Engineering Mathematics Miscellaneous Easy Questions and Answers | Page - 37

Engineering Mathematics Miscellaneous

At least one eigenvalue of a singular matrix is

1. Positive
1. zero
1. negative
1. imaginary

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Atleast one eigen value of a singular matrix is zero.

Correct Option: B

Atleast one eigen value of a singular matrix is zero.

Consider a 3 × 3 real symmetric matrix S such that two of its eigenvalues are a ≠ 0, b ≠ 0 with respective eigenvectors

if a ≠ b then x₁y₁ + x₂y₂ + x₃y₃ equals

1. a
1. b
1. ab
1. 0

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We know that the Eigen vectors corresponding to distinct Eigen values of real symmetric matrix are orthogonal.

Correct Option: D

We know that the Eigen vectors corresponding to distinct Eigen values of real symmetric matrix are orthogonal.

One of the eigenvectors of matrix -5 2 is
-9 6

1. - 1
  1
1. - 2
  9
1. 2
  -1
1. 1
  1

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Eigen values of the matrix -5 -9 are 4, – 3
2 6

⇒ the eigen vector corresponding to eigen vector λ is Ax = λx (verify the options)
1
1

is eigen vector corresponding to eigen value λ = 3

Correct Option: D

Eigen values of the matrix -5 -9 are 4, – 3
2 6

⇒ the eigen vector corresponding to eigen vector λ is Ax = λx (verify the options)
1
1

is eigen vector corresponding to eigen value λ = 3

Choose the CORRECT set of functions, which are linearly dependent.

1. sin x, sin² x and cos² x
1. cos x, sin x and tan x
1. cos 2x, sin² x and cos² x
1. cos 2x, sin x and cos x

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Going by options,
cos 2x = 2cos² x – 1 =2 cos² x – (sin² x + cos² x)
cos 2x = cos²x – sin² x

Correct Option: C

Going by options,
cos 2x = 2cos² x – 1 =2 cos² x – (sin² x + cos² x)
cos 2x = cos²x – sin² x

The eigen values of a symmetric matrix are all

1. complex with non-zero positive imaginary part
1. complex with non-zero negative imaginary part
1. real
1. pure imaginary

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Suppose the eigen value of matrix A is n
(= λ + i B_B) say.
and the eigen vector is n whereas the conjugate pair of eigen value and eigen vector is λ and n
So, An = λ n ...(1)
And An = λ n ...(2)
Taking transpose of equation
x^-TA^T = x^-Tλ ...(3)
⇒ x^-TA^T = x^-T⇒ x^-TAn = x^-T⇒ λ = λ
⇒ α + iβ = α - iβ
⇒ 2iβ = 0
⇒ β = 0
Hence eigen value of a symmetric matrix are real

Correct Option: C

Suppose the eigen value of matrix A is n
(= λ + i B_B) say.
and the eigen vector is n whereas the conjugate pair of eigen value and eigen vector is λ and n
So, An = λ n ...(1)
And An = λ n ...(2)
Taking transpose of equation
x^-TA^T = x^-Tλ ...(3)
⇒ x^-TA^T = x^-T⇒ x^-TAn = x^-T⇒ λ = λ
⇒ α + iβ = α - iβ
⇒ 2iβ = 0
⇒ β = 0
Hence eigen value of a symmetric matrix are real