Practice question On Eigen values and Eigen vectors Linear Algebra | Matrix algebra | GATE | UPSC | IES | Engineering

Linear Algebra

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Question no. 1

What are the eigen values of ?

looks_one 1,4

looks_two 2,3

looks_3 0,5

looks_4 1,5

Option looks_3 0,5

Solution :

( 4 - λ)(1 - λ) - 4 = 0
λ² - 5λ + 4 - 4 = 0
λ² - 5λ = 0
λ(λ - 5) = 0
λ = 0 , 5

Question no. 2

What is the eigenvalues of the matrix ?

looks_one 1, 4 ,4

looks_two 1 , 4 ,-4

looks_3 3 , 3 ,3

looks_4 1 , 2 ,6

option looks_one 1, 4 ,4

Solution :

( 3 - λ )[(3-λ)² -1] + 1(-3 + λ -1 ) - 1(1+3 -λ) = 0
( 3 - λ )(9 + λ² - 6λ -1 ) + 2( λ - 4 ) = 0
( 3 - λ )( λ² - 6λ + 8 ) + 2( λ - 4 ) = 0
( 3 - λ )( λ - 4)( λ - 2) + 2( λ - 4 ) = 0
(λ - 4) [( 3 - λ )( λ - 2) + 2 ] = 0
(λ - 4) (λ² - 5λ + 4) = 0
(λ-4)(λ-4)(λ -1) = 0
λ = 1 , 4 ,4

Question no. 3

What is the sum of eigenvalues of ?

looks_one 8

looks_two 10

looks_3 4

looks_4 5

Option looks_one 8

Solution :
sum of eigen values = Sum of principal diagonal element
sum of eigen values = 1 + 5 + 2 = 8

Question no. 4

What is the value of x and y if and eigenvalues of A are 4 and 8 ?

looks_one x= 3 , y = 2

looks_two x= 2 , y = 4

looks_3 x = 4 , y = 2

looks_4 x= 2 , y = 3

Option looks_4 x= 2 , y = 3

Solution :
Sum of eigen values = 4 + 8 = 12
Sum of principal diagonal element = x + 10
Sum of eigen values = Sum of principal diagonal element
x + 10 = 12
x = 2
Determinant = product of eigen values
10x + 4y = 32
10*2 + 4y = 32
4y = 12
y = 3

Question no. 5

What are the eigen values of ?

looks_one 1 ,-1

looks_two 1 , i

looks_3 i , -i

looks_4 0 , 1

Option looks_3 i , -i

Solution :

λ² + 1 = 0
λ² = -1
λ² = i²
λ = -i , i

Question no. 6

What are the values of x and y , if and eigen values of A are 3 , 4 and 1 ?

looks_one x= 2 , y = 2

looks_two x = 3 , y = 2

looks_3 x = 2 , y = 3

looks_4 x= 4 , y = 1

option looks_4 x= 4 , y = 1

Solution :
Sum of eigen value = Sum of principal Diagonal elements
3 + x + y = 8
x + y = 5 .....[1]
Determinant = product of eigen values
3xy = 12
xy = 4 ......[2]
Solve [1] and [2]
x = 4 , y = 1

Question no. 7

The Eigenvalues of the martix

looks_one 1 and 4

looks_two -1 and 2

looks_3 0 and 5

looks_4 can't be determined

Option looks_3 0 and 5

Solution :

( 4 - λ)(1 - λ) - 4 = 0
λ² - 5λ + 4 - 4 = 0
λ² - 5λ = 0
λ(λ - 5) = 0
λ = 0 , 5

Question no. 8

What are the eigenvalues of the matrix ?

looks_one -1 , 1

looks_two 1 ,6

looks_3 2 , 5

looks_4 4 , -1

option looks_two 1 ,6

Solution :

( 2 - λ)(5 - λ) - 4 = 0
λ² - 7λ + 6 = 0
λ² - 6λ - λ + 6 = 0
λ(λ - 6) - 1( λ - 6) = 0
λ = 1 , 6

Question no. 9

For a given matrix , one of the eigenvalues is 3 . The other two eigenvalues are

looks_one 2, -5

looks_two 3 , -5

looks_3 2 , 5

looks_4 3 , 5

Option looks_two 3 , -5

Solution :
Sum of eigen value = Sum of principal Diagonal elements
Let the other two eigen values are x and y
Sum of eigen value = x+ y + 3
Sum of principal Diagonal elements = 1
x+ y + 3 = 1
x + y = - 2
Option 2 is satisfied the above equation.

Question no. 10

The eigenvalues of the matrix are 5 and 1 . What are the eigenvalues of the matrix S² ?

looks_one 1 , 25

looks_two 6 , 4

looks_3 5 , 1

looks_4 2 , 10

option looks_one 1 , 25

Solution :
If S is a matrix , λ is the eigen values
If S² is a matrix , λ² is the eigen values
S has eigen values 5 and 1
S² will have 5² and 1

Question no. 11

The minimum and maximum given eigen values of the matrix are -2 and 6 respectively. What is the other eigenvalue ?

looks_one 5

looks_two3

looks_3 1

looks_4 -1

option looks_two3

Solution :
Sum of eigen value = Sum of principal Diagonal elements
Let the third eigen value is x
Sum of eigen value = x -2 + 6 = x + 4
Sum of principal Diagonal elements = 7
x+4 = 7
x = 3

Question no. 12

If a square matrix A is real and symmetric , then eigenvalues

looks_one are always real

looks_two are always real and positive

looks_3 are always real and negative

looks_4 occur in complex conjugate pairs

option looks_one are always real

Solution :
If a square matrix A is real and symmetric , then eigenvalues are always real

Question no. 13

The characteristics equation of a (3 ✕ 3) matrix P is defined as
a(λ) = |Iλ - P | = λ³ + λ² + 2λ + 1 = 0
If I denotes identity matrix , then the inverse of matrix P will be

looks_one P² + P + 2I

looks_two P² + P + I

looks_3 -( P² + P + 2I )

looks_4 - (P² + P + I)

option looks_3 -( P² + P + 2I )

Solution :
λ³ + λ² + 2λ + 1 = 0
Matrix P will also satisfy the characteristic equation , therefore
P³ + P² + 2P + I = 0
I = -( P³ + P² + 2P)
P^-1I = -P^-1(P³ + P² + 2P)
P^-1 = -( P² + P + 2I )

Question no. 14

The matrix has one eigenvalue equal to 3, the sum of the two eigen value is

looks_one p

looks_two p-1

looks_3 p-2

looks_4 p-3

option looks_3 p-2

Solution :
Sum of eigen value = Sum of principal Diagonal elements
Let the other two eigen values are x and y
Sum of eigen value = x+ y + 3
Sum of principal Diagonal elements = 1 + p
x+ y + 3 = 1+p
x+y = p - 2

Question no. 15

The eigenvalues of the matrix are

looks_one -7 , 8

looks_two -6 , 5

looks_3 3 , 4

looks_4 1 , 2

Option looks_two -6 , 5

Solution :

( 4 - λ)(-5 - λ) - 10 = 0
λ² + λ - 30 = 0
λ² + 6λ - 5λ - 30 = 0
λ(λ + 6) - 5( λ + 5) = 0
λ = 5 , -6