Systems of Linear Equations, Matrix Algebra & Transform Theory- 1 - Free

(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)

Final Determination

• Statement 1 is true, as the product results in a 1x1 matrix, which is a scalar.
• Statement 2 is false, as the product is not defined due to dimension mismatch.

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

Option B is correct.

Let A = [[1 2]; [5 7]]. Compute the determinant: ad - bc = 1 × 7 - 2 × 5 = -3.

The formula for the inverse of a 2×2 matrix is A^-1 = 1/(ad - bc) [[d, -b]; [-c, a]].

Substituting a = 1, b = 2, c = 5, d = 7 gives A^-1 = 1/(-3) [[7, -2]; [-5, 1]].

Equivalently, multiplying the scalar -1 inside the bracket gives A^-1 = 1/3 [[-7, 2]; [5, -1]].

The matrix in Option B is the same as this result, so Option B is the correct choice.

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

Correct option: D

Statement 1 is not necessarily true.

Let A and B be n×n real symmetric matrices and let X be an n×k real matrix. Then X^T is k×n, so the product X^T A^T B A X is of size k×k.

This product is a scalar (a 1×1 matrix) only when k = 1, i.e., when X is a column vector. Therefore, without the additional assumption that X is a vector, the expression need not be a scalar.

Statement 2 is not always true.

For conformable sizes, let X be n×n and Y be n×m; then Y^T X Y is m×m. Taking transpose gives (Y^T X Y)^T = Y^T X^T Y.

Hence Y^T X Y is symmetric iff X = X^T (i.e., X is symmetric). Without assuming X is symmetric, the product need not be symmetric.

Conclusion: Both statements are false in general. Choose option D.