Test: Linear Algebra - Civil Engineering (CE) MCQ

• If λ₁ ,λ₂ ,λ₃ ....λ_n are the eigen values of a non–singular matrix A, then A^-1 has the eigen values  1/λ₁ ,1/λ₂ ,1/λ₃ ....1/λ_n 
• Thus eigen values of A^-1are 1/2, 1/3, -1/3

• If λ₁ ,λ₂ ,λ₃ ....λ_n are the eigen values of  a matrix A, then A² has the eigen values  λ₁² ,λ₂² ,λ₃² ....λ_n² 
• So, eigen values of A² are 1, 4, 9.

• Since the sum of the eigenvalues of an n–square matrix is equal to the trace of the matrix (i.e. sum of the diagonal elements)
• So, required sum = 8 + 5 + 5  = 18

The eigenvalues of a matrix A are the roots of the characteristic equation of A, which are invariant under the transpose operation. This means that the matrix A and its transpose A^T have the same eigenvalues.

Given that 2 and 4 are the eigenvalues of A, the eigenvalues of A^T will also be 2 and 4.

• Since 1 and 3 are the eigenvalues of A so the characteristic equation of A is:
• Also, by Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation. So:
  

• Since A + B is defined, A and B are matrices of the same type, say m x n. Also, AB is defined.
• So, the number of columns in A must be equal to the number of rows in B i.e. n = m.
• Hence, A and B are square matrices of the same order.

Inverse matrix is defined for square matrix only.

The characteristic equation ∣A − λD = 0

Corresponding to λ = 4, we have

= 0

 = 0 which gives only one independent equation,

-9x + 2y = 0

∴ x/2 = y/9 gives eigen vector (2,9)

Corresponding to λ = −3,

which gives -x + y = 0 (only one independent equation)

∴ x/1 = y/1 which gives (1,1)

So, the eigen vectors are 

Determinant of a skew-symmetric even ordered matrix A should be a perfect square.

All Eigen values of  are positive

2 > 0

∴ 2 x 2 leading minor must be greater than zero

No. of non zero rows = 2
rank = 2