Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.2: Systems Of Linear Equations In Two Variables
Problem 41E
Related questions
Question
![**Solving a System of Equations using Cramer's Rule**
In this example, we will demonstrate how to use Cramer's Rule to solve a system of linear equations.
Consider the following system of equations:
\[
\begin{cases}
4x_1 + 4x_2 - x_3 = 4 \\
-3x_1 - 4x_2 + x_3 = -2 \\
-x_1 - 2x_2 + x_3 = -4
\end{cases}
\]
Let's solve this system step-by-step using Cramer's Rule.
##### Step 1: Write down the coefficient matrix \(A\).
\[
A = \begin{pmatrix}
4 & 4 & -1 \\
-3 & -4 & 1 \\
-1 & -2 & 1
\end{pmatrix}
\]
##### Step 2: Calculate the determinant of \(A\), denoted as \( \text{det}(A) \).
For the given 3x3 matrix:
\[
\text{det}(A) = 4 \begin{vmatrix} -4 & 1 \\ -2 & 1 \end{vmatrix} - 4 \begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -3 & -4 \\ -1 & -2 \end{vmatrix}
\]
Solving these 2x2 determinants individually:
\[
\begin{vmatrix} -4 & 1 \\ -2 & 1 \end{vmatrix} = (-4)(1) - (1)(-2) = -4 + 2 = -2
\]
\[
\begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} = (-3)(1) - (1)(-1) = -3 + 1 = -2
\]
\[
\begin{vmatrix} -3 & -4 \\ -1 & -2 \end{vmatrix} = (-3)(-2) - (-4)(-1) = 6 - 4 = 2
\]
Therefore:
\[
\text{det}(A) = 4(-2) - 4(-2) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63ea7450-a183-49e4-8fe5-4699c8139185%2F8ecdb04a-55f4-45c6-8ed2-f7228b192832%2F9jxpmbp.png&w=3840&q=75)
Transcribed Image Text:**Solving a System of Equations using Cramer's Rule**
In this example, we will demonstrate how to use Cramer's Rule to solve a system of linear equations.
Consider the following system of equations:
\[
\begin{cases}
4x_1 + 4x_2 - x_3 = 4 \\
-3x_1 - 4x_2 + x_3 = -2 \\
-x_1 - 2x_2 + x_3 = -4
\end{cases}
\]
Let's solve this system step-by-step using Cramer's Rule.
##### Step 1: Write down the coefficient matrix \(A\).
\[
A = \begin{pmatrix}
4 & 4 & -1 \\
-3 & -4 & 1 \\
-1 & -2 & 1
\end{pmatrix}
\]
##### Step 2: Calculate the determinant of \(A\), denoted as \( \text{det}(A) \).
For the given 3x3 matrix:
\[
\text{det}(A) = 4 \begin{vmatrix} -4 & 1 \\ -2 & 1 \end{vmatrix} - 4 \begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -3 & -4 \\ -1 & -2 \end{vmatrix}
\]
Solving these 2x2 determinants individually:
\[
\begin{vmatrix} -4 & 1 \\ -2 & 1 \end{vmatrix} = (-4)(1) - (1)(-2) = -4 + 2 = -2
\]
\[
\begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} = (-3)(1) - (1)(-1) = -3 + 1 = -2
\]
\[
\begin{vmatrix} -3 & -4 \\ -1 & -2 \end{vmatrix} = (-3)(-2) - (-4)(-1) = 6 - 4 = 2
\]
Therefore:
\[
\text{det}(A) = 4(-2) - 4(-2) +
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