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mikau

Member

Registered: 2005-08-22

Posts: 1,504

matrice inverse, determinant

this is weird.

given the matrice A =

|(b + 8c) , (2c - 2b), (4b - 4c)|
|(4c - 4a) , (c + 8b), (2a - 2c)|
|(2b - 2a), (4a - 4b), (a + 8b)|

and P =
|0 1 2|
|2 0 1|
|1 2 0|

find P^(-1)AP and then determine determinant of A.

by the rule det(A*B) = det(A)*det(B) we have

det(P^-1AP) = det(P^-1)det(A)det(P) = det(A)det(P^-1)det(P) = det(A*P^-1*P) = det(A). 

based on the problem, i'm assuming the determinant of
P^(-1)AP is going to be easy to compute, and according to the solution of the problem, it ends up being a triangular matrix, so you can just take the product of the diagonals. Okay, thats fine, but the first step is to compute P^-1, the books answer states:

P^-1 =
|-2 4 1|
|1 -2 4|
|4 1 -2|

oddly, this is the exact answer i got except each term was over 9. (in otherwords, multiply the above matrice by a scaler of 1/9 and thats the inverse i obtained) moreover, according to the solution, the product P^-1AP results in
|9a 0 0|
|0 9b 0|
|0 0 9c|

wereby the determinant ends up being 9^3abc. if they had used the inverse i found, they would have ended up with neat and tidy determinant of abc. What this looks like is an ALMOST brilliantly contrived problem.

Am i missing something?

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A logarithm is just a misspelled algorithm.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: matrice inverse, determinant

Nope, the book's wrong. They got as far as transposing the matrix of cofactors and then forgot to divide the whole thing by detA.

An easy way to prove to yourself that the book is wrong would be to work out A[sup]-1[/sup]A.
Using the book's answer, this works out to be:

(9 0 0)
(0 9 0)
(0 0 9).

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Why did the vector cross the road?
It wanted to be normal.