M is a \(2\times2\) matrix with eigenvalues 4 and 9. The eigenvalues of \(M^{2}\) are
Concept: If \(\lambda\) is an eigenvalue of a matrix \(M\), then \(\lambda^k\) is an eigenvalue of the matrix \(M^k\).
Calculation: Given eigenvalues of \(M\) are \(\lambda_1 = 4\) and \(\lambda_2 = 9\). The eigenvalues of \(M^2\) are \(\lambda_1^2\) and \(\lambda_2^2\). \[\lambda(M^2) = \{4^2, 9^2\} = \{16, 81\}\] Ans- D
The partial differential equation \[\frac{\partial^{2}u}{\partial t^{2}}-C^{2}\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right)=0; \text{ where } C \ne 0\] is known as
The given equation can be rearranged as: \[\frac{\partial^{2}u}{\partial t^{2}}=C^{2}\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right)\] This is the standard form of the two-dimensional (2-D) Wave Equation, which involves a second-order derivative with respect to time and the Laplacian (spatial derivatives). Ans- B
Which one of the following functions is analytic in the region \(|z|\le1\)?
A function is analytic in a region if it has no singularities (poles) within that region. The region is the closed unit disk \(|z|\le1\).
We check the poles (roots of the denominator) for each option:
Option A: Pole at \(z=0\). \(|z|=0 \le 1\). Pole is inside. (Not analytic).
Option B: Pole at \(z=-2\). \(|z|=|-2|=2\). Since \(2 > 1\), the pole is outside the region. (Analytic)
Option C: Pole at \(z=0.5\). \(|z|=0.5 \le 1\). Pole is inside. (Not analytic).
Option D: Pole at \(z=-j0.5\). \(|z|=|-j0.5|=0.5 \le 1\). Pole is inside. (Not analytic).
Ans- B
If \(f=2x^{3}+3y^{2}+4z,\) the value of line integral \(\int_{C} \text{grad} f \cdot d\mathbf{r}\) evaluated over contour \(C\) formed by the segments \((-3,-3,2)\rightarrow(2,-3,2)\rightarrow(2,6,2)\rightarrow(2,6,-1)\) is .
The integral is of the form \(\int_{C} \nabla f \cdot d\mathbf{r}\). This is a line integral of a gradient vector field, which is path-independent because the vector field \(\nabla f\) is conservative. The value of the integral depends only on the potential function \(f\) at the final and initial points of the contour \(C\). \[\int_{C} \nabla f \cdot d\mathbf{r} = f(\text{Final Point}) - f(\text{Initial Point})\]
Potential function: \(f(x, y, z) = 2x^3 + 3y^2 + 4z\). Initial Point: \(P_i = (-3, -3, 2)\). Final Point: \(P_f = (2, 6, -1)\).
1. Calculate \(f(P_f)\): \[f(2, 6, -1) = 2(2^3) + 3(6^2) + 4(-1)\] \[f(P_f) = 2(8) + 3(36) - 4 = 16 + 108 - 4 = 120\]
2. Calculate \(f(P_i)\): \[f(-3, -3, 2) = 2(-3)^3 + 3(-3)^2 + 4(2)\] \[f(P_i) = 2(-27) + 3(9) + 8 = -54 + 27 + 8 = -19\]
3. Calculate the Integral: \[\int_{C} \nabla f \cdot d\mathbf{r} = f(P_f) - f(P_i) = 120 - (-19) = 139\] Ans- 139
The rank of the matrix, \[M=\begin{pmatrix}0&1&1\\ 1&0&1\\ 1&1&0\end{pmatrix}\] is .
The rank of a matrix is the order of the highest ordered non-zero minor. For an \(n \times n\) matrix, if the determinant is non-zero, the rank is \(n\).
