Engineering Mathematics Miscellaneous


Engineering Mathematics Miscellaneous

Engineering Mathematics

  1. Consider the matrix A =
    50
    70
    7080

    whose eigenvectors corresponding to eigenvalues λ1 and λ2 are

    respectively. The value of x1 Tx2 is ________.









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    A =
    50
    70
    7080

    Eigen values of A are λ1 , λ2
    λ1 + λ2 = 130
    λ1λ2 = - 900
    Given that X1 =
    70
    λ1-50

    X2 =
    λ2
    -80
    70

    X1TX2 = [70λ1 - 50]
    λ2
    -80
    70

    = 70 λ2 –5600 + 70 λ1 – 3500
    = 70 (λ1 + λ2) – 9100
    = 70 (130) – 9100 = 0

    Correct Option: C

    A =
    50
    70
    7080

    Eigen values of A are λ1 , λ2
    λ1 + λ2 = 130
    λ1λ2 = - 900
    Given that X1 =
    70
    λ1-50

    X2 =
    λ2
    -80
    70

    X1TX2 = [70λ1 - 50]
    λ2
    -80
    70

    = 70 λ2 –5600 + 70 λ1 – 3500
    = 70 (λ1 + λ2) – 9100
    = 70 (130) – 9100 = 0


  1. The product of eigenvalues of the matrix P is
    P =
    2
    0
    1
    4-33
    02-1









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    Product of eigen value = | p |

    =
    2
    0
    1
    = 2
    4-33
    02-1

    Correct Option: B

    Product of eigen value = | p |

    =
    2
    0
    1
    = 2
    4-33
    02-1



  1. The number of linearly independent eigenvectors of matrix
    A =
    2
    0
    1
    is ____
    020
    003









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    Consider A - λI =
    2 - λ
    1
    0
    02 - λ0
    003 - λ

    ∴ characteristics equation is | A - λI |
    λ = 2, 2, 3
    There are only two linearly independent eigen vectors.

    Correct Option: A

    Consider A - λI =
    2 - λ
    1
    0
    02 - λ0
    003 - λ

    ∴ characteristics equation is | A - λI |
    λ = 2, 2, 3
    There are only two linearly independent eigen vectors.


  1. The condition for which the eigenvalues of the
    matrix A =
    2
    1
    are positive, is
    1K









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    All eigen values of A =
    2
    1
    are position 2 > 0
    1k

    ∴ 2 × 2 leading minor will be greater then zero.
    2
    1
    > 0
    1k

    2 k –1 > 0
    2 K > 1
    k >
    1
    .
    2

    Correct Option: A

    All eigen values of A =
    2
    1
    are position 2 > 0
    1k

    ∴ 2 × 2 leading minor will be greater then zero.
    2
    1
    > 0
    1k

    2 k –1 > 0
    2 K > 1
    k >
    1
    .
    2



  1. The lowest eigenvalue of the 2 × 2 matrix
    4
    2
    is _______.
    13









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    4 - λ
    2
    = 0
    13 - λ

    (4 – λ) (3 – λ) –2 = 0
    λ2 –7 λ+ 12 – 2 = 0
    λ2 –7 λ+ 10 = 0
    λ2 –5 λ– 2 λ+ 10 = 0
    λ(λ – 5) – 2 (λ – 5) = 0
    λ12 = 2,5
    The lowest eigenvalve of the matrix is 2.

    Correct Option: A

    4 - λ
    2
    = 0
    13 - λ

    (4 – λ) (3 – λ) –2 = 0
    λ2 –7 λ+ 12 – 2 = 0
    λ2 –7 λ+ 10 = 0
    λ2 –5 λ– 2 λ+ 10 = 0
    λ(λ – 5) – 2 (λ – 5) = 0
    λ12 = 2,5
    The lowest eigenvalve of the matrix is 2.