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The product of non-zero eigen values of the matrix is..... 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


Chetana. 9th Sep, 2019
Answer (1)
anshika.verma2019 20th Nov, 2020

Dear user,

Let, the given matrix be A.

1             0             0             0             1

0             1             1             1             0

A   =       0             1             1             1             0

0             1             1             1             0

1            0             0             0             1


To find eigen value,

Put, A    = (lambda) (I) 5

=> A – (lambda) X (I) 5 = 0



|1 0 0 0 1|                           |1 0 0 0 0|

|0 1 1 1 0|                           |0 1 0 0 0|

|0 1 1 1 0| - (X)     |0 0 1 0 0|  = 0

|0 1 1 1 0|                           |0 0 0 1 0|

|1 0 0 0 1|                           |0 0 0 0 1|

Thus,

1-X         0             0             0             1

0             1-X         1             1             0

0             1             1-X         1             0        = 0

0             1             1             1-X         0

1            0             0             0             1-X


Now, take determinant value

=> -x 5 + 5x 4 – 6x 3 = 0

=> -x 3 (x 2 - 5x + 6) = 0

=> x 3 (x 2 – 3x - 2x + 6) = 0

=> x 3 ((x - 3) -2(x - 3)) = 0

Thus,

=> x 3 (x-3)(x-2) = 0

=> x = 0, x = 3, x=2

Now, since x is not = 0 as per question

Thus, valid eigen values are: 2,3.

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