Given \(|A| = 3\), we start with:
\[\left| \text{adj} \left( -4 \, \text{adj} - 3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|\]
Step 1: Simplify the innermost expression:
\[= \left| -4 \, \text{adj} \left( -3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|^2\]
Step 2: Expand the outer term:
\[= 4^5 \, \left| \text{adj} \left( -3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|^2\]
Step 3: Replace the outermost adj with its expression:
\[= 2^{12} \cdot 3^{12} \cdot \left| 3 \, \text{adj} \left( 2A^{-1} \right) \right|^8\]
Step 4: Simplify the term inside the absolute value:
\[= 2^{12} \cdot 3^{12} \cdot 3^8 \cdot \left| \text{adj} \left( 2A^{-1} \right) \right|^8\]
Step 5: Use the property of adjugates:
\[= 2^{12} \cdot 3^{20} \cdot \left| 2A^{-1} \right|^{16}\]
Step 6: Replace \(\left| 2A^{-1} \right|^{16}\) with its determinant form:
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{|2A|^{16}}\]
Step 7: Substitute \(|2A|^{16} = 2^{16} \cdot |A|^{16}\):
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{2^{48} \cdot |A|^{16}}\]
Step 8: Replace \(|A| = 3\):
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{2^{48} \cdot 3^{16}}\]
Step 9: Simplify powers of 2 and 3:
\[= \frac{2^{12}}{2^{48}} \cdot \frac{3^{20}}{3^{16}} = \frac{1}{2^{36}} \cdot 3^4\]
Step 10: Further simplify:
\[= 2^{-36} \cdot 3^4\]
Step 11: Combine the terms:
\[m = -36, \quad n = 20\]
Step 12: Final result:
\[m + 2n = 4\]
Three students, Neha, Rani, and Sam go to a market to purchase stationery items. Neha buys 4 pens, 3 notepads, and 2 erasers and pays ₹ 60. Rani buys 2 pens, 4 notepads, and 6 erasers for ₹ 90. Sam pays ₹ 70 for 6 pens, 2 notepads, and 3 erasers.
Based upon the above information, answer the following questions:
(i) Form the equations required to solve the problem of finding the price of each item, and express it in the matrix form \( A \mathbf{X} = B \).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: