A) \[CB+A\]
B) \[BAC\]
C) \[C(A+B)\]
D) \[C(A+B)\]
Correct Answer: D
Solution :
Given that\[O(A)=2\times 3,\text{ }0(B)=3\times 2\]and \[O(C)=3\times 3\] \[\Rightarrow \] \[O(A)=3\times 2,O(B)=2\times 3\] (a) \[CB+A\] Now order of CB = (order of C) (order of B) = (order of C is\[3\times 3\]) (order of B is\[3\times 2\]) = order of CB is\[3\times 2\] Since, \[O(A)=3\times 2\] \[\therefore \]Matrix\[CB+A\]can be determined. (b) \[O(BA)=3\times 3\] and \[O(C)=3\times 3\] \[\therefore \]Matrix BAC can be determined. (c) \[C(A+B)\] \[O(A+B)=2\times 3\] \[\Rightarrow \] \[O(A+B)=3\times 2\] and \[O(C)=3\times 3\] \[\therefore \]Matrix\[C(A+B)\]can be determined. (d) \[C(A+B)\] \[O(A+B)=2\times 3\] and \[O(C)=3\times 3\] \[\therefore \]Matrix\[C(A+B)\]cannot be determined \[\therefore \]Option is correct.You need to login to perform this action.
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