If A is a square matrix of order 3 and |adj A| = 49, then |7A^-1|^2

The correct answer is Option (2) → \(7^4\)

For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 3:

    \(|A|^2 = 49\)

For a matrix of order 3, \(|kB| = k^3|B|\), and \(|A^{-1}| = \frac{1}{|A|}\):

    \( \displaystyle |7A^{-1}| = 7^3 \cdot \frac{1}{|A|} \)

Squaring:

    \( \displaystyle |7A^{-1}|^2 = \frac{7^6}{|A|^2} \)

    \( \displaystyle = \frac{7^6}{7^2} = 7^4 \)

Squaring removes the sign, so it does not matter whether |A| = 7 or −7.

Tip: A scalar comes out of a determinant as \(k^n\), not k: \(|7A^{-1}|\) is \(7^3|A^{-1}|\) for a 3 × 3 matrix.