Question

If \[ A(\operatorname{adj}A)= \begin{bmatrix} 2026 & 0 & 0\\ 0 & 2026 & 0\\ 0 & 0 & 2026 \end{bmatrix}, \] then the value of \[ \left|\operatorname{adj}A\right| \] is equal to \[ \_\_\_\_. \]

• A. \( 2026 \)
• B. \( (2026)^{-1} \)
• C. \( (2026)^{-2} \)
• D. \( (2026)^2 \)

Hint:

For any scalar matrix \( A(\text{adj } A) = kI_n \), the determinant value is \( |A| = k \), and the determinant of its adjoint is always \( k^{n-1} \). Here, \( n=3 \), so the answer is immediately \( k^2 \).

Answer 1

The Correct Option is D

Concept: This question is based on the following properties of the adjoint of a square matrix:
• \[ A(\operatorname{adj}A)=|A|I_n \]
• \[ \left|\operatorname{adj}A\right|=|A|^{\,n-1}, \] where \(n\) is the order of the square matrix.

Step 1: Express the given matrix in terms of the identity matrix.
The given matrix equation is \[ A(\operatorname{adj}A)= \begin{bmatrix} 2026 & 0 & 0\\ 0 & 2026 & 0\\ 0 & 0 & 2026 \end{bmatrix}. \] Factoring out \(2026\), \[ A(\operatorname{adj}A) = 2026 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} = 2026I_3, \] where \(I_3\) is the identity matrix of order \(3\).

Step 2: Find the determinant of \(A\).
Using the property \[ A(\operatorname{adj}A)=|A|I_3, \] and comparing it with \[ A(\operatorname{adj}A)=2026I_3, \] we obtain \[ |A|=2026. \]

Step 3: Find \(\left|\operatorname{adj}A\right|\).
Since \(A\) is a \(3\times3\) matrix, \[ \left|\operatorname{adj}A\right| = |A|^{3-1} = |A|^2. \] Substituting \(|A|=2026\), \[ \left|\operatorname{adj}A\right| = (2026)^2. \]

Hence,

\[ \boxed{\left|\operatorname{adj}A\right|=(2026)^2.} \]