11. If A¹ = [1 27 4 6 12. If A¹ = -1 solve AX = 1 2 0] 3 1 2 4 1 [3] for X. -2 solve AX = 1 for X. 4
11. If A¹ = [1 27 4 6 12. If A¹ = -1 solve AX = 1 2 0] 3 1 2 4 1 [3] for X. -2 solve AX = 1 for X. 4
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
number 11, 12, 14, and 15.
![### Problem 11
Given:
\[ A^{-1} = \begin{bmatrix} 1 & 2 \\ 4 & 6 \end{bmatrix} \]
To solve:
\[ AX = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \]
Find the matrix \(X\).
### Problem 12
Given:
\[ A^{-1} = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 1 \\ 2 & 4 & 1 \end{bmatrix} \]
To solve:
\[ AX = \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} \]
Find the matrix \(X\).
### Explanation of Matrices
In these problems, you're given the inverse of a matrix \(A\) and need to solve for vector \(X\) in the equation \(AX = B\). To solve for \(X\), you can use the property that multiplying both sides of the equation by \(A^{-1}\) gives \(X = A^{-1}B\).
In Problem 11, you work with a 2x2 system, while Problem 12 involves a 3x3 system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9d8e919-6c53-43b3-96c1-5f9438fdaf13%2F659f5a69-a298-44d3-a9b2-e6609b0e43bf%2Fdjz3i8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 11
Given:
\[ A^{-1} = \begin{bmatrix} 1 & 2 \\ 4 & 6 \end{bmatrix} \]
To solve:
\[ AX = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \]
Find the matrix \(X\).
### Problem 12
Given:
\[ A^{-1} = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 1 \\ 2 & 4 & 1 \end{bmatrix} \]
To solve:
\[ AX = \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} \]
Find the matrix \(X\).
### Explanation of Matrices
In these problems, you're given the inverse of a matrix \(A\) and need to solve for vector \(X\) in the equation \(AX = B\). To solve for \(X\), you can use the property that multiplying both sides of the equation by \(A^{-1}\) gives \(X = A^{-1}B\).
In Problem 11, you work with a 2x2 system, while Problem 12 involves a 3x3 system.
![**Instruction:**
In Problems 15 – 19, use an inverse matrix to find the solution to the systems.
**Problem 15:**
Solve the following system of equations:
\[
\begin{align*}
x + 3y - 6z &= 7 \\
2x - y + 2z &= 0 \\
x + y + 2z &= -1
\end{align*}
\]
Solution: \((1, 0, -1)\)
**Problem 16:**
Solve the following system of equations:
\[
\begin{align*}
x + 2y + z &= 5 \\
2x + y - 3z &= -2 \\
3x + y + 4z &= -5
\end{align*}
\]
Solution: \((-3, 4, 0)\)
**Problem 17:**
Solve the following system of equations:
\[
\begin{align*}
x + 2y &= \\
2x + 3y &= \\
-x - 2 &=
\end{align*}
\]
Note: Problem 17 is incomplete; additional information is needed to solve.
**Explanation:**
For each system of equations, you can represent it in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. Using the inverse of the matrix \(A\), the solution can be found by calculating \(X = A^{-1}B\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9d8e919-6c53-43b3-96c1-5f9438fdaf13%2F659f5a69-a298-44d3-a9b2-e6609b0e43bf%2Frv8dct4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Instruction:**
In Problems 15 – 19, use an inverse matrix to find the solution to the systems.
**Problem 15:**
Solve the following system of equations:
\[
\begin{align*}
x + 3y - 6z &= 7 \\
2x - y + 2z &= 0 \\
x + y + 2z &= -1
\end{align*}
\]
Solution: \((1, 0, -1)\)
**Problem 16:**
Solve the following system of equations:
\[
\begin{align*}
x + 2y + z &= 5 \\
2x + y - 3z &= -2 \\
3x + y + 4z &= -5
\end{align*}
\]
Solution: \((-3, 4, 0)\)
**Problem 17:**
Solve the following system of equations:
\[
\begin{align*}
x + 2y &= \\
2x + 3y &= \\
-x - 2 &=
\end{align*}
\]
Note: Problem 17 is incomplete; additional information is needed to solve.
**Explanation:**
For each system of equations, you can represent it in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. Using the inverse of the matrix \(A\), the solution can be found by calculating \(X = A^{-1}B\).
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