Engineering Maths - Linear Algebra

21. The vector

  
 
1
2
-1
 
     

is an eigen vector of

A=
 
-22-3
21-6
-1-20
 

one of the eigen values of A is

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23. Given: Matrix A =

 
1 6 1
1 2 0
0 0 3
 

the Largest eigen value is

  • Option : D
  • Explanation : Now

     
    1-λ11
    11-λ1
    111-λ
     

    =0

     

     ⇒
     
    3-λ3-λ3-λ
    11-λ1
    111-λ
     
    =0
     ⇒3-λ
    111
    11-λ1
    111-λ
    =0

     

     ⇒3-λ
    100
    11
    10λ
    =0

    Hence eigen values are 0,0,3

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24. Corresponding to highest eigen value the eigenvector is Corresponding to highest eigen value the eigenvector is

  • Option : C
  • Explanation : Repeating above procedure, we get successively RepeaRepeating above procedure, we get successively
ting

    It follows that the largest eigen value is 4 and corresponding eigenvector is It follows that the largest eigen value is 4 and corresponding eigenvector is

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25. Given Matrix A=

The eigen-values are

 

Given Matrix A=The eigen-values are

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X=
 
0100
0010
0001
0001
 
Xare
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