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The product of the non-zero eigenvalues of the matrix is ____

$\begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix}$

8 Answers

0 votes
0 votes
Rank is less then n so det=o

Trace=5

So non-zero eigen values r 2 and 3
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0 votes
Convert the matrix into lower triangular matrix by row transformations , then we know that principal diagonal elements of this matrix will be Eigen values,  out of these 5 EIgen values three will be 0 and Two eigenvalues  will be 1 .so product of non zero eigenvalues values =1X1 =1 ans.
0 votes
0 votes
Answer: (C)

Explanation:

let B be the given matrix.

=> B^2 = 2 0 0 0 2

                 0 3 3 3 0

                 0 3 3 3 0

                 0 3 3 3 0

                 2 0 0 0 2

Characteristic polynomial of B is given by:

                x^5 + c1*x^4 + c2*x^3 + c3*x^2 + c4*x^1 + c5

where c2 is sum of product of eigenvalues taken 2 at a time.

further we have a relation:

               c2 = 1/2 * (t1^2 - t2)

where t1 is trace of B and t2 is trace of B^2

As rank of B is 2 => there are 2 non-zero eigenvalues of B

let it be e1 and e2.

As the product of anything with 0 is 0

hence c2 = e1 * e2 (c2 is sum of product of eigenvalues taken 2 at a time)

=> e1 * e2 = 1/2 * (t1^2 - t2)

here t1 = 5 and t2 = 13

=> e1 * e2 = 1/2 * (5^2 - 13)

=> e1 * e2 = 6
–2 votes
–2 votes
The characteristic equation is : | A - zI | = 0 , where I is an identity matrix of order 5. i.e. determinant of the below shown matrix to be 0. 1-z 0 0 0 1 0 1-z 1 1 0 0 1 1-z 1 0 0 1 1 1-z 0 1 0 0 0 1-z Now solve this equation to find values of z. Steps to solve : 1) Expand the matrix by 1st row. (1-z) [ ( 1-z , 1 , 1, 0 ) (1 , 1-z, 1, 0) (1, 1, 1-z, 0) (0, 0, 0, 1-z ) ] + 1. [ ( 0, 1-z, 1, 1 ) ( 0, 1, 1-z, 1) ( 0, 1, 1, 1-z )( 1, 0, 0, 0) ] Note: ( matrix is represented in brackets, row wise ) 2) Expand both of the above 4x4 matrices along the last row. (1-z)(1-z) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] + 1.(-1) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] 3) Apply row transformations to simplify above matrices ( both the matrices are same). C1 <- C1 + C2 + C3 R2 <- R2- R1 R3 <- R3 - R1 result is : (1-z)(1-z) [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] - 1. [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] 4) Solve the matrix by expanding 1st column. result is : (1-z)(1-z)(z)(z) - (3-z)(z)(z) = 0 solve further to get : z^3 ( 3-z ) ( z-2 ) = 0 hence z = 0 , 0 , 0 , 3 , 2 Therefore product of non zero eigenvales is 6.
Answer:

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