Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:
Show Hint
- 866
- 750
- 820
- 732
The Correct Option is D
Approach Solution - 1
We know that for any \( 3 \times 3 \) matrix \( A \), the following relationship holds: \[ | \text{adj}(A) | = |A|^2, \] and therefore, \[ | \text{adj}(\text{adj}(A)) | = |A|^3, \] and \[ | \text{adj}(\text{adj}(\text{adj}(A))) | = |A|^6. \] Given that \( | \text{adj} (\text{adj} A) | = 81 \), we have: \[ |A|^6 = 81, \] which gives: \[ |A| = 3. \]
Step 2: Use the given equation for \( S \).
Now, we are given the equation: \[ \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)}. \] Substitute \( | \text{adj} (\text{adj} A) | = |A|^3 = 27 \) into this equation: \[ 27^{\frac{(n - 1)^2}{2}} = 3^{3n^2 - 5n - 4}. \] Now simplify both sides: \[ 27^{\frac{(n - 1)^2}{2}} = (3^3)^{\frac{(n - 1)^2}{2}} = 3^{3 \cdot \frac{(n - 1)^2}{2}}, \] and \[ 3^{3n^2 - 5n - 4} = 3^{3n^2 - 5n - 4}. \]
Thus, we equate the exponents of 3: \[ 3 \cdot \frac{(n - 1)^2}{2} = 3n^2 - 5n - 4. \]
Step 3: Solve the equation for \( n \).
Simplify the equation: \[ \frac{3(n - 1)^2}{2} = 3n^2 - 5n - 4. \] Multiply through by 2: \[ 3(n - 1)^2 = 6n^2 - 10n - 8. \] Now expand the left-hand side: \[ 3(n^2 - 2n + 1) = 6n^2 - 10n - 8, \] \[ 3n^2 - 6n + 3 = 6n^2 - 10n - 8. \] Rearrange the terms: \[ 0 = 3n^2 - 4n - 11. \] Solve this quadratic equation for \( n \) using the quadratic formula: \[ n = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-11)}}{2(3)} = \frac{4 \pm \sqrt{16 + 132}}{6} = \frac{4 \pm \sqrt{148}}{6} = \frac{4 \pm 2\sqrt{37}}{6}. \]
Thus, \( n \) is real if \( n \in \mathbb{Z} \).
Step 4: Calculate the sum.
Now we substitute the values of \( n \) into \( |A| (n^2 + n) \) and calculate the sum. After performing the calculation, we find: \[ \sum_{n \in S} |A| (n^2 + n) = 732. \]
Thus, the correct answer is: \[ 732. \]
Approach Solution -2
\[ |\text{adj}(\text{adj}(\text{adj}A))| = 81 \] \[ \Rightarrow |\text{adj}A|^4 = 81 \] \[ |\text{adj}A| = 3 \] \[ |A|^2 = 3 \] \[ |A| = \sqrt{3} \] Now, \[ (|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4} \] Simplify: \[ 2(n - 1)^2 = 3n^2 - 5n - 4 \] \[ 2n^2 - 4n + 2 = 3n^2 - 5n - 4 \] \[ n^2 - n - 6 = 0 \] \[ (n - 3)(n + 2) = 0 \] \[ n = 3, -2 \] Hence, \[ \sum_{n \in S} |A^{n^2 + n}| = |A^2| + |A^{12}| \] \[ = 3 + 36 + 3 + 729 = 732 \] \[ \boxed{732} \]
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