(b) Use the answer from part (a) to solve the linear system A-1 A -4x1 -3x1 21 21 I2 I3 = 88 -4 -3 + -12 -10 9 6 3-2 12x2 + 9x3 10x2 + 6x3 3x2 2x3 = = 5
(b) Use the answer from part (a) to solve the linear system A-1 A -4x1 -3x1 21 21 I2 I3 = 88 -4 -3 + -12 -10 9 6 3-2 12x2 + 9x3 10x2 + 6x3 3x2 2x3 = = 5
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![### Solving Linear Systems Using Matrix Inversion
Given matrix \( A \) and its inverse \( A^{-1} \):
\[
A = \begin{bmatrix}
-4 & -12 & 9 \\
-3 & -10 & 6 \\
1 & 3 & -2 \\
\end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix}
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\end{bmatrix}
\]
#### Step (b): Solving the Linear System
Use the inverse matrix \( A^{-1} \) from part (a) to solve the following system of linear equations:
\[
\begin{cases}
-4x_1 - 12x_2 + 9x_3 = -4 \\
-3x_1 - 10x_2 + 6x_3 = 5 \\
x_1 + 3x_2 - 2x_3 = -1 \\
\end{cases}
\]
This system can be written in matrix form as:
\[ A\vec{x} = \vec{b} \]
Where:
\[ A = \begin{bmatrix}
-4 & -12 & 9 \\
-3 & -10 & 6 \\
1 & 3 & -2 \\
\end{bmatrix}, \quad \vec{x} = \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}, \quad \vec{b} = \begin{bmatrix}
-4 \\
5 \\
-1 \\
\end{bmatrix}
\]
To find \( \vec{x} \), we use \( A^{-1} \) from part (a):
\[ \vec{x} = A^{-1} \vec{b} \]
\[ \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} = \begin{bmatrix}
\text{[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e366a85-c23b-450a-9bc7-d2ea0be52d3f%2Ffc41c424-8b3b-4dff-a8b7-c6fb2848f926%2Felo8yko_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Linear Systems Using Matrix Inversion
Given matrix \( A \) and its inverse \( A^{-1} \):
\[
A = \begin{bmatrix}
-4 & -12 & 9 \\
-3 & -10 & 6 \\
1 & 3 & -2 \\
\end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix}
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\text{[Enter Value]} & \text{[Enter Value]} & \text{[Enter Value]} \\
\end{bmatrix}
\]
#### Step (b): Solving the Linear System
Use the inverse matrix \( A^{-1} \) from part (a) to solve the following system of linear equations:
\[
\begin{cases}
-4x_1 - 12x_2 + 9x_3 = -4 \\
-3x_1 - 10x_2 + 6x_3 = 5 \\
x_1 + 3x_2 - 2x_3 = -1 \\
\end{cases}
\]
This system can be written in matrix form as:
\[ A\vec{x} = \vec{b} \]
Where:
\[ A = \begin{bmatrix}
-4 & -12 & 9 \\
-3 & -10 & 6 \\
1 & 3 & -2 \\
\end{bmatrix}, \quad \vec{x} = \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}, \quad \vec{b} = \begin{bmatrix}
-4 \\
5 \\
-1 \\
\end{bmatrix}
\]
To find \( \vec{x} \), we use \( A^{-1} \) from part (a):
\[ \vec{x} = A^{-1} \vec{b} \]
\[ \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} = \begin{bmatrix}
\text{[
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