(2) Now, use your answer to part (a) to solve the following system of equations for X. 1 AX = 10 X = 27 (3) Finally, solve the following system of equations for Y. AY =| 21 Y = 40
(2) Now, use your answer to part (a) to solve the following system of equations for X. 1 AX = 10 X = 27 (3) Finally, solve the following system of equations for Y. AY =| 21 Y = 40
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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Question
Solve part 2 and 3 please with explanation
![```
(2) Now, use your answer to part (a) to solve the following system of equations for \( X \).
\[
AX = \begin{bmatrix} 1 \\ 10 \\ 27 \end{bmatrix}, \quad X = \begin{bmatrix} \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \end{bmatrix}
\]
(3) Finally, solve the following system of equations for \( Y \).
\[
AY = \begin{bmatrix} 5 \\ 21 \\ 40 \end{bmatrix}, \quad Y = \begin{bmatrix} \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \end{bmatrix}
\]
```
This layout involves solving systems of linear equations using the matrix form, where \( AX \) and \( AY \) are given column matrices, and the goal is to determine the unknown matrices \( X \) and \( Y \). The boxes represent areas where you would input or calculate the missing values.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36a79c17-bf73-44c3-9503-d2f30f855206%2F629abbd2-8754-4237-84b5-a406b1867695%2Fur2yi7c_processed.png&w=3840&q=75)
Transcribed Image Text:```
(2) Now, use your answer to part (a) to solve the following system of equations for \( X \).
\[
AX = \begin{bmatrix} 1 \\ 10 \\ 27 \end{bmatrix}, \quad X = \begin{bmatrix} \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \end{bmatrix}
\]
(3) Finally, solve the following system of equations for \( Y \).
\[
AY = \begin{bmatrix} 5 \\ 21 \\ 40 \end{bmatrix}, \quad Y = \begin{bmatrix} \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \\ \boxed{\phantom{s}} \end{bmatrix}
\]
```
This layout involves solving systems of linear equations using the matrix form, where \( AX \) and \( AY \) are given column matrices, and the goal is to determine the unknown matrices \( X \) and \( Y \). The boxes represent areas where you would input or calculate the missing values.
![## Part 1
Rework problem 26 from section 6.2 of your text, involving a matrix \( A \) and two matrix equations. Use the following matrix instead of the one given in your text.
\[
A = \begin{bmatrix}
1 & -2 & 3 \\
1 & -1 & 5 \\
-3 & 9 & -2
\end{bmatrix}
\]
1. First, find the inverse of \( A \).
\[
A^{-1} = \begin{bmatrix}
-43 & 23 & -7 \\
-13 & 7 & -2 \\
6 & -3 & 1
\end{bmatrix}
\]
### Explanation:
This exercise involves finding the inverse of a 3x3 matrix \( A \). The matrix \( A \) provided is:
\[
\begin{bmatrix}
1 & -2 & 3 \\
1 & -1 & 5 \\
-3 & 9 & -2
\end{bmatrix}
\]
The inverse matrix \( A^{-1} \) is calculated as:
\[
\begin{bmatrix}
-43 & 23 & -7 \\
-13 & 7 & -2 \\
6 & -3 & 1
\end{bmatrix}
\]
Each element in \( A^{-1} \) corresponds to specific calculations involving determinants and cofactors from the original matrix \( A \). Finding the inverse of a matrix is useful in solving systems of linear equations, among other applications in linear algebra.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36a79c17-bf73-44c3-9503-d2f30f855206%2F629abbd2-8754-4237-84b5-a406b1867695%2F6gpi53_processed.png&w=3840&q=75)
Transcribed Image Text:## Part 1
Rework problem 26 from section 6.2 of your text, involving a matrix \( A \) and two matrix equations. Use the following matrix instead of the one given in your text.
\[
A = \begin{bmatrix}
1 & -2 & 3 \\
1 & -1 & 5 \\
-3 & 9 & -2
\end{bmatrix}
\]
1. First, find the inverse of \( A \).
\[
A^{-1} = \begin{bmatrix}
-43 & 23 & -7 \\
-13 & 7 & -2 \\
6 & -3 & 1
\end{bmatrix}
\]
### Explanation:
This exercise involves finding the inverse of a 3x3 matrix \( A \). The matrix \( A \) provided is:
\[
\begin{bmatrix}
1 & -2 & 3 \\
1 & -1 & 5 \\
-3 & 9 & -2
\end{bmatrix}
\]
The inverse matrix \( A^{-1} \) is calculated as:
\[
\begin{bmatrix}
-43 & 23 & -7 \\
-13 & 7 & -2 \\
6 & -3 & 1
\end{bmatrix}
\]
Each element in \( A^{-1} \) corresponds to specific calculations involving determinants and cofactors from the original matrix \( A \). Finding the inverse of a matrix is useful in solving systems of linear equations, among other applications in linear algebra.
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