If $A, B$ are square matrices of the same order, then prove that $a d j ((A B) = (a d j B) (a d j A) .$

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Solution

Solution:-
Given:- A, B are square matrices of the same order
To prove:- $a d j (A B) = (a d j (B)) (a d j (A))$
Proof:-
R.H.S.-
$(a d j (B)) (a d j (A))$
L.H.S.-
As we know that,
$a d j (A B) = d e t (A B) {(A B)}^{- 1} ⟶ (i) {d e t (A B) = determinant of (A B)}$
We also know that,
$d e t (A B) = d e t (A) . d e t (B) & {(A B)}^{- 1} = B^{- 1} A^{- 1}$
Substituting these values in ${e q}^{n} (i)$ , we have
$∴ a d j (A B) = d e t (A) . d e t (B) B^{- 1} A^{- 1} ⟶ (i i)$
As we also know that if X is a square matrix then,
$X^{- 1} = \frac{a d j (A)}{d e t (A)}$
$∴ a d j (A B) = d e t (A) . d e t (B) . \frac{a d j (B)}{d e t (B)} . \frac{a d j (A)}{d e t (A)} {∵ determinant of any sq. matrix is a numerical value}$
$\Rightarrow a d j (A B) = a d j (B) . a d j (A)$
Hence proved

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