/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Use Cramers rule to solve those ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 42

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{array}{l} 5 x-3 y-z=16 \\ 2 x+y-3 z=5 \\ 3 x-2 y+2 z=5 \end{array}\right.$$

Short Answer

Expert verified
The solution is \(x = -\frac{11}{13}\), \(y = -4\), \(z = -\frac{28}{13}\).

Step by step solution

01

Write the System in Matrix Form

First, we write the system of equations as a matrix equation of the form **AX = B**, where **A** is the coefficient matrix, **X** is the column matrix of variables, and **B** is the constants' matrix. Given system: \[\begin{align*}5x - 3y - z &= 16 \2x + y - 3z &= 5 \3x - 2y + 2z &= 5\end{align*}\]Matrix form: \[A = \begin{bmatrix} 5 & -3 & -1 \2 & 1 & -3 \3 & -2 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} 16 \ 5 \ 5 \end{bmatrix}\]
02

Calculate the Determinant of the Coefficient Matrix

Calculate the determinant \(D\) of the coefficient matrix \(A\). Using the formula for the determinant of a 3x3 matrix:\[D = \begin{vmatrix} 5 & -3 & -1 \2 & 1 & -3 \3 & -2 & 2 \end{vmatrix}\]To compute this:\[= 5 \begin{vmatrix} 1 & -3 \-2 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 2 & -3 \3 & 2 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \3 & -2 \end{vmatrix}\]Solving these:\[= 5[(1)(2) - (-3)(-2)] + 3[(2)(2) - (-3)(3)] - [(2)(-2) - (1)(3)]\]\[= 5(2 - 6) + 3(4 + 9) - (-4 - 3)\]\[= 5(-4) + 3(13) + 7\]\[= -20 + 39 + 7\]\[= 26\]Since \(D = 26 eq 0\), we can use Cramer's Rule.
03

Apply Cramer's Rule to Solve for Each Variable

Cramer's Rule states that each variable \(x\), \(y\), and \(z\) can be found using:\[x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}\]We need to calculate \(D_x\), \(D_y\), \(D_z\).
04

Calculate \(D_x\)

Replace the first column of \(A\) with \(B\) to calculate \(D_x\).\[D_x = \begin{vmatrix} 16 & -3 & -1 \5 & 1 & -3 \5 & -2 & 2 \end{vmatrix}\]Solving this determinant:\[= 16(1)(2) - (-3)(5)(2) - 1(5)(-2) + (-3)(3)(5) - 1(5) - 2(16)(-2)\]\[= 32 + 30 - 10 - 15 + 5 - 64\]\[= -22\]
05

Calculate \(D_y\)

Replace the second column of \(A\) with \(B\) to calculate \(D_y\).\[D_y = \begin{vmatrix} 5 & 16 & -1 \2 & 5 & -3 \3 & 5 & 2 \end{vmatrix}\]Solving this determinant:\[= 5(5)(2) - (16)(3)(2) - 1(2)(5) + (16)(3)(-1) - 3(5) - (-1)(5)\]\[= 50 - 96 - 10 - 48 + 0 + 0\]\[= -104\]
06

Calculate \(D_z\)

Replace the third column of \(A\) with \(B\) to calculate \(D_z\).\[D_z = \begin{vmatrix} 5 & -3 & 16 \2 & 1 & 5 \3 & -2 & 5 \end{vmatrix}\]Solving this determinant:\[= 5 \begin{vmatrix} 1 & 5 \-2 & 5 \end{vmatrix} - (-3) \begin{vmatrix} 2 & 5 \3 & 5 \end{vmatrix} + 16 \begin{vmatrix} 2 & 1 \3 & -2 \end{vmatrix}\]Calculate these small matrices:\[= 5(1)(5) - 5(-2) + 3(2)(5) - 3(3)(1) + 16(2)(-2) - 16(1)(3)\]\[= 25 + 10 + 30 - 9 - 64 - 48\]\[= -56\]
07

Compute the Variables using Cramer's Rule

Now solve for \(x\), \(y\), and \(z\) using Cramer's rule:\[x = \frac{D_x}{D} = \frac{-22}{26} = -\frac{11}{13}\]\[y = \frac{D_y}{D} = \frac{-104}{26} = -4\]\[z = \frac{D_z}{D} = \frac{-56}{26} = -\frac{28}{13}\]
08

Conclusion

Thus, the solution to the system of equations is \(x = -\frac{11}{13}\), \(y = -4\), and \(z = -\frac{28}{13}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Determinant of a Matrix
The determinant of a matrix is a special number that can be calculated from its elements. It's often denoted as \(D\) for a matrix \(A\).
In solving systems of linear equations using Cramer's Rule, the determinant plays a crucial role by helping us understand if the system has a unique solution. This is because, for Cramer's Rule to be applicable, the determinant of the coefficient matrix must be non-zero \((D eq 0)\).

