) = 0 \]This simplifies to:\[(x - 1)^2 = 0\] indicating that \( x - 1 = 0 \). Thus, \( x = 1 \).

Key Concepts

Cofactor Expansion

When solving determinants, especially for matrices that are 3x3 or larger, cofactor expansion is a common technique. It's a method that simplifies a larger determinant into smaller, more manageable parts. By expanding along a row or a column, we break the determinant into a combination of smaller matrices, specifically 2x2 matrices. This is helpful because calculating 2x2 determinants is straightforward. To perform cofactor expansion, you pick a row or column to expand along. Here, we expanded along the first row. Each element in the chosen row or column is multiplied by its cofactor. The cofactor is the determinant of a smaller matrix, found by removing the row and column of the chosen element, with alternating signs applied. This method helps structure the process, making it easier to handle complex matrices.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, generally in the form of \(ax^2 + bx + c = 0\). These equations can represent a variety of problems in mathematics, physics, and engineering. Solving these equations can reveal important values, like roots or solutions, that satisfy the equation.There are several methods to solve quadratic equations: factoring, completing the square, or using the quadratic formula. In this particular exercise, the quadratic equation \(x^2 - 2x + 1 = 0\) was achieved after expansion and simplification. This can be factored into \((x-1)(x-1) = 0\), demonstrating that the equation has an exact root or repeated solution. Understanding quadratic equations is essential, as they are a fundamental building block in algebra.

Solving for Variables

At the core of many algebraic problems is the task of solving for variables. This involves isolating the variable on one side of the equation to determine its value. In our exercise, the variable in question is \(x\), and solving for it involves several steps.Initially, we used the determinant, expanded it using cofactor expansion, calculated smaller determinants, and finally arrived at a quadratic equation. For the quadratic \(x^2 - 2x + 1 = 0\), we found it could be factored neatly, allowing us to solve for \(x\).Remember the key steps when solving for variables:

• Isolate the variable on one side of the equation.
• Simplify the equation as much as possible.
• Use algebraic principles like factoring or applying the quadratic formula to find solutions.

Solving equations is crucial for many fields, from scientific research to everyday problem-solving.