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Chapter 8: Problem 52

Solve for \(x\) $$\left|\begin{array}{rr} x & 4 \\ -1 & x \end{array}\right|=20$$

Short Answer

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

Step by step solution

01

Calculate the determinant of the 2x2 matrix

The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) can be calculated by using the formula \(ad - bc\). In this case, the matrix is \(\begin{bmatrix} x & 4 \\ -1 & x \end{bmatrix}\), so the determinant is \(x*x - (-1*4) = x^2 + 4\).
02

Set up the equation

According to the exercise, the determinant equals to 20, so the equation to solve is \(x^2 + 4 = 20\). It can be simplified to \(x^2 = 20 - 4 = 16\).
03

Solve the equation for \(x\)

Since \(x^2\) equals to a positive number, there are two solutions to the equation \(x^2 = 16\). These are \(x = \sqrt{16} = 4\) and \(x = -\sqrt{16} = -4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

2x2 matrix
A 2x2 matrix is one of the simplest forms of a matrix, consisting of two rows and two columns of numbers. Matrices are useful mathematical tools for a variety of applications, including solving systems of linear equations and performing transformations in coordinate spaces. A matrix with this size looks like this: \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \] Here, each element, \(a, b, c,\) and \(d\), represents a number at their respective positions. In the exercise given, our matrix is \(\begin{bmatrix} x & 4 \ -1 & x \end{bmatrix}\).
  • The first row contains \(x\) and 4
  • The second row contains -1 and \(x\)
Matrices are foundational elements in linear algebra, and understanding how to manipulate them is crucial for mastering the subject.
Quadratic equations
Quadratic equations are polynomial equations of the second degree, and they usually take the form \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, and \(x\) represents the variable or unknown that we solve for. These equations are called "quadratic" from the Latin word "quadratus," meaning "square," because the variable is squared. In our simplified determinant equation, we have \(x^2 = 16\), which is a special case of a quadratic equation where the coefficients \(a = 1\), \(b = 0\), and \(c = -16\).
  • When solving a quadratic equation, one aims to find the values of \(x\) that make the equation true.
  • Quadratic equations can typically have two solutions because they are expressed by parabolic graphs that can intersect the \(x\)-axis at zero, one, or two points.
Exploring quadratic equations can help you understand various real-world phenomena where relationships are squared.
Solving equations
Solving equations involves finding the value(s) of the unknown variable(s) that satisfy the equation. Once we reach a point where the equation is simplified, we proceed to isolate the variable. In our exercise, we simplified the determinant equation to become \(x^2 = 16\). To solve for \(x\), perform the following steps:
  • Recognize that since \(x^2\) equals a positive number, there are two potential solutions due to the principle of taking a square root.
  • Take the square root of both sides of the equation to get \(x = \pm \sqrt{16}\).
  • Calculate \(x = 4\) and \(x = -4\).
This gives us two values of \(x\) that satisfy the original equation given by the matrix determinant expression. Understanding these steps allows you to tackle quadratic equations and determinants in similar problems efficiently.

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Most popular questions from this chapter

The sums have been evaluated. Solve the given system for \(a\) and \(b\) to find the least squares regression line for the points. Use a graphing utility to confirm the results. $$\left\\{\begin{array}{l} 5 b+10 a=20.2 \\ 10 b+30 a=50.1 \end{array}\right.$$

Find the general form of the equation of the line that passes through the two points. \((3,-3),(10,-1)\)

Find the domain of the function and identify any horizontal or vertical asymptotes. $$f(x)=\frac{x-4}{x^{2}+16}$$

Find the domain of the function and identify any horizontal or vertical asymptotes. $$f(x)=\frac{5}{x-6}$$

Find the point of equilibrium of the demand and supply equations. The point of equilibrium is the price \(p\) and the number of units \(x\) that satisfy both the demand and supply equations. Demand \(\quad\) Supply \(p=400-0.0002 x \quad p=225+0.0005 x\)
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