Question

If $A$ and $B$ are non-singular square matrices of the same order, then $adj(AB)$ is equal to [AMU 1999]
A. $(adjA)(adjB)$
B. $(adjB)(adjA)$
C. $(adj\,{{B}^{-1}})(adj\,{{A}^{-1}})$
D. $(adj\,{{A}^{-1}})(adj\,{{B}^{-1}})$

Answer 1

Hint:
Check whether the options satisfy the properties of a matrix's adjoint. If you multiply a matrix $A$ by its adjoint, you get a diagonal matrix whose diagonal entries are the determinant $det(A)$. $I$ is an identity matrix, and $ Aadj(A)=adj(A)A=det(A) I$.

Formula Used:
According to the properties of the Inverse and Adjoint Matrix (say A):
$adj(A) = adj(A) A = |A|I$

Complete step-by-step answer:
$A$ and $B$ are non-singular, so $AB$ is non-singular.
Hence, $AB |A B|=|A||B|$
$(A B)(\operatorname{adj} . A B)=|A B|$
$(A B)(adj B . adj A)=\left|A B / I_n\right| $
$=A(B, adjB) adj A $
$=\left(A\left|B / I_n\right| adj A\right)$
$=|B|(A .adj A) $
$=|B||A|$
Therefore,
$\Rightarrow(A B)(a d j(A B))=(B . adj B)(A \cdot adjA) $
$\Rightarrow(A B)(a d jA B)=(AB)(adjB . adj A) $
$\Rightarrow adj(AB)=adj(B) adj(A)$

Note:
Keep in mind that multiplying the individual adjoint of each matrix in reverse order produces the same result as multiplying the two matrices in the adjoint.

Additional Information:
Properties of Invertible Matrices
Assume that $A$ and $B$ are $n \times n$ invertible matrices. Then:
$AB$ is invertible $(AB)^{-1}=B^{-1}A^{-1}. $
$A^{-1}$ is invertible $ (A^{-1})^{-1}=A.$
$kA$ is invertible for any nonzero scalar k $(kA)^{-1}=\dfrac{1}{k}A^{-1}$.
If A is a diagonal matrix, with diagonal entries $d_{1},\: d_{2},\cdots, d_{n}$ where none of the diagonal entries is 0, then $A^{−1}$ exists and is a diagonal matrix. Furthermore, the diagonal entries of $A^{−}1$ are $1/d_{1},\: 1/d_{2},\cdots , 1/d_{n}.$
If product $AB$ is not invertible, then $A $ or $B$ is not invertible.
If $A$ or $B$ is not invertible, then $AB$ are not invertible.