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• #32406
Aniruddha Bardhan
Participant

see the attachement.

My approach: B=adj(A) and C=adj(B). so det (B)= A^2=2^2 and det (C)=B^2=2^4.

so det(2AB^TC)= 2^8 . I did not get any option with 2^8

May be I am wrong kindly check.

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#34638
Jatin Kr Dey
Participant

I think you should have to apply the following property :

$$\displaystyle det(cA) = c^{order(A)} det(A)$$

#34646
Jatin Kr Dey
Participant

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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