[Jatin Kr Dey]

Hello Ramesh, your question is not clear to us. Can you please send any photo of it?

[Jatin Kr Dey]

Hello Amit, your question is not clear to us. Can you please send any photo of it?

[Jatin Kr Dey]

Hello Amit, your question is not clear to us. Can you please send any photo of it?

[Jatin Kr Dey]

Try to use $$ \displaystyle (a^2+b^2+c^2-ab-bc-ca)= \frac{1}{2} [(a-b)^2+(b-c)^2+(c-a)^2] $$

 

then proceed by the RMS - AM   ineqality .

[Jatin Kr Dey]

Given, $$ cofactors(b_{ij})= c_{ij}  \Rightarrow Adj(B)= C^T $$ \

$$ cofactors(a_{ij})= b_{ij} \Rightarrow Adj(A)= B^T $$  \

$$ det(A) = 2 $$

and the order of the matrices is 3 .

We have to apply the followings :

$$ |Adj(A) = (|A|)^{order(A) - 1} $$

$$ |A| = |A^T| $$

$$ |ABC| = |A| |B| |C| $$

$$ |cA| = c^{order(A)}|A| $$

So now $$ |Adj(A)| = 2^{3-1}=|B^T|=|B| $$ \

$$ |Adj(B)| = (2^{3-1})^{3-1}=|C^T|=|C| = 2^4$$

$$ |2A| = 2^{3}|A|  = 2^4 $$

Therefore , $$ \displaystyle |2AB^TC| = |2A| |B^T| |C| = 2^4 . 2^2 . 2^4 = \sum_{r=1}^{11} {10 \choose r-1} $$

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