26 апр. 2017 г. · Divide the third row by 2, swap columns 1 and 2 and multiply the third column by 2, you get det(M−tI)=−det[31−t031−t121202−t40261−t]=−det[ABCD],.

23 июл. 2020 г. · Determinant is linear in each column, so a 1−λ factor comes out of third column, then reduces to a 3×3 determinant.

6 февр. 2015 г. · I have to find the characteristic polynomial to find Jordan normal form. I chose to solve this via column expansion on the first determinant, and then row ...

4 апр. 2022 г. · I am trying to find the one for 4x4 matrix. It looks like this,. λ4−Tλ3+T2−Tr(A2)2λ2+(something)λ+D=0. But I'm not sure what the coefficient ...

20 янв. 2017 г. · The best, and very short way: First step. Subtract to the rows 2, 3 and 4, the first one. Second step. Add to the first column the sum of the columns 2, 3 and ...

16 нояб. 2015 г. · Thus, the characteristic polynomial is p(λ)=(λ−1700)2(λ−1275)(λ−850).

30 июл. 2016 г. · A matrix can be diagonalizable if its characteristic polynomial and minimal polynomial are the same. Take, for instance, the 3×3 diagonal matrix with diagonal ...

22 янв. 2021 г. · I've found that because A is singular, 0 is an eigenvalue to A. And because |A−2I|=0, |2I−A|=0 and 2 is also an eigenvalue.

2 июн. 2021 г. · I don't see any shortcut. The characteristic polynomial of that matrix isλ4−24λ3+216λ2−864λ+1296,. which turns out to be equal to (λ−6)4.

6 апр. 2020 г. · For an n×n matrix A the best formula to use is det(λI−A))=λn+n∑i=1βiλn−i. where βi=(−1)i sum of principal minors of order i.