Problem 28 - A Graphical Approach to Precalculus with Limits

Question

Solve each determinant equation for \(x\). $$\operatorname{det}\left[\begin{array}{ll}x & 3 \\\x & x\end{array}\right]=4$$

Step-by-Step Solution

Answer

The solutions are \( x = 4 \) and \( x = -1 \).

Step 1: Understand the Determinant Formula for a 2x2 Matrix

The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as \( ad - bc \). First, apply this formula to the given matrix: \( \begin{bmatrix} x & 3 \ x & x \end{bmatrix} \).

Step 2: Express the Determinant Using Elements

For the matrix \( \begin{bmatrix} x & 3 \ x & x \end{bmatrix} \), we use the formula from Step 1. Here, \( a = x \), \( b = 3 \), \( c = x \), and \( d = x \). Therefore, the determinant is: \( (x)(x) - (3)(x) \). Simplifying, we get: \( x^2 - 3x \).

Step 3: Set the Determinant Equal to the Given Value

The problem states that the determinant of the matrix equals 4. So, we have the equation: \( x^2 - 3x = 4 \).

Step 4: Rearrange the Equation Into Standard Quadratic Form

To solve this quadratic equation, rearrange it to the standard form \( ax^2 + bx + c = 0 \). This gives us \( x^2 - 3x - 4 = 0 \).

Step 5: Solve the Quadratic Equation

To solve \( x^2 - 3x - 4 = 0 \), we can factor it. We look for two numbers whose product is \(-4\) and sum is \(-3\). These numbers are \(-4\) and \(1\). So, the equation factors to: \((x - 4)(x + 1) = 0 \).

Step 6: Find the Solutions for \(x\)

Solve each factor equal to zero: \( x - 4 = 0 \) gives \( x = 4 \), and \( x + 1 = 0 \) gives \( x = -1 \). Thus, the solutions are \( x = 4 \) and \( x = -1 \).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that has the highest variable exponent of 2, meaning it can be written in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. Quadratic equations have many practical applications in physics, engineering, and more. By turning a problem into a quadratic equation, we can find solutions using different methods.
Some common methods for solving quadratic equations include:

• Factoring: Expressing the equation as a product of its linear factors. If it works, it's usually the simplest method.
• Completing the square: Rewriting the equation as a perfect square trinomial, which allows for easier solving.
• Quadratic formula: This formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) works for any quadratic equation.

In the context of the given problem, we transformed the determinant equation into the quadratic form \( x^2 - 3x - 4 = 0 \), which we then solved by factoring.

2x2 Matrix

A 2x2 matrix is a simple mathematical structure consisting of four elements arranged in two rows and two columns. Matrices are widely used in many areas like physics, computer science, and economics. A matrix can represent a variety of things such as transformations in space, a system of equations, or even just data.
For a 2x2 matrix written as \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), each element can have practical significance depending on the context it is used in. For example:

• Matrix transformations: A 2x2 matrix might represent linear transformations in a 2D space.
• Systems of equations: They can represent and solve systems with two linear equations.

In our problem, the 2x2 matrix \( \begin{bmatrix} x & 3 \ x & x \end{bmatrix} \) was a part of the determinant calculation. Each element of the matrix was substituted into the determinant formula specific to 2x2 matrices to solve for \( x \).

Determinant Calculation

Calculating a determinant is an essential operation in linear algebra, used to evaluate certain properties of matrices, such as invertibility. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated using the formula \( ad - bc \).
This determinant calculation is crucial because:

• Matrix Invertibility: If a matrix has a non-zero determinant, it is invertible, meaning it has an inverse matrix.
• Volume and area calculations: In geometry, determinants can reflect scale changes in linear transformations.

The exercise involves using the formula \( ad - bc \) to set up the equation \( x^2 - 3x = 4 \), obtained by substituting \( x \) for \( a \) and \( c \), and 3 for \( b \), in the matrix \( \begin{bmatrix} x & 3 \ x & x \end{bmatrix} \). Understanding how each component of this determinant contributes to the overall formula helps solve the problem step by step. By equating it to 4, we found our quadratic equation that solved for the possible values of \( x \).