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3 and 4 .Determinants and Matrices
normal
Which of the following statement is correct about two square matrices $A$ and $B$ of same order
A
trace $(Adj(AB))$ = $Adj(trace (AB).I)$
B
trace $((A + B) (A -B)) \neq trace (A^2) -trace(B^2)$
C
trace $(Adj(|A| |B| AB))$ -trace $(Adj(|AB| BA))$ = $0$
D
If $A$ is symmetric matrix and $B$ is skew symmetric matrix then trace $(AB' -BA')\neq 0$
Solution
$\operatorname{trace}(\operatorname{adj\,}(|\mathrm{A} \| \mathrm{B}| \mathrm{AB}))$
$=\operatorname{trace}\left(|\mathrm{A}|^{\mathrm{n}-1}|\mathrm{B}|^{\mathrm{n}-1}(\mathrm{Adj\,} \mathrm{B})(\mathrm{Adj\,} \mathrm{A})\right)$
$\operatorname{trace}(\operatorname{Adj\,}(|\mathrm{AB}| \mathrm{BA}))$
$=\operatorname{trace}\left(|A|^{n-1}|B|^{n-1}(A d j \,A) \cdot(A d j \,B)\right)$
Standard 12
Mathematics