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Tagged: Determinant

- This topic has 2 replies, 2 voices, and was last updated 7 months, 1 week ago by Jatin Kr Dey.

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- August 5, 2019 at 10:19 am #32406Aniruddha BardhanParticipant
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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You must be logged in to view attached files.August 28, 2019 at 11:46 am #34638Jatin Kr DeyParticipantI think you should have to apply the following property :

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

- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.

August 28, 2019 at 1:18 pm #34646Jatin Kr DeyParticipantGiven, $$ cofactors(b_{ij})= c_{ij} \Rightarrow Adj(B)= C^T $$ \

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

$$ det(A) = 2 $$

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

- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.
- This reply was modified 7 months, 1 week ago by Jatin Kr Dey.

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