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Chapter 8: Problem 16

Use Cramer's rule to solve each system of equations, if possible. $$\begin{array}{r} x-4 y=-7 \\ 3 x+8 y=19 \end{array}$$

Short Answer

Expert verified
The solution is \((x, y) = (1, 2)\).

Step by step solution

01

Write the system in matrix form

The given system of equations can be written in the form of a matrix equation \ \( AX = B \) \, where \ \( A \) is the coefficient matrix, \ \( X \) is the column matrix of variables, and \ \( B \) is the column matrix of constants. In this case, \[A = \begin{bmatrix} 1 & -4 \ 3 & 8 \end{bmatrix},\ X = \begin{bmatrix} x \ y \end{bmatrix},\ B = \begin{bmatrix} -7 \ 19 \end{bmatrix}\]
02

Calculate the determinant of matrix A

The determinant of matrix \ \( A \) is calculated as follows: \\[\det(A) = \begin{vmatrix} 1 & -4 \ 3 & 8 \end{vmatrix} = (1)(8) - (3)(-4) = 8 + 12 = 20\]Since the determinant is non-zero, Cramer's rule can be applied.
03

Find the determinant for x - \\(\det(A_x)\\)

To find \ \( x \), replace the first column of \ \( A \) with matrix \ \( B \) and calculate the determinant. \[A_x = \begin{bmatrix} -7 & -4 \ 19 & 8 \end{bmatrix}\]Then,\[\det(A_x) = (-7)(8) - (19)(-4) = -56 + 76 = 20\]
04

Find the determinant for y - \\(\det(A_y)\\)

To find \ \( y \), replace the second column of \ \( A \) with matrix \ \( B \) and calculate the determinant. \[A_y = \begin{bmatrix} 1 & -7 \ 3 & 19 \end{bmatrix}\]Then,\[\det(A_y) = (1)(19) - (3)(-7) = 19 + 21 = 40\]
05

Apply Cramer's Rule

Using Cramer's Rule, the solutions for \ \( x \) and \ \( y \) are given by:\[x = \frac{\det(A_x)}{\det(A)} = \frac{20}{20} = 1\]\[y = \frac{\det(A_y)}{\det(A)} = \frac{40}{20} = 2\]
06

Solution of the System

The solution to the system of equations is \ \( x = 1 \) and \ \( y = 2 \). Thus, the point \( (x, y) = (1, 2) \) satisfies both equations.

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Key Concepts

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

Understanding Determinants
The determinant is a special number that you can calculate from a square matrix. It provides important properties of the matrix, such as whether it is invertible or not. For a 2x2 matrix like \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix},\] the determinant is calculated as:\[\text{det}(A) = ad - bc.\]Understanding determinants is crucial because
  • If the determinant of a matrix is zero, it means that the system of equations has no unique solutions.
  • If the determinant is non-zero, the system can be uniquely solved using methods like Cramer's Rule.
In the example provided, the determinant of matrix A is 20. This non-zero value confirms that Cramer's Rule can be applied to find a unique solution to the system of equations. Remember, the determinant acts as a gateway 鈥� only when it is non-zero can you proceed with solving using Cramer's Rule or finding the inverse of the matrix.
Decoding Systems of Equations
A system of equations involves finding the values of variables that satisfy all the given equations simultaneously. In our example, we have:\[\begin{array}{r}x - 4y = -7 \3x + 8y = 19\end{array}\]These two linear equations form a system that we need to solve. Here's why understanding systems of equations is important:
  • They are fundamental in mathematics, used in various fields to represent relationships between quantities.
  • Simplifying or solving them allows us to interpret real-world situations effectively.
Our task is to determine the values of \(x\) and \(y\) that satisfy both equations. Using methods like substitution, elimination, or Cramer's Rule helps us find these solutions efficiently. For this exercise, Cramer's Rule provides a structured approach by leveraging matrices.
Matrix Form Representation
Converting a system of equations into matrix form is a powerful technique that simplifies solving complex equations. The matrix form \[AX = B\]represents these relationships:
  • \(A\) is the coefficient matrix containing the coefficients of the variables.
  • \(X\) is the variable matrix, representing the unknowns of the system.
  • \(B\) is the constant matrix containing the constants from the equations.
For our specific problem:\[A = \begin{bmatrix} 1 & -4 \ 3 & 8 \end{bmatrix}, \X = \begin{bmatrix} x \ y \end{bmatrix}, \B = \begin{bmatrix} -7 \ 19 \end{bmatrix}.\] By using matrix notation, we create a unified form which can be manipulated using various algebraic rules. This allows techniques like Cramer's Rule to be applied seamlessly, as matrix operations become streamlined and systematic.

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Most popular questions from this chapter

In Exercise \(58,\) you were asked to solve this system of equations using an augmented matrix. $$\begin{array}{rr} 2 x+z+y= & -3 \\ 2 y-z+x= & 0 \\ x+y+2 z= & 5 \end{array}$$ A graphing calculator or graphing utility can be use to solve systems of linear equations by entering the coefficients of the matrix. Solve this system and confirm your answer with the calculator's answer.

Explain the mistake that is made. $$\text { Multiply }\left[\begin{array}{ll} 3 & 2 \\\1 & 4\end{array}\right]\left[\begin{array}{ll}-1 & 3 \\\\-2 & 5\end{array}\right]$$ Solution: Multiply corresponding elements. Simplify. $$\left[\begin{array}{ll}3 & 2 \\\1 & 4 \end{array}\right]\left[\begin{array}{ll}-1 & 3 \\\\-2 & 5\end{array}\right]=\left[\begin{array}{ll} (3)(-1) & (2)(3) \\\\(1)(-2) & (4)(5)\end{array}\right]$$This is incorrect. What mistake was made?

Ginger talks Gary into putting less money in the money market and more money in the stock (see Exercise 95 ). They place \(10,000\) of their savings into investments. They put some in a money market account earning \(3 \%\) interest, some in a mutual fund that has been averaging \(7 \%\) a year, and some in a stock that rose \(10 \%\) last year. If they put \( 3,000\) more in the stock than in the mutual fund and the mutual fund and stock have the same growth in the next year as they did in the previous year, they will earn \(\$ 840\) in a year. How much money did they put in each of the three investments?

Determine whether each of the following statements is true or false: When a system of linear equations is represented by a square augmented matrix, the system of equations always has a unique solution.

Explain the mistake that is made. Find the inverse of \(A\) given that \(A=\left[\begin{array}{rr}2 & 5 \\ 3 & 10\end{array}\right]\) Solution: $$A^{-1}=\frac{1}{A} \quad A^{-1}=\frac{1}{\left[\begin{array}{rr}2 & 5 \\\3 & 10\end{array}\right]}$$ Simplify. $$A^{-1}=\left[\begin{array}{ll} \frac{1}{2} & \frac{1}{5} \\\\\frac{1}{3} & \frac{1}{10}\end{array}\right]$$ This is incorrect. What mistake was made?
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