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Two matrices A and B have in total 6 different elements (none

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 $A B$ is defined.

4 (6!)^{2}

3 (6!)

12 (6!)

8 (6!)

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Benjamin Discussed

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

A B

is defined.

10 mins ago

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10 mins ago

Text solutionVerified

Let

a \times b

be the order of first matrix then order of second matrix should be

b \times c .

Also given

(a \times b) + (b \times c) = 6,

a, b, c > 0

\Rightarrow b \times (a + c) = 6

\Rightarrow 6 = 3.2 = 6.1

\Rightarrow

b = 3,

a + c = 2

a = 1, c = 1

\Rightarrow

a + c = 3,

b = 2

No. of solutions

=^{3 - 1} C_{2 - 1} =^{2} C_{1} = 2

\Rightarrow

b = 6,

a + c = 1

But

a, c > 0

\to a + c \neq = 1

\Rightarrow

a + c = 6,

b = 1

No. of solutions

=^{6 - 1} C_{2 - 1} =^{5} C_{1} = 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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Question Text	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 $A B$ is defined.
Answer Type	Text solution:1
Upvotes	150

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.

Text solutionVerified

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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 $A B$ is defined.