Question

If $A$ is a square matrix of order $n$ such that $|\text{adj}(\text{adj} A)|=|A|^{25}$, then $n$ is equal to

• A. 6
• B. 4
• C. 5
• D. 3
• E. 7

Hint:

Logic Tip: Memorize the adjoint determinant hierarchy for a matrix $A$ of order $n$: 1. $|\text{adj} A| = |A|^{n-1}$ 2. $|\text{adj}(\text{adj} A)| = |A|^{(n-1)^2}$ 3. $\text{adj}(\text{adj} A) = |A|^{n-2} A$

Answer 1

The Correct Option is A

Concept:
The determinant of the adjoint of a matrix follows a specific power rule based on the order $n$ of the square matrix $A$. The fundamental property is $|\text{adj} A| = |A|^{n-1}$. By applying this property recursively to the adjoint of an adjoint, we get: $$|\text{adj}(\text{adj} A)| = |A|^{(n-1)^2}$$
Step 1: Set up the equation based on determinant properties.
We are given the relation: $$|\text{adj}(\text{adj} A)| = |A|^{25}$$ Substitute the property formula into the left side of the equation: $$|A|^{(n-1)^2} = |A|^{25}$$
Step 2: Equate the exponents.
Since the bases are the same (and assuming $|A| \neq 0, 1, -1$ to avoid trivial identities), the exponents must be equal: $$(n - 1)^2 = 25$$
Step 3: Solve for the order n.
Take the square root of both sides: $$n - 1 = \pm 5$$ Since $n$ represents the order (dimensions) of a matrix, it must be a positive integer ($n \ge 1$). Therefore, we discard the negative root: $$n - 1 = 5$$ $$n = 6$$