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let A be n*n matrix with characteristic polynomial of the form f(x)=(x-c1)^d1.....(x-ck)^dk, that is having root cj with multiplicity dj and j=1,2...k then, c1d1+c2d2+.....ckdk will be equal to ? a) determinant of A b)rank of A c)trace of A d) none i want the explanation also.
let A be n*n matrix with characteristic polynomial of the form f(x)=(x-c1)^d1.....(x-ck)^dk, that is having root cj with multiplicity dj and j=1,2...k then, c1d1+c2d2+.....ckdk will be equal to ?a) determinant of A b)rank of Ac)trace of Ad) nonei want the explanation also.

```
6 years ago

Arun Kumar
IIT Delhi
256 Points
```							If we see the$c_{1}d_{1}+c_{2}d_{2}+.....c_{k}d_{k}$is sum of the roots of f(x)In the expansion of the $det(xI-A)$about the first column there is a term $(x-a_{11})(x-a_{22})(x-a_{33})...(x-a_{nn})$Which contributes in having coefficient of $x^{n-1}$No other term in the expansion has this because other terms has coefficients untill$x^{n-2}$if take sum of coefficients that's trace of the matrix.so BArun KumarIIT DelhiAskiitians Faculty
```
6 years ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions