Given that, ‌‌‌O(A)=2×3,O(B)=3×2, ‌O(C)=3×3 ‌⇒O(A′)=3×2,O(B′)=2×3 ‌‌ (a) ‌C(A+B′) ‌‌ Now, ‌O(A+B′)=2×3 ‌‌ and ‌O(C)=3×3 ‌‌ So, matrix ‌C(A+B′)‌ cannot be determined. ‌ ‌‌ (b) ‌C(A+B′)′ ‌O(A+B′)=2×3 ‌⇒O(A+B′)′=3×2‌ and ‌O(C)=3×3 Therefore, matrix C(A+B′)′ can be determined. (c) O(BA)=3×3 and O(C)=3×3 Therefore, matrix BAC can be determined. (d) CB+A′ Now, order of CB= (order of C ) (order of B) =( order of C is 3×3 ) (order of B is 3×2 ) = order of CB is 3×2 Since, O(A′)=3×2 Therefore, matrix CB+A′ can be determined.