A and B are square matrices of order 3. C is a unit matrix of order 3. D=(AB−C)−1 ⇒D and AB−C are inverse of each other
‌D×(AB−C)=I⇒|D|×|AB−C|=1
‌⇒|AB−C|=‌
1
|D|
or |D|=‌
1
|AB−C|
Now, |BC−C|×|BDA|=‌
1
|D|
×|BDA| ‌=‌
|B|×|D|×|A|
|D|
‌=|B|×|A|=|BA| If D and AB−C are inverse of each other, then (AB−C)×D=D×(AB−C)=I ‌ABD−D‌=DAB−D=I ⇒‌ABD‌=DAB=D+I ∴‌ABD‌=DAB Hence, both statements are true.