# GATE | GATE-CS-2014-(Set-2) | Question 57

The product of the non-zero eigenvalues of the matrix

```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```

is ______
(A) 4
(B) 5
(C) 6
(D) 7

Explanation: 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 4×4 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.

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