Test: Engineering Mathematics- 3 - Computer Science Engineering (CSE) MCQ

Sum of the eigen values of matrix is = Sum of diagonal values present in the matrix

∴ 1 + 0 + P = 3 + λ₂ + λ₃

⇒ P + 1 = 3 + λ₂ + λ₃

⇒ λ₂ + λ₃ = P + 1 – 3 = P – 2

Two eigen value of X is 4 and (2 + 7i)

∴ (2 - 7i) (conjugate roots) must be the third root

Determinant of P = product of eigenvalues

Δ = 4 × (2 + 7i) × (2 - 7i) 

Δ = 212

For a given matrix A if V is the eigen vector corresponding to the eigen value λ, then:

AV = λV

∴ {α (−4, 2, 1) |α ≠ 0, αϵR} are the corresponding eigenvectors.

• The eigenvalue of the triangular (lower or upper) matrix are just the diagonal elements of matrix.
• The product of the eigenvalue of a matrix is equal to its determinants.
• If λ is the eigenvalue of matrix, then 1/λ is the eigenvalue of its inverse since orthogonal is equal to inverse matrix then it has 1/λ as its eigenvalue

​Example:

Eigenvalues are 1, 4 and 6 (diagonal elements)

Product of eigen value = determinants = 1 × 4 × 6 = 24

Orthogonal matrix and Inverse of given matrix have eigenvalues: 1,1/4 and 1/6

W.K.T.∑λ_i = ∑a_ii
∴ - 1 + 7 = 1 + a ⇒ a = 5

Also π λ_i = |A|

∴ - 1 × 7 = a – 4b

 - 7 = 5 – 4b ⇒ b = 3

Given Matrix:

Characteristic Equation:

(2−λ) (4−λ) −3 = 0

λ² − 6λ + 5 = 0

λ = 5 or λ = 1

Eigen vectors are also called invariant vectors, characteristic vectors, or latent vectors. Eigenvalues are also called characteristic roots or latent roots.

(A - λI)X = 0

Given λ = 1

⇒ (A - I) X = 0

⇒ x = 2y + 3x = 0, 10x – 5y + 5z = 0, 5x – 4y + 5z = 0

By solving above equations, we get x = 

 i.e. one linearly independent latent vector.

Characteristic equation is given

 

λ³ − 5λ² = 0

λ = 0 or λ = 5

Therefore largest value is 5.

Let λ1 and λ2 be the two eigen values

Product of eigen value is equal to determinant of matrix

Sum of eigen value is equal to its trace
 

a + d = λ1 + λ2 = 7

λ1 × (7 − λ1) = 6

λ1² − 7λ + 6 = 0

λ1 = 6 or λ2 = 1

|λ1 − λ2| = 5

(5 - λ) (1 - λ) + 4 = 0

5 – 5λ - λ + λ² + 4 = 0

λ² - 6λ + 9 = 0

(λ - 3)² = 0

λ = 3

for eigen vector

So, only one independent eigen vector.