Problem 81 - Precalculus with Limits

Question

In Exercises 77-84, solve for \(x\). \(\left| \begin{array}{c} x-1 & 2 \\ 3 & x-2 \end{array} \right| = 0\)

Step-by-Step Solution

Answer

The solutions for the equation are \(x = 4\) and \(x = -1\)

Step 1: Find the determinant

The determinant of a 2x2 matrix \(\left| \begin{array}{c} a & b \ c & d \end{array} \right|\) is calculated as \(ad-bc\). Apply this formula to our matrix: \(x-1)(x-2)-2*3\). Thus, the determinant is \(x^2 -3x + 2 - 6\)

Step 2: Set the determinant equal to zero

Equating the given determinant to zero, \(x^2 -3x + 2 - 6 = 0\), simplifying it we get \(x^2 - 3x - 4 = 0\).

Step 3: Solve the quadratic equation

This is a quadratic equation in the form \(ax^2 + bx + c = 0\). It can be solved by factoring, using the quadratic formula, or by completing the square. Here, the equation can be factored as \((x-4)(x+1) = 0\). Thus, the solutions for the equation are \(x = 4\) and \(x = -1\)

Key Concepts

Determinants of Matrices

The determinant of a matrix plays a crucial role in various mathematical computations and can often determine whether a system of linear equations has a unique solution. When talking about a 2x2 matrix, determining its determinant is quite straightforward. You take a matrix \begin{array}{c} a & b \: c & d \: \: \: \: \: \: \: end{array}\ and compute the determinant by multiplying the values in the leading diagonal and subtracting the product of the values in the opposite diagonal, represented by the formula \(ad - bc\). For instance, if our matrix is: \(\begin{array}{cc} x-1 & 2 \ 3 & x-2 \: end{array}\), the determinant is calculated as \((x-1)(x-2) - (2*3)\).The importance of the determinant cannot be overstated, as it can indicate whether a set of linear equations is consistent, or if a matrix is invertible. Zero determinant, as in our exercise, implies that the matrix does not have an inverse and that there may be no unique solutions to the corresponding linear equations.

Quadratic Formula

The quadratic formula is a powerful tool that can be employed to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is given by the expression \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation and \(\sqrt{b^2 - 4ac}\) is called the discriminant. If the discriminant is positive, we have two real solutions; if it is zero, there is one real solution; and if it is negative, there are two complex solutions.In the context of the exercise, using the quadratic formula isn't necessary as the quadratic equation is easily factorable. However, knowing this formula is essential when you encounter equations that cannot be factored simply. Applying the formula allows you to solve for \(x\) even in more complex situations.

Factoring Quadratic Equations

Factoring is a method used to solve quadratic equations that are written in the standard form \(ax^2 + bx + c = 0\) by expressing them as the product of binomials. This method works by finding two numbers that multiply together to give \(ac\) (the product of the coefficient of \(x^2\) and the constant term) and add together to give \(b\) (the coefficient of \(x\)). When these two numbers are found, the quadratic equation can be rewritten as \((x - p)(x - q) = 0\), where \(p\) and \(q\) are the solutions for \(x\).Take for instance our problem, which simplifies to the quadratic equation \(x^2 - 3x - 4 = 0\). We look for two numbers that multiply to -4 and add to -3, which are -4 and 1. Thus, we can factor our equation to \((x - 4)(x + 1) = 0\), giving us two possible solutions for \(x\): 4 and -1.

Absolute Value Equations

Absolute value equations involve an expression inside absolute value bars, and this expression can take on different values depending on the nature of the numbers involved. The absolute value of a number is its distance from zero on a number line, regardless of the direction. To solve equations with absolute values, one must consider both the positive and negative scenarios of the expression within the bars.In our original exercise, the absolute value bars were part of the determinant of a matrix. Usually, absolute value equations look like \(|ax + b| = c\), and we'd solve them by considering both \(ax + b = c\) and \(ax + b = -c\). However, in the context of the determinant, the absolute value is actually set to zero, which simplifies solving: it tells us that the determinant itself, without considering absolute value, must be zero, thus leading us directly to solving the quadratic equation formed by the determinant.