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Question

Two matrices A and B have in total 6 different elements (none repeated). How many different matrices A and B are possible such that product AB is defined.

A
4(6!)2
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B
3(6!)
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C
12(6!)
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D
8(6!)
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Solution

The correct option is D 8(6!)
Let a×b be the order of first matrix then order of second matrix should be b×c.
Also given (a×b)+(b×c)=6, a,b,c>0
b×(a+c)=6
6=3.2=6.1
If b=3, a+c=2
a=1,c=1
If a+c=3, b=2
No. of solutions =31C21=2C1=2
If b=6, a+c=1
But a,c>0 a+c1
If a+c=6, b=1
No. of solutions =61C21=5C1=5
Total =1+2+5=8.
Also the 6 elements can be rearranged in 6! ways.
Required =8(6!).
Hence, the answer is 8(6!).

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