Test: Systems of Linear Equations, Matrix Algebra & Transform Theory- 1 - Civil Engineering (CE) MCQ

(b) This can be written as AX = B Where A

Angemented matrix 

rank(A) ≠ rank(). The system is inconsistant .So system has no solution.

(a)

∴ rank of() = rank of(A) = 3

∴ The system has unique solution.

(b)

∴ rank (A) = rank () = 3
∴ The system has unique solution

(c )

∴ rank(A) = 2 ≠ 3 = rank() .

∴ The system is inconsistent and has no solution.

(c)
 Given  

A non-homogeneous system of linear equations representing an over-determined system (where there are more equations than unknowns) can generally be described as follows:

• If the system is consistent, it means there exists at least one solution.
• If the system is inconsistent, it means there are no solutions.

Given that the system is over-determined (more equations than unknowns), it is more likely to be inconsistent because it is harder for all equations to be satisfied simultaneously. However, if it is consistent, it would typically have a unique solution since the extra constraints (more equations) usually reduce the solution space to a single point.

Thus, the correct answer is: Inconsistent having no solution.

given set of equations are x₁+2x₂+x₃+4x₄=2 , 3x₁+6x₂+3x₃+12x₄=6

consider AB = 

⇒ R₂ → R₂ - 3R₁

AB = 

P(A)=1; P(AB)=1;n=4

⇒ P(A) =P(B) < no. of variables

⇒ Infinitely many solutions ⇒multiple non-trivial solution

The eigenvalues of a skew-symmetric matrix have specific characteristics:

• They are either zero or pure imaginary.
• This is because a skew-symmetric matrix has the property that its transpose is equal to the negative of itself.
• As a result, any real eigenvalue must be zero.
• Non-zero eigenvalues appear in conjugate imaginary pairs.

(b)

Let A = 

Then C = AB = 

Then det (AB) = 0.

Then also every minor
of order 2 is also zero.
∴ rank(C) =1.

(a) (P) Singular matrix → Determinant is zero
(Q) Non-square matrix → Determinant is not defined
(R) Real symmetric → Eigen values are always real
(S) Orthogonal → Determinant is always one

(a)

(b)
(PQ) ^-1 = P Q^-1P^-1P = Q^-1

(b)

=  

Concept:

Non-homogeneous equation of type AX = B has infinite solutions;
if ρ(A | B) = ρ(A) < Number of unknowns
Calculation:

Given:

2x + 3y = 1

4x + 6y = λ

The augmented matrix is given by:

Answer: B
Solution:
The given system of equations is linearly dependent, meaning all the equations are scalar multiples of each other. This implies there are infinitely many solutions as the system represents the same line in three-dimensional space. Thus, the system has infinitely many solutions.

Concept:

The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.

• If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
• If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
• If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

Calculation:

The augmented matrix for the system of equations is

Performing: R₃ → R₃ – R₂

      …

If λ = 6 and μ ≠ 20 then

Rank (A | B) = 3 and Rank (A) = 2

∵ Rank (A | B) ≠ Rank (A)

∴ Given the system of equations has no solution for λ = 6 and μ ≠ 20

Statement 1: "The matrix product [X^T ][A^T ][B][ A] [X] is a scalar."

• Matrix [A] and [B] are symmetric. This means [A^T ]= [A] and [B^T ]= [B].
• The expression becomes [X^T ][A ][B][ A] [X
• Since A and B are symmetric matrices, and X^T is the transpose of X, this product will be a scalar because the result is a 1x1 matrix. This is because, in a matrix product of this form, the final result is a scalar if it satisfies the form [X^T ][A ][B][ A] [X], as the dimensions are properly aligned and it results in a scalar (a 1x1 matrix).

Thus, Statement 1 is true.

Statement 2: "The matrix product [Y^T ][X][ Y] is always symmetric."

• The matrix product Y^T X Y will be symmetric if [Y^T ][X][ Y] = [Y^T ][X^T][ Y].
• In general, this product is not guaranteed to be symmetric unless specific properties hold for the matrices X and Y. In general, for any arbitrary matrices X and Y, the product Y^T X Y is not necessarily symmetric.

Thus, Statement 2 is false.

Conclusion:

• Statement 1 is true, and Statement 2 is false.

The correct answer is: A: Statement 1 is true but 2 is false.