Test: Mathematical Physics - 2 - Physics MCQ

Therefore, 
Therefore, rate of change with distance in the direction of 

Note:  . then vector function is conservative and hence line integral (work clone) along a closed path is zero)  

Since. U is unitary matrix.


or (A + iB) (A - iB) = I  (since, A and B are Hermitian = B)
A² + B² + i(BA - AB) = I
Now. equating real and imaginary
Part. A² + B² = I and BA - AB = 0

We know that, det (e^A) = e^trace A

D² - 3D + 2 = 0
(D - 2) (D - 1) = 0 
Therefore, complementary function Be^x + Ce^2x and

A' will be a diagonal matrix, whose diagonal elements are the eigenvalues of A.
Therefore, to find eigenvalues of A.

Therefore, eigenvalues are λ = - 2, - 2, 4

Fourier transfonn of f' (t)

 where g is analytic at z₀ and g(z₀) ≠ 0
 where h(z) = (z - z₀) g'(z) - mg(z) is analytic at z = z₀ and h(z₀) = - mg (z₀) ≠ 0
∴ f' (z) has a pole of order (m + 1) at z = z₀.

Using partial fraction.

Poles are at z = + 2
Therefore, residue at 2 = 2

Now, y (1) = -1, put x = 1, y = -1


∴ for y to be defined log x must be defined i.e.. x > 0
Also when log x - 1 = 0, log x = 1 
So in the range 0 < x < 3 (option (b)). there will be a point x = 2.73 at whichy is not defined.
Also at x = e : y → ∞ (blow up).

∴ 

Fourier transform of f (x)

 (integrating by parts)

Since A is upper trinagular and B is lower triangular matrix. Hence, their eigenvalues are the principle diagonal elements.
Therefore, eigenvalues of A are 1, 0, 2 and eigenvalues of B are -1,1, 3
Since eigenvalues of A and B are all distinct, hence their corresponding eigenvectors will be linearly independent.
Hence, both A and B are digonalizable.