If A = then det(A) Thus, -7 -91 3 A-1 -102 and det(A1) O is Nonzero 3 -34 34 [图] 3 7 -34 -102 x
If A = then det(A) Thus, -7 -91 3 A-1 -102 and det(A1) O is Nonzero 3 -34 34 [图] 3 7 -34 -102 x
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
![**Problem Statement:**
Find the area of a parallelogram with corner points at (1,2), (3,5), (8,5), and (6,2).
**Solution:**
To find the area of a parallelogram given its vertices, you can use the vector method or the shoelace formula. Here, the corner points are (1,2), (3,5), (8,5), and (6,2). Check coordinates and apply the appropriate method to compute the area.
1. **Vector Method (Cross Product):**
- Consider vectors formed by these points. For example, vectors \( \mathbf{v}_1 = (3-1, 5-2) = (2, 3) \) and \( \mathbf{v}_2 = (8-1, 5-2) = (7, 3) \).
- Compute the cross product magnitude: \( |\mathbf{v}_1 \times \mathbf{v}_2| = |2 \times 3 - 3 \times 7| = |6 - 21| = 15 \).
2. **Shoelace Formula:**
- Use the formula \( \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \).
- Apply the coordinates:
\[
\frac{1}{2} \left| 1 \cdot 5 + 3 \cdot 5 + 8 \cdot 2 + 6 \cdot 2 - (2 \cdot 3 + 5 \cdot 8 + 5 \cdot 6 + 2 \cdot 1) \right|
\]
- Compute the expression:
\[
\frac{1}{2} \left| 5 + 15 + 16 + 12 - (6 + 40 + 30 + 2) \right| = \frac{1}{2} \left| 48 - 78 \right| = \frac{1}{2} \times 30 = 15
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18574973-f25e-4ab8-b7d6-6007b5b87fc4%2F69f21f38-97e4-4cbc-80ee-debf90fe495c%2Fy4gqyl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area of a parallelogram with corner points at (1,2), (3,5), (8,5), and (6,2).
**Solution:**
To find the area of a parallelogram given its vertices, you can use the vector method or the shoelace formula. Here, the corner points are (1,2), (3,5), (8,5), and (6,2). Check coordinates and apply the appropriate method to compute the area.
1. **Vector Method (Cross Product):**
- Consider vectors formed by these points. For example, vectors \( \mathbf{v}_1 = (3-1, 5-2) = (2, 3) \) and \( \mathbf{v}_2 = (8-1, 5-2) = (7, 3) \).
- Compute the cross product magnitude: \( |\mathbf{v}_1 \times \mathbf{v}_2| = |2 \times 3 - 3 \times 7| = |6 - 21| = 15 \).
2. **Shoelace Formula:**
- Use the formula \( \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \).
- Apply the coordinates:
\[
\frac{1}{2} \left| 1 \cdot 5 + 3 \cdot 5 + 8 \cdot 2 + 6 \cdot 2 - (2 \cdot 3 + 5 \cdot 8 + 5 \cdot 6 + 2 \cdot 1) \right|
\]
- Compute the expression:
\[
\frac{1}{2} \left| 5 + 15 + 16 + 12 - (6 + 40 + 30 + 2) \right| = \frac{1}{2} \left| 48 - 78 \right| = \frac{1}{2} \times 30 = 15
\]
![If \( A = \begin{bmatrix} -7 & -9 \\ 9 & 3 \end{bmatrix} \),
then \(\text{det}(A) = -102\) is Nonzero.
Thus,
\[ A^{-1} = \begin{bmatrix} \frac{1}{34} & \frac{3}{34} \\ \frac{3}{34} & \frac{7}{-102} \end{bmatrix} \]
and \(\text{det}(A^{-1}) = \_\_\_\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18574973-f25e-4ab8-b7d6-6007b5b87fc4%2F69f21f38-97e4-4cbc-80ee-debf90fe495c%2F0ps00m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:If \( A = \begin{bmatrix} -7 & -9 \\ 9 & 3 \end{bmatrix} \),
then \(\text{det}(A) = -102\) is Nonzero.
Thus,
\[ A^{-1} = \begin{bmatrix} \frac{1}{34} & \frac{3}{34} \\ \frac{3}{34} & \frac{7}{-102} \end{bmatrix} \]
and \(\text{det}(A^{-1}) = \_\_\_\).
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