Solve the matrix equation AX = B for X. 16 448 B= -54 A = 48 X= - 16 12
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
![**Matrix Problem Explanation**:
To solve the given matrix equation \( AX = B \) for \( X \), we need the inverse of matrix \( A \).
Given matrices:
\[
A = \begin{bmatrix} 1 & 6 \\ -5 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -16 \\ 12 \end{bmatrix}
\]
We are looking for matrix \( X \) such that:
\[
X = \begin{bmatrix} \square \\ \square \end{bmatrix}
\]
**Steps to Solve:**
1. **Check for the Inversibility of Matrix \( A \)**:
- Calculate the determinant of \( A \):
\[
\text{det}(A) = (1)(4) - (-5)(6) = 4 + 30 = 34
\]
- Since the determinant is non-zero, \( A \) is invertible.
2. **Find the Inverse of Matrix \( A \)**:
- The inverse \( A^{-1} \) can be found using the formula:
\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
- For matrix \( A \):
\[
A^{-1} = \frac{1}{34} \begin{bmatrix} 4 & -6 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{34} & \frac{-6}{34} \\ \frac{5}{34} & \frac{1}{34} \end{bmatrix}
\]
3. **Calculate Matrix \( X \)**:
- Multiply \( A^{-1} \) by \( B \):
\[
X = A^{-1}B = \begin{bmatrix} \frac{4}{34} & \frac{-6}{34} \\ \frac{5}{34} & \frac{1}{34} \end{bmatrix} \begin{bmatrix} -16 \\ 12 \end{bmatrix}
\]
- Simplify the matrix multiplication to find \( X \).
By following](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F917d2425-81cd-4e01-b117-3301ec41a936%2F0a34d991-7bc3-42d3-984a-a454e16f8e31%2F0s03dt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Matrix Problem Explanation**:
To solve the given matrix equation \( AX = B \) for \( X \), we need the inverse of matrix \( A \).
Given matrices:
\[
A = \begin{bmatrix} 1 & 6 \\ -5 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -16 \\ 12 \end{bmatrix}
\]
We are looking for matrix \( X \) such that:
\[
X = \begin{bmatrix} \square \\ \square \end{bmatrix}
\]
**Steps to Solve:**
1. **Check for the Inversibility of Matrix \( A \)**:
- Calculate the determinant of \( A \):
\[
\text{det}(A) = (1)(4) - (-5)(6) = 4 + 30 = 34
\]
- Since the determinant is non-zero, \( A \) is invertible.
2. **Find the Inverse of Matrix \( A \)**:
- The inverse \( A^{-1} \) can be found using the formula:
\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
- For matrix \( A \):
\[
A^{-1} = \frac{1}{34} \begin{bmatrix} 4 & -6 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{34} & \frac{-6}{34} \\ \frac{5}{34} & \frac{1}{34} \end{bmatrix}
\]
3. **Calculate Matrix \( X \)**:
- Multiply \( A^{-1} \) by \( B \):
\[
X = A^{-1}B = \begin{bmatrix} \frac{4}{34} & \frac{-6}{34} \\ \frac{5}{34} & \frac{1}{34} \end{bmatrix} \begin{bmatrix} -16 \\ 12 \end{bmatrix}
\]
- Simplify the matrix multiplication to find \( X \).
By following
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