Question

If \( A = \begin{bmatrix} 1 & 2 3 & 4 \end{bmatrix} \), then \( \text{adj}(\text{adj}(\text{adj } A)) \)

• A. \( A \)
• B. \( A^{-1} \)
• C. \( |A|A^{-1} \)
• D. \( \frac{A^{-1}}{|A|} \)

Hint:

For any \( 2 \times 2 \) matrix (\(n = 2\)), the double adjoint operation always returns the original matrix itself (\(\text{adj}(\text{adj }A) = A\)). This means you can instantly cancel out pairs of adjacent "adj" operators, saving massive derivation time on any exam!

Answer 1

The Correct Option is C

Concept: For a non-singular square matrix \( A \) of order \( n \times n \), there are useful formulas that describe repetitive operations of the adjoint matrix:
• \( \text{adj}(\text{adj } A) = |A|^{n-2} A \)
• \( \text{adj}(k \cdot A) = k^{n-1} \cdot \text{adj } A \) where \( k \) is a scalar coefficient.
• Matrix inverse identity: \( A^{-1} = \frac{1}{|A|} \text{adj } A \implies \text{adj } A = |A|A^{-1} \)

Step 1: Determining the order \(n\) and the determinant \(|A|\) of the given matrix.
The matrix \( A \) is: \[ A = \begin{bmatrix} 1 & 2 3 & 4 \end{bmatrix} \] The order of this square matrix is \( 2 \times 2 \), so \( n = 2 \). Now, let us find its determinant value \(|A|\): \[ |A| = (1)(4) - (2)(3) = 4 - 6 = -2 \]

Step 2: Simplifying the target expression using general adjoint identities.
Let us substitute the nested property formula \( \text{adj}(\text{adj } A) = |A|^{n-2} A \). Since \( n = 2 \) here: \[ \text{adj}(\text{adj } A) = |A|^{2-2} A = |A|^0 A = 1 \cdot A = A \quad \cdots (1) \] Now, let's look at the full three-fold composition expression from the question: \[ \text{adj}\Big( \text{adj}(\text{adj } A) \Big) \] Using our result from equation (1) to replace the inner double adjoint group with \(A\): \[ \text{adj}\Big( \text{adj}(\text{adj } A) \Big) = \text{adj}(A) \quad \cdots (2) \]

Step 3: Matching the simplified expression with the provided option structures.
From equation (2), the expression simplifies down to simply \(\text{adj } A\). Let us express \(\text{adj } A\) in terms of the matrix inverse using the definition: \[ A^{-1} = \frac{\text{adj } A}{|A|} \quad \Rightarrow \quad \text{adj } A = |A|A^{-1} \] This perfectly matches the algebraic form given in option (C).