If A and B are invertible matrices of same order, then which one of t…

The correct answer is Option (2) → \(\mathrm{adj}\,(AB) = (\mathrm{adj}\,A)(\mathrm{adj}\,B)\)

The adjoint of a product follows the same reversal law as the inverse. Using \(A^{-1} = \frac{\mathrm{adj}\,A}{|A|}\):

    \( \displaystyle \frac{\mathrm{adj}\,(AB)}{|AB|} = (AB)^{-1} = B^{-1}A^{-1} = \frac{(\mathrm{adj}\,B)(\mathrm{adj}\,A)}{|B|\,|A|} \)

Since \(|AB| = |A|\,|B|\):

    \(\mathrm{adj}\,(AB) = (\mathrm{adj}\,B)(\mathrm{adj}\,A)\)

Matrix multiplication is not commutative, so \((\mathrm{adj}\,A)(\mathrm{adj}\,B)\) is not equal to this in general. Option (2) is NOT true.

The other three are standard results:

    \((AB)^{-1} = B^{-1}A^{-1}\): the reversal law for inverses.
\(\mathrm{adj}\,A^T = (\mathrm{adj}\,A)^T\): the cofactors of \(A^T\) are those of A, transposed.
\(|A^{-1}| = (|A|)^{-1}\): from \(AA^{-1} = I\), \(|A|\,|A^{-1}| = 1\).