Question

if A=\(\frac{1}{5! 6! 7!}\begin{bmatrix} 5! & 6! & 7!\\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}\), then |adj(adj(2A))| is equal t

• A. \(2^{20}\)
• B. \(2^{16}\)
• C. \(2^{12}\)
• D. \(2^{8}\)

Answer 1

The Correct Option is B

Given the matrix \( A \), the determinant is calculated as:

\[ |A| = \frac{1}{5!6!7!} \begin{vmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{vmatrix} \]

The row transformations are applied as follows:

First, perform the operation \( R_3 \to R_3 - R_2 \) and then \( R_2 \to R_2 - R_1 \), resulting in:

\[ \begin{vmatrix} 1 & 8 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{vmatrix} \]

The determinant of the modified matrix is:

\[ |A| = 2 \]

Next, the adjugate of the matrix \( 2A \) is calculated. The formula for the adjugate of a matrix \( A \) is given by:

\[ \text{adj}(2A) = 2A^{(n-1)^2} \]

For a \( 3 \times 3 \) matrix, \( n = 3 \), so the formula becomes:

\[ \text{adj}(2A) = 2A^4 \]

The determinant of the adjugate is calculated as:

\[ | \text{adj}(2A) | = 2 A^4 \]

Substituting the previously calculated determinant \( |A| = 2 \), we get:

\[ = (2^3 |A|)^4 = 2^{12} |A|^4 = 2^{12} \times 2^4 = 2^{16} \]