Systems of Linear Equations, Matrix Algebra & Transform Theory- 2 - 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.

(b)

If a = 0 then rank (A) = rank() = 2. Therefore the system is consistant
∴ The system has solⁿ .

Ans.(d)
Consider the matrix A =   ,Now det (A) = 0
So byCramer's Rule the system has no solution

(d)

 =


For infinite solution of the system
α − 2 = 0 and β − 7 = 0
⇒ α = 2 and β = 7.

(b)
We know , rank (A) + Solution space X(A) = no. of unknowns.
⇒2 + X(A) = 3 . [Solution space X(A)= No. of linearly independent vectors]
⇒ X(A) =1.

(b). Highest possible rank of A= 2 ,as Ax = b is an inconsistent system.

The correct option is A (2 × 2)
Given, X_(4×3)′ Y_(4×3)′ P_(2×3)

None of the given matrices is square, hence (AB)⁻¹ = B⁻¹A⁻¹ does not hold for the above given matrices.

Now, Order of X^TY i,e.

XT_(3×4)Y_(4×3) = (3×3)

⇒ Order of (X^TY)⁻¹=(3×3)

Order of P(X^TY)⁻¹=(2×3)

Order of P(X^TY)⁻¹PT=(2×2)

⇒ Order [P(X^TY)⁻¹P^T]=(2×2)

(c)

∴Rank(A) = 2

Ans. (b)False
Laplace transform of  

B is correct.

The Laplace transform is defined by the integral L{f(t)} = ∫₀^∞ e^-st f(t) dt where the integral converges.

For f(t) = sin(at), evaluation of the integral (or using the standard transform table) gives ∫₀^∞ e^-st sin(at) dt = a / (s² + a²) for Re(s) > 0.

Therefore L{sin(at)} = a / (s² + a²), which matches option B.

Concept:

The order of highest ordered non zero minor is called the rank of a matrix.

Example:

If the square matrix is of order n × n then find the determinant of it if the value is non zero then its rank is n if the value comes out to be zero then find the determinant of  (n - 1) order of all sub-matrix if the value comes out to be non zero of any sub-matrix then the rank is n if the value is zero then the same processes continues further and the order at which value of the determinant is non zero is the rank of a matrix, Use this concept only for order  3 × 3 or lower order for more than 3 × 3 order we use the concept of Echelon form matrix.

Calculation:

The determinant of A is, |A| = ps – qr

Determinant of B is, |B| = (p² + q²) (r² + s²) – (pr + qs)²

|B| = (p²s² + q²r² – 2pqrs)

|B| = (ps - qr)²

Now, it is clear that |A|² = |B|

If the rank of A, ρ(A) = 1, that means the determinant of the matrix is zero, and hence the determinant of the matrix also zero. Therefore, the rank of the matrix B is 1

If the rank of A, ρ(A) = 2, that means the determinant of the matrix is non-zero, and hence the determinant of the matrix also non-zero. Therefore, the rank of the matrix B is 2.

Therefore, ρ(B) = ρ(A) = N

Concept:

From the properties of a matrix,

The rank of m × n matrix is always ≤ min {m, n}

If the rank of matrix A is ρ(A) and rank of matrix B is ρ(B), then the rank of matrix AB is given by

ρ(AB) ≤ min {ρ(A), ρ(B)}

If n × n matrix is singular, the rank will be less than ≤ n

Calculation:

Given:

AX = B

Where A is 2 × 4 matrices and b ≠ 0

The order of A^T is 4 × 2

The order of A^TA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (A^T) ≤ min (2, 4) = 2

Rank (A^TA) ≤ min (2, 2) = 2

As the matrix A^TA is of order 4 × 4, to have a unique solution the rank of A^TA should be 4.

Therefore, the unique solution of this equation is not possible.

Concept:

RANK OF A MATRIX:

The rank of a matrix is said to be R if:

(a) It has at least one non-zero minor of order R.

(b) Every minor of A of order higher than R is zero.

[Note: Non-zero row is that row in which all the elements are not zero.]

Calculation:

Given:

Now R₂ → R₂ – 2R₁

Now R₃ → R₃ – 4R₁

Now R₃ – (3/2)R₂

Number of non-zero rows = rank of the matrix = 2

∴ The rank of the following matrix is 2.

Ans.(c)   

Ans. (d)

Ans.(a)

To find the rank of a matrix of order n , first , complete its determinant ( in the case of a square matrix). if it is not 0 , then its rank = n . if it is 0. then see weather there is any non-zero minor of order n-1. if such minor exists, then the rank of the matrix = n-1

Correct option is C. 
Laplace inverse of [(sI−A)⁻¹]
e^At = L⁻¹[sI−A]⁻¹

Ans. (d)

I implies II means ≡ I → II

If |X| ≠ 0 then X^-1

|

If X^-1 then |X| ≠ 0 also |Adj X| = |X|^{n - 1} then |Adj X| ≠ 0

If X^-1 then |X| ≠ 0

I implies II and II implies I

∴ Both I and II are equivalent

Note:

X^-1 means X is invertible

|X| determinant of X

Ans. (c)
f(t) = u(t – ζ)
L{f(t)} = L{u(t – ζ)}

The fundamental period of x(t) = 2 sin(mt) + 3 sin(3mt) is determined by finding the least common multiple (LCM) of the individual periods of the sine components.

Periods of individual terms:

For sin(mt), the period is T₁ = 2π / m.

For sin(3mt), the period is T₂ = 2π / (3m).

LCM calculation:

T₂ is a third of T₁. The LCM of T₁ and T₂ is T₁, as T₁ is already a multiple of T₂.

Substituting m = π (assuming m is chosen such that the period simplifies to a numerical value):

T₁ = 2π / π = 2 seconds.

Thus, the fundamental period is 2 seconds.

Ans. (c)

Ans. (c)

Ans. (c)

Ans. (c)