Engineering Mathematics Miscellaneous
-
Consider the matrix A = 50 70 70 80
whose eigenvectors corresponding to eigenvalues λ1 and λ2 are
respectively. The value of x1 Tx2 is ________.
-
View Hint View Answer Discuss in Forum
A = 50 70 70 80
Eigen values of A are λ1 , λ2
λ1 + λ2 = 130
λ1λ2 = - 900Given 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 = 0Correct Option: C
A = 50 70 70 80
Eigen values of A are λ1 , λ2
λ1 + λ2 = 130
λ1λ2 = - 900Given 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
- The product of eigenvalues of the matrix P is
P = 2 0 1 4 -3 3 0 2 -1
-
View Hint View Answer Discuss in Forum
Product of eigen value = | p |
= 2 0 1 = 2 4 -3 3 0 2 -1 Correct Option: B
Product of eigen value = | p |
= 2 0 1 = 2 4 -3 3 0 2 -1
- The number of linearly independent eigenvectors of matrix
A = 2 0 1 is ____ 0 2 0 0 0 3
-
View Hint View Answer Discuss in Forum
Consider A - λI = 2 - λ 1 0 0 2 - λ 0 0 0 3 - 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 0 2 - λ 0 0 0 3 - λ
∴ characteristics equation is | A - λI |
λ = 2, 2, 3
There are only two linearly independent eigen vectors.
- The condition for which the eigenvalues of the
matrix A = 2 1 are positive, is 1 K
-
View Hint View Answer Discuss in Forum
All eigen values of A = 2 1 are position 2 > 0 1 k
∴ 2 × 2 leading minor will be greater then zero.2 1 > 0 1 k
2 k –1 > 0
2 K > 1k > 1 . 2 Correct Option: A
All eigen values of A = 2 1 are position 2 > 0 1 k
∴ 2 × 2 leading minor will be greater then zero.2 1 > 0 1 k
2 k –1 > 0
2 K > 1k > 1 . 2
- The lowest eigenvalue of the 2 × 2 matrix
4 2 is _______. 1 3
-
View Hint View Answer Discuss in Forum
4 - λ 2 = 0 1 3 - 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 1 3 - 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.