Let \( A,B \) be square matrices of order \( 3\times 3 \) . Then \( |A.\,\mathrm{adj}(A).B| \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( |A|^3.|B| \)
The determinant of a product is the product of the determinants:
\( \displaystyle |A\,\mathrm{adj}(A)\,B|=|A|\,|\mathrm{adj}(A)|\,|B| \)
For a matrix of order n, \( |\mathrm{adj}(A)|=|A|^{n-1} \). Here n = 3, so \( |\mathrm{adj}(A)|=|A|^2 \):
\( \displaystyle |A\,\mathrm{adj}(A)\,B|=|A|\cdot|A|^2\cdot|B| \)
\( \displaystyle =|A|^3|B| \)
Shortcut: Since \( A\,\mathrm{adj}(A)=|A|\,I \) and a scalar k taken out of a 3 × 3 determinant gives \( k^3 \),
\( \displaystyle |A\,\mathrm{adj}(A)\,B|=\big||A|\,I\big|\,|B| \)
\( \displaystyle =|A|^3|B| \)
Why the other options are incorrect
Option (1) \( |A|^2.|B| \) — Incorrect. This counts only \( |\mathrm{adj}(A)|=|A|^2 \) and misses the factor |A| from A itself.
Option (3) \( |A|^2.|B|^2 \) — Incorrect. B appears only once in the product, so |B| is not squared; the power of |A| is 3, not 2.
Option (4) \( |A|.|B|^3 \) — Incorrect. |B| has power 1 and |A| has power 3, not the other way round.