We start with the given determinant expression:
\[ |3 \, \text{adj}(2 \, \text{adj}(|A|A))| \]
Now simplifying step-by-step:
\[ |3 \, \text{adj}(2|A|^2 \, \text{adj}A)| \]
\[ |3 \cdot 2^2 \cdot |A|^4 \, \text{adj}(\text{adj}A)| \]
We know that \( |\text{adj}A| = |A|^{n-1} \). Hence, substituting this property, we get:
\[ 3^3 \cdot 2^6 \cdot |A|^{12} \cdot |A|^2 = 3^{-13} \cdot 2^{-10} \]
From this,
\[ |A|^{16} = 3^{-16} \cdot 2^{-16} \]
\[ |A| = \frac{1}{6} \]
Now consider:
\[ |3 \, \text{adj}(2A)| = |3 \cdot 2^2 \, \text{adj}A| = 3^3 \cdot 2^6 |A|^2 \]
\[ = 3 \cdot 2^4 \]
Therefore,
\[ M = 4, \quad n = 1 \]
Hence,
\[ |3m + 2n| = 14 \]
Step 1: Simplify \( |3\operatorname{adj}(2\operatorname{adj}(|A|A))| \)
Using determinant properties:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = |3| \times |\operatorname{adj}(2)| \times |\operatorname{adj}(|A|A)|. \]
1. Simplify \( |\operatorname{adj}(|A|A)| \):
Using the property \( |\operatorname{adj}(A)| = |A|^{n-1} \), where \( n = 4 \):
\[ |\operatorname{adj}(|A|A)| = |A|^4. \]
2. Simplify \( |\operatorname{adj}(2)| \):
Using \( \operatorname{adj}(kA) = k^{n-1}\operatorname{adj}(A) \):
\[ |\operatorname{adj}(2)| = 2^{n-1} = 2^3. \]
Thus:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = 3^3 \times 2^3 \times |A|^4 \times |A|^4 = 3^3 \times 2^3 \times |A|^8. \]
Given:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = 2^{-10} \times 3^{-13}. \]
Equating powers of \( 2 \) and \( 3 \):
\[ 2^3 \times |A|^8 = 2^{-10}, \quad 3^3 \times |A|^8 = 3^{-13}. \]
Solve for \( |A| \):
\[ |A| = 2^{-1} \times 3^{-1}. \]
Step 2: Simplify \( |3\operatorname{adj}(2A)| \)
\[ |3\operatorname{adj}(2A)| = |3| \times |\operatorname{adj}(2A)| = 3^2 \times |\operatorname{adj}(2)|^2 \times |A|^2. \] \[ |3\operatorname{adj}(2A)| = 3^2 \times 2^2 \times (2^{-1} \times 3^{-1})^2 = 2^4 \times 3^1. \]
Step 3: Compute \( |3m + 2n| \)
Given \( |A| = 2^m \cdot 3^n \), substitute \( m = -4 \), \( n = -1 \):
\[ 3m + 2n = 3(-4) + 2(-1) = -12 - 2 = -14. \] \[ |3m + 2n| = 14. \]
Let \[ R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix} \text{ be a non-zero } 3 \times 3 \text{ matrix, where} \]
\[ x = \sin \theta, \quad y = \sin \left( \theta + \frac{2\pi}{3} \right), \quad z = \sin \left( \theta + \frac{4\pi}{3} \right) \]
and \( \theta \neq 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \). For a square matrix \( M \), let \( \text{trace}(M) \) denote the sum of all the diagonal entries of \( M \). Then, among the statements:
Which of the following is true?
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)