I. Consider the following linear system: 3x1 + 9x2 + 3x3 = 6 2x1 -2x1 – 4x2 - x3 = 12 = -8 We can view this system as a matrix equation: 3 9 3 6 2 0 -1 -2 -4 12 Which has the form: Ax X = B To solve the original system, we can solve this equation for X by multiplying by A-1: A-1 × (A × X) = A-1 × B (A-1 × A) × X = A¯1 × B Iz x X = A-1 × B x = A-1 x B Use the bottom equation to solve for X. Find A-1 for the matrix A and solve the original system by using A-1 and B in the (1) equation below. X = A-1 x B
I. Consider the following linear system: 3x1 + 9x2 + 3x3 = 6 2x1 -2x1 – 4x2 - x3 = 12 = -8 We can view this system as a matrix equation: 3 9 3 6 2 0 -1 -2 -4 12 Which has the form: Ax X = B To solve the original system, we can solve this equation for X by multiplying by A-1: A-1 × (A × X) = A-1 × B (A-1 × A) × X = A¯1 × B Iz x X = A-1 × B x = A-1 x B Use the bottom equation to solve for X. Find A-1 for the matrix A and solve the original system by using A-1 and B in the (1) equation below. X = A-1 x B
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Matrix Equation and System of Linear Equations**
I. **Consider the following linear system:**
\[
\begin{align*}
3x_1 + 9x_2 + 3x_3 &= 6 \\
2x_1 - x_3 &= 12 \\
-2x_1 - 4x_2 &= -8 \\
\end{align*}
\]
We can view this system as a matrix equation:
\[
\begin{bmatrix}
3 & 9 & 3 \\
2 & 0 & -1 \\
-2 & -4 & 0 \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix}
=
\begin{bmatrix}
6 \\
12 \\
-8 \\
\end{bmatrix}
\]
Which has the form: \( A \times X = B \)
To solve the original system, we can solve this equation for \( X \) by multiplying by \( A^{-1} \):
\[
\begin{align*}
A^{-1} \times (A \times X) &= A^{-1} \times B \\
(A^{-1} \times A) \times X &= A^{-1} \times B \\
I_3 \times X &= A^{-1} \times B \\
X &= A^{-1} \times B \\
\end{align*}
\]
Use the bottom equation to solve for \( X \).
[1] **Find \( A^{-1} \) for the matrix \( A \) and solve the original system by using \( A^{-1} \) and \( B \) in the equation below.**
\[ X = A^{-1} \times B \]
[2] **Use the matrix \( A^{-1} \) you just found to solve the system:**
\[
\begin{align*}
3x_1 + 9x_2 + 3x_3 &= 12 \\
2x_1 - x_3 &= 9 \\
-2x_1 - 4x_2 &= -3 \\
\end{align*}
\]
Show your work here.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa9a9ea7-0a60-40f2-88c8-00f846edd995%2Feac341ea-65ee-4000-815d-8255393bf258%2Fac0g16v_processed.png&w=3840&q=75)
Transcribed Image Text:**Matrix Equation and System of Linear Equations**
I. **Consider the following linear system:**
\[
\begin{align*}
3x_1 + 9x_2 + 3x_3 &= 6 \\
2x_1 - x_3 &= 12 \\
-2x_1 - 4x_2 &= -8 \\
\end{align*}
\]
We can view this system as a matrix equation:
\[
\begin{bmatrix}
3 & 9 & 3 \\
2 & 0 & -1 \\
-2 & -4 & 0 \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix}
=
\begin{bmatrix}
6 \\
12 \\
-8 \\
\end{bmatrix}
\]
Which has the form: \( A \times X = B \)
To solve the original system, we can solve this equation for \( X \) by multiplying by \( A^{-1} \):
\[
\begin{align*}
A^{-1} \times (A \times X) &= A^{-1} \times B \\
(A^{-1} \times A) \times X &= A^{-1} \times B \\
I_3 \times X &= A^{-1} \times B \\
X &= A^{-1} \times B \\
\end{align*}
\]
Use the bottom equation to solve for \( X \).
[1] **Find \( A^{-1} \) for the matrix \( A \) and solve the original system by using \( A^{-1} \) and \( B \) in the equation below.**
\[ X = A^{-1} \times B \]
[2] **Use the matrix \( A^{-1} \) you just found to solve the system:**
\[
\begin{align*}
3x_1 + 9x_2 + 3x_3 &= 12 \\
2x_1 - x_3 &= 9 \\
-2x_1 - 4x_2 &= -3 \\
\end{align*}
\]
Show your work here.
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