Problem 9 Solve the given system of equati... [FREE SOLUTION]

Chapter 8: Problem 9

Solve the given system of equations by Cramer's rule. $$ \begin{array}{r} u+2 v+\quad w=8 \\ 2 u-2 v+2 w=7 \\ u-4 v+3 w=1 \end{array} $$

Short Answer

Expert verified

$ u = 3, v = \frac{9}{8}, w = \frac{3}{4} $.

Step by step solution

Write the System of Equations in Matrix Form

Convert the given system of equations into a matrix form: $ A\mathbf{x} = \mathbf{b} $, where $ A $ is the coefficient matrix, $ \mathbf{x} $ is the column matrix of variables, and $ \mathbf{b} $ is the column matrix of constants.So, the system:\[\begin{align*}1u + 2v + 1w &= 8, \2u - 2v + 2w &= 7, \1u - 4v + 3w &= 1,\end{align*}\]can be written as:\[A = \begin{bmatrix} 1 & 2 & 1 \ 2 & -2 & 2 \ 1 & -4 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} u \ v \ w \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 8 \ 7 \ 1 \end{bmatrix}.\]

Compute the Determinant of the Coefficient Matrix

Calculate $ \det(A) $, the determinant of the matrix $ A $:\[\begin{vmatrix} 1 & 2 & 1 \ 2 & -2 & 2 \ 1 & -4 & 3 \end{vmatrix} = 1((-2)(3) - (-4)(2)) - 2((2)(3) - (1)(2)) + 1((2)(-4) - (1)(2)).\]Solving this gives:\[\det(A) = 1(-6 + 8) - 2(6 - 2) + 1(-8 - 2) = 2 - 8 - 10 = -16.\]

Compute Determinant for Variables

For each variable, compute the determinant by replacing the corresponding column of $ A $ with $ \mathbf{b} $.

Step 3.1: Compute Determinant for u

Replace the first column of $ A $ with $ \mathbf{b} $:\[A_u = \begin{bmatrix} 8 & 2 & 1 \ 7 & -2 & 2 \ 1 & -4 & 3 \end{bmatrix}.\]Compute $ \det(A_u) $:\[= 8((-2)(3) - (-4)(2)) - 2((7)(3) - (1)(2)) + 1((7)(-4) - (-2)(1)).\]Solving this gives:\[\det(A_u) = 8(-6 + 8) - 2(21 - 2) + 1(-28 + 2) = 16 - 38 - 26 = -48.\]

Step 3.2: Compute Determinant for v

Replace the second column of $ A $ with $ \mathbf{b} $:\[A_v = \begin{bmatrix} 1 & 8 & 1 \ 2 & 7 & 2 \ 1 & 1 & 3 \end{bmatrix}.\]Compute $ \det(A_v) $:\[= 1((7)(3) - (1)(2)) - 8((2)(3) - (1)(2)) + 1((2)(1) - (7)(1)).\]Solving this gives:\[\det(A_v) = 1(21 - 2) - 8(6 - 2) + 1(2 - 7) = 19 - 32 - 5 = -18.\]

Step 3.3: Compute Determinant for w

Replace the third column of $ A $ with $ \mathbf{b} $:\[A_w = \begin{bmatrix} 1 & 2 & 8 \ 2 & -2 & 7 \ 1 & -4 & 1 \end{bmatrix}.\]Compute $ \det(A_w) $:\[= 1((-2)(1) - (-4)(7)) - 2((2)(1) - (1)(7)) + 8((2)(-4) - (1)(-2)).\]Solving this gives:\[\det(A_w) = 1(-2 + 28) - 2(2 - 7) + 8(-8 + 2) = 26 + 10 - 48 = -12.\]

Solve for Each Variable Using Cramer's Rule

Using Cramer's Rule, solve each variable:\[u = \frac{\det(A_u)}{\det(A)} = \frac{-48}{-16} = 3,\]\[v = \frac{\det(A_v)}{\det(A)} = \frac{-18}{-16} = \frac{9}{8},\]\[w = \frac{\det(A_w)}{\det(A)} = \frac{-12}{-16} = \frac{3}{4}.\]

Conclusion: Write the Solution

The solution to the system of equations is:\[ u = 3, \quad v = \frac{9}{8}, \quad w = \frac{3}{4}.\]

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.

