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Chapter 9: Problem 80

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{aligned} &3 x+2 y=-4\\\ &2 x-y=-5 \end{aligned}$$

Short Answer

Expert verified
( x, y ) = ( -2, 1 )

Step by step solution

01

Write the system in matrix form

First, write the system of equations as a matrix equation. The system given is: \[ \begin{aligned} 3x + 2y &= -4 \ 2x - y &= -5 \end{aligned} \] Bring it to the form \[ A \mathbf{x} = \mathbf{b}\] with: \[ A = \begin{pmatrix} 3 & 2 \ 2 & -1 \end{pmatrix}, \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \mathbf{b} = \begin{pmatrix} -4 \ -5 \end{pmatrix} \]
02

Find the determinant of matrix A ( D )

Calculate the determinant of matrix A : \[ A = \begin{pmatrix} 3 & 2 \ 2 & -1 \end{pmatrix} \] The determinant D of A is found using formula: \[ D = det(A) = (3)(-1)-(2)(2) = -3 - 4 = -7 \]
03

Find the determinants of matrices A_x and A_y

Form matrices A_x and A_y by replacing the respective columns of A with vector b : \[ A_x = \begin{pmatrix} -4 & 2 \ -5 & -1 \end{pmatrix}, A_y = \begin{pmatrix} 3 & -4 \ 2 & -5 \end{pmatrix} \] Compute det(A_x) : \[ det(A_x) = (-4)(-1)-(2)(-5) = 4 + 10 = 14 \] Compute det(A_y) : \[ det(A_y) = (3)(-5)-(2)(-4) = -15 + 8 = -7 \]
04

Apply Cramer's rule to find solutions for x and y

Using Cramer's rule: \[ x = \frac{det(A_x)}{D} = \frac{14}{-7} = -2 \] \[ y = \frac{det(A_y)}{D} = \frac{-7}{-7} = 1 \] Thus, the solution to the system is: \( x = -2 \) and \( y = 1 \)

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Key Concepts

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

Determinant
The determinant is a special number that can be calculated from a square matrix. For a 2脳2 matrix, the determinant is calculated as follows:
\ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \ det(A) = ad - bc
In the given exercise, we have the matrix A = \begin{pmatrix} 3 & 2 \ 2 & -1 \end{pmatrix}. Here, to compute the determinant D, we proceed as: \[ D = 3(-1) - 2(2) = -3 - 4 = -7 \] This determinant, -7, is crucial as it is used in Cramer's rule to find the solution of systems of equations when D is not zero. If the determinant was zero, it means the matrix is singular and we would need to use other methods to solve the equations.
System of Equations
A system of equations is a collection of two or more equations with the same set of variables. In this exercise, we need to solve the following system of equations:
3x + 2y = -4
2x - y = -5
Using Cramer's rule entails writing the system in matrix form as \[A\mathbf{x} = \mathbf{b}\], where
  • A is the coefficient matrix \(\begin{pmatrix} 3 & 2 \ 2 & -1 \end{pmatrix}\)
  • \(\mathbf{x}\) is the variable matrix \(\begin{pmatrix} x \ y \end{pmatrix}\)
  • \(\mathbf{b}\) is the constant matrix \(\begin{pmatrix} -4 \ -5 \end{pmatrix}\)
Then, using the determinant of matrix A, we can find the values of x and y by forming and using determinants of modified matrices derived from A.
Matrix
A matrix is a rectangular array of numbers arranged in rows and columns. It is a fundamental concept in linear algebra. For example:
A = \begin{pmatrix} a & b \ c & d \end{pmatrix}
Matrices are used to solve linear equations efficiently. In the provided exercise, we start with the matrix A:\[A = \begin{pmatrix} 3 & 2 \ 2 & -1 \end{pmatrix}\]
We then used two other matrices, \(A_x\) and \(A_y\), where columns of A are replaced with the constants from the equations:
  • \(A_x = \begin{pmatrix} -4 & 2 \ -5 & -1 \end{pmatrix}\)
  • \(A_y = \begin{pmatrix} 3 & -4 \ 2 & -5 \end{pmatrix}\)
Finally, we find the determinants of these matrices and apply Cramer's rule to solve the system of equations.

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Most popular questions from this chapter

Solve each system for \(x\) and y using Cramer's rule. Assume a and b are nonzero constants. $$\begin{array}{l} b^{2} x+a^{2} y=b^{2} \\ a x+b y=a \end{array}$$

Find the equation of the circle passing through the given points. $$(2,1),(-1,0), \text { and }(3,3)$$

Find the inverse, if it exists, for each matrix. $$\left[\begin{array}{rrrr} 3 & 2 & 0 & -1 \\ 2 & 0 & 1 & 2 \\ 1 & 2 & -1 & 0 \\ 2 & -1 & 1 & 1 \end{array}\right]$$

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{aligned} &3 x+2 y=-4\\\ &5 x-y=2 \end{aligned}$$

$$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right], \quad B=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right], \quad \text { and } \quad C=\left[\begin{array}{ll} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array}\right] $$ where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \((A B) C=A(B C)\) (associative property)
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