Question

Let A be a $n \times n$ matrix such that $| A |=2$ If the determinant of the matrix$\operatorname{Adj}\left(2 \cdot \operatorname{Adj}\left(2 A ^{-1}\right)\right) \cdot$ is $2^{84}$, then $n$ is equal to ____

Answer 1

Correct Answer: 27

Let A be a \( n \times n \) matrix such that \( |A| = 2 \). If the determinant of the matrix \( \operatorname{Adj}\left(2 \cdot \operatorname{Adj}\left(2 A^{-1}\right)\right) \) is \( 2^{84} \), then \( n \) is equal to ___.

Step 1: Understanding properties of the adjugate matrix
The adjugate of a matrix \( A \), denoted as \( \operatorname{Adj}(A) \), satisfies the property: \[ A \cdot \operatorname{Adj}(A) = |A| \cdot I \] where \( I \) is the identity matrix and \( |A| \) is the determinant of \( A \).

Step 2: Apply to \( A^{-1} \)
The adjugate of \( A^{-1} \) is given by: \[ \operatorname{Adj}(A^{-1}) = |A^{-1}| \cdot A = \frac{1}{|A|} \cdot A \] Since \( |A| = 2 \), we have \( \operatorname{Adj}(A^{-1}) = \frac{1}{2} A \).

Step 3: Analyze \( 2 \cdot \operatorname{Adj}(2 A^{-1}) \)
Now consider \( 2 \cdot \operatorname{Adj}(2 A^{-1}) \). We use the property of adjugate for scalar multiplication: \[ \operatorname{Adj}(c A) = c^{n-1} \cdot \operatorname{Adj}(A) \] Therefore: \[ \operatorname{Adj}(2 A^{-1}) = 2^{n-1} \cdot \operatorname{Adj}(A^{-1}) = 2^{n-1} \cdot \frac{1}{2} A = 2^{n-2} A \] Thus: \[ 2 \cdot \operatorname{Adj}(2 A^{-1}) = 2 \cdot 2^{n-2} A = 2^{n-1} A \]

Step 4: Determinant of \( \operatorname{Adj}(2 \cdot \operatorname{Adj}(2 A^{-1})) \)
Now we find the determinant of the matrix \( \operatorname{Adj}(2 \cdot \operatorname{Adj}(2 A^{-1})) \): \[ \operatorname{Adj}(2 \cdot \operatorname{Adj}(2 A^{-1})) = \operatorname{Adj}(2^{n-1} A) \] Using the adjugate property for scalar multiplication again: \[ \operatorname{Adj}(2^{n-1} A) = (2^{n-1})^{n-1} \cdot \operatorname{Adj}(A) = 2^{(n-1)(n-1)} \cdot \operatorname{Adj}(A) \] Therefore: \[ \operatorname{Adj}(2^{n-1} A) = 2^{(n-1)(n-1)} \cdot A^{n-1} \] Taking the determinant of both sides: \[ \text{det}(\operatorname{Adj}(2^{n-1} A)) = 2^{(n-1)(n-1)} \cdot |A|^{n-1} \] Since \( |A| = 2 \), we have: \[ \text{det}(\operatorname{Adj}(2^{n-1} A)) = 2^{(n-1)(n-1)} \cdot 2^{2(n-1)} = 2^{(n-1)(n-1) + 2(n-1)} \]

Step 5: Equate to \( 2^{84} \)
We are given that the determinant is \( 2^{84} \). Thus: \[ 2^{(n-1)(n-1) + 2(n-1)} = 2^{84} \] Therefore: \[ (n-1)(n-1) + 2(n-1) = 84 \] Simplifying the equation: \[ (n-1)^2 + 2(n-1) = 84 \] Let \( y = n-1 \), so the equation becomes: \[ y^2 + 2y = 84 \] Solving the quadratic equation: \[ y^2 + 2y - 84 = 0 \] Using the quadratic formula: \[ y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-84)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 336}}{2} = \frac{-2 \pm \sqrt{340}}{2} \] \[ y = \frac{-2 \pm 2\sqrt{85}}{2} = -1 \pm \sqrt{85} \] Therefore, \( y = 9 \), so \( n-1 = 9 \) and \( n = 10 \).

Final Answer:
\( n = 10 \)