System of Equations

A system of equations is a collection of equations that you deal with all at once. Each equation has multiple variables, and the solution is the set of variable values that make all the equations true simultaneously. In our example, we have three equations with variables: $ u, v, $ and $ w $. These equations together form a system:\[\begin{array}{r}u+2 v+ w=8 \2 u-2 v+2 w=7 \u-4 v+3 w=1 \\end{array}\]Such systems can often be solved using various algebraic methods, and in this case, Cramer's rule provides an efficient tactic when the system meets certain criteria.

Determinant Calculation

Determining the solution involves calculating the determinant, a special number that can be computed from a square matrix. It plays a crucial role in solving a system of linear equations using Cramer's rule.

For a 3脳3 matrix $ A $, the determinant $ \det(A) $ is calculated based on its elements. For our matrix:\[A = \begin{bmatrix} 1 & 2 & 1 \2 & -2 & 2 \1 & -4 & 3 \end{bmatrix}\]data- Calculate it as follows: - Multiply and subtract products in a specific pattern: - Take the primary diagonal product and subtract the other diagonals.
After calculation, we find $ \det(A) = -16 $. This determinant is vital because if it were zero, Cramer's rule wouldn鈥檛 apply.

Matrix Algebra

Matrix algebra is a fundamental part of linear algebra. It refers to a set of operations that can be performed on matrices - ordered grids of numbers. In this context, matrices provide a compact way to handle linear equations.

For our system, the matrix form is:

Coefficient matrix $ A $: $ \begin{bmatrix} 1 & 2 & 1 \2 & -2 & 2 \1 & -4 & 3 \end{bmatrix} $
Variable column matrix $ \mathbf{x} $: $ \begin{bmatrix} u \v \w \end{bmatrix} $
Constants matrix $ \mathbf{b} $: $ \begin{bmatrix} 8 \7 \1 \end{bmatrix} $

Matrix calculations enable us to handle multiple equations efficiently, and applications like Cramer's rule rely on these operations to find solutions.

Linear Algebra Concepts

Linear algebra covers the study of vectors and matrices, which are central to numerous mathematical and real-world applications, including systems of equations.

Cramer's rule, a part of linear algebra, is a straightforward method for solving square systems of linear equations鈥攚here the number of equations equals the number of unknowns when the determinant is non-zero.

Each variable in the system corresponds to a modified version of the original matrix, with one column swapped for the constants.

In this exercise, determinants from modified matrices were calculated to find the solution variables $ u, v, $ and $ w $. This approach helps understand the intrinsic relationship between matrices and their determinants in solving equations.

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

In Problems, the given matrix $\mathbf{A}$ is symmetric. Find an orthogonal matrix $\mathbf{P}$ that diagonalizes $\mathbf{A}$ and the diagonal matrix $\mathbf{D}$ such that $\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}$ $$ \left(\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{array}\right) $$

Find the least squares line for the given data. $$ (2,1),(3,2),(4,3),(5,2) $$

Determine whether the given message is a code word in the Hamming $(7,4)$ code. If it is, decode it. If it is not, correct the single error and decode the corrected message. $$ \left.\begin{array}{llllllll} (1 & 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right) $$

In Problems, determine whether the given matrix $\mathbf{A}$ is diagonalizable. If so, find the matrix $\mathbf{P}$ that diagonalizes $\mathbf{A}$ and the diagonal matrix $\mathbf{D}$ such that $\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}$. $$ \left(\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 3 & -2 \\ -5 & 3 & 8 \end{array}\right) $$

Find the least squares line for the given data. $$ (1,1),(2,1.5),(3,3),(4,4.5),(5,5) $$

See all solutions

Recommended explanations on Physics Textbooks

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视