To find the determinant of a 3x3 matrix, you use the formula involving cross multiplication of its elements:
  • Select the first row element and multiply it with the determinant of the smaller 2x2 matrix formed without the first row and column where this element is.
  • Repeat for the other two elements of the first row, changing signs as you go along: first positive, second negative, third positive.
  • Add these three results.
When solving problems, calculating determinants correctly is key, as any mistake will affect subsequent calculations.
Gaussian Elimination
Gaussian elimination is a straightforward method for solving systems of linear equations. It transforms a system into a simpler equivalent system from which solutions can be directly obtained.

Here's how it works:
  • The technique involves using row operations to change the original matrix into an upper triangular matrix, where all elements below the diagonal are zeros.
  • Once in this form, known as "row echelon form," you can easily find the variable values through back-substitution.
  • It is especially useful when the determinant of the matrix is zero \((D = 0)\), meaning Cramer's Rule cannot be applied.
Although Gaussian elimination can handle systems where Cramer's Rule falls short, care must be taken with row operations to maintain equivalent systems.
System of Linear Equations
A system of linear equations is a set of equations with multiple variables where each equation is linear.
Simply put, each term is either a constant or the product of a constant and a single variable.

To solve a system means finding values for the variables that satisfy all included equations simultaneously. There are a few key characteristics of these systems:
  • They are represented in the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constants matrix.
  • They can have a unique solution, infinitely many solutions, or no solution at all, depending on the relationships between the equations.
  • Techniques such as Cramer's Rule, Gaussian elimination, and matrix methods are employed to find solutions efficiently.
Understanding the structure and behavior of these systems is crucial for effectively using them in various mathematical and engineering contexts.
Matrix Methods
Matrix methods offer a powerful approach for solving systems of linear equations. They are grounded in the idea of representing equations in a condensed form using matrices.
These methods enable systematic solution strategies:
  • For a given system, first express it as the matrix equation \(AX = B\) using the coefficient, variable, and constant matrices.
  • Methods like Cramer's Rule and Gaussian Elimination are utilized to manipulate these matrices to find a solution.
  • If \(D eq 0\), Cramer's Rule provides a direct way to solve for variables by replacing columns of the matrix with the constant matrix and calculating the determinant.
  • For \(D = 0\), or for larger systems, other techniques like Gaussian Elimination might be more practical.
Mastering these methods requires practice but opens up efficient ways to handle complex systems, making them indispensable tools in algebra.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Use a graphing utility to approximate the solutions \((x, y)\) of each system. Zoom in on the relevant intersection points until you are sure of the first two decimal places of each coordinate. (b) In Exercises \(23-28\) only, also use an algebraic method of solution. Round the answers to three decimal places and check to see that your results are consistent with the graphical estimates obtained in part (a). $$\left\\{\begin{array}{l}y=\sqrt[3]{x} \\\y=\ln x\end{array}\right.$$

Find all values of \(x\) for which \(\left|\begin{array}{ccc}x-4 & 0 & 0 \\ 0 & x+4 & 0 \\ 0 & 0 & x+1\end{array}\right|=0\)

The sum of two numbers is \(64 .\) Twice the larger number plus five times the smaller number is \(20 .\) Find the two numbers. (Let \(x\) denote the larger number and let \(y\) denote the smaller number.)

A function \(f\) is defined as follows. The domain of \(f\) is the set of all \(2 \times 2\) matrices (with real entries). If \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) then \(f(A)=a d-b c\) (a) Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right),\) and \(B=\left(\begin{array}{cc}3 & -1 \\ 5 & 8\end{array}\right) .\) Compute \(f(A), f(B),\) and \(f(A B) .\) Is it true, in this case, that \(f(A) \cdot f(B)=f(A B) ?\) (b) Let \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(B=\left(\begin{array}{ll}e & f \\ g & h\end{array}\right) .\) Show that \(f(A) \cdot f(B)=f(A B)\).

In a two-digit number, the sum of the digits is \(14 .\) Twice the tens digit exceeds the units digit by one. Find the number.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.