Calculation of Determinant: \[\det(M) = 0\begin{vmatrix}0&1\\ 1&0\end{vmatrix} - 1\begin{vmatrix}1&1\\ 1&0\end{vmatrix} + 1\begin{vmatrix}1&0\\ 1&1\end{vmatrix}\] \[\det(M) = 0 - 1(0 - 1) + 1(1 - 0)\] \[\det(M) = 0 - 1(-1) + 1(1) = 1 + 1 = 2\] Since \(\det(M) = 2 \ne 0\), the matrix \(M\) is full rank. \[\text{Rank}(M) = 3\] Ans- 3
Consider a \(2 \times 2\) matrix \(M = [\mathbf{v}_1 \ \mathbf{v}_2]\), where \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are the column vectors. Suppose \(M^{-1} = \begin{pmatrix} \mathbf{u}_1^T \\ \mathbf{u}_2^T \end{pmatrix}\), where \(\mathbf{u}_1^T\) and \(\mathbf{u}_2^T\) are the row vectors. Consider the following statements:
Statement 1: \(\mathbf{u}_1^T \mathbf{v}_1 = 1\) and \(\mathbf{u}_2^T \mathbf{v}_2 = 1\) Statement 2: \(\mathbf{u}_1^T \mathbf{v}_2 = 0\) and \(\mathbf{u}_2^T \mathbf{v}_1 = 0\)
By definition of the inverse matrix, the product \(M^{-1}M\) must equal the identity matrix \(I\): \[M^{-1} M = \begin{pmatrix} \mathbf{u}_1^T \\ \mathbf{u}_2^T \end{pmatrix} [\mathbf{v}_1 \ \mathbf{v}_2] = \begin{pmatrix} \mathbf{u}_1^T \mathbf{v}_1 & \mathbf{u}_1^T \mathbf{v}_2 \\ \mathbf{u}_2^T \mathbf{v}_1 & \mathbf{u}_2^T \mathbf{v}_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]
Comparing the elements:
The diagonal elements correspond to Statement 1: \(\mathbf{u}_1^T \mathbf{v}_1 = 1\) and \(\mathbf{u}_2^T \mathbf{v}_2 = 1\). (True)
The off-diagonal elements correspond to Statement 2: \(\mathbf{u}_1^T \mathbf{v}_2 = 0\) and \(\mathbf{u}_2^T \mathbf{v}_1 = 0\). (True)
Ans- C
The closed-loop line integral \[\oint_{|z|=5} \frac{z^3 + z^2 + 8}{z+2} dz\] evaluated Counter-clockwise, is
This integral is evaluated using Cauchy’s Integral Formula: \[\oint_C \frac{f(z)}{z-a} dz = 2\pi i f(a)\]
1. Identify Components:
\(f(z) = z^3 + z^2 + 8\), which is analytic everywhere.
The singular point (pole) is at \(a = -2\).
The contour \(C\) is the circle \(|z|=5\). Since \(|-2| < 5\), the pole is inside the contour.
2. Calculate \(f(a)\): \[f(-2) = (-2)^3 + (-2)^2 + 8\] \[f(-2) = -8 + 4 + 8 = 4\]
3. Apply the Formula: \[\oint_{|z|=5} \frac{z^3 + z^2 + 8}{z+2} dz = 2\pi i f(-2)\] \[\oint_{|z|=5} \frac{z^3 + z^2 + 8}{z+2} dz = 2\pi i (4) = 8\pi i\] Ans- A
A periodic function \(f(t)\), with a period of \(2\pi\), is represented as its Fourier series. \[f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos nt + \sum_{n=1}^{\infty} b_n \sin nt\] If \(f(t)=\begin{cases} A \sin t, & 0 \le t \le \pi \\ 0, & \pi < t < 2\pi \end{cases}\), the Fourier series coefficients \(a_1\) and \(b_1\) of \(f(t)\) are:
The period is \(T=2\pi\). The coefficients \(a_1\) and \(b_1\) are: \[a_1 = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos(t) dt \quad ; \quad b_1 = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin(t) dt\]
1. Calculate \(a_1\): \[a_1 = \frac{1}{\pi} \int_0^{\pi} (A \sin t) \cos t dt\] Using the identity \(\sin t \cos t = \frac{1}{2} \sin(2t)\): \[a_1 = \frac{A}{2\pi} \int_0^{\pi} \sin(2t) dt = \frac{A}{2\pi} \left[ -\frac{\cos(2t)}{2} \right]_0^{\pi}\] \[a_1 = \frac{A}{4\pi} [-\cos(2\pi) + \cos(0)] = \frac{A}{4\pi} [-1 + 1] = 0\]
2. Calculate \(b_1\): \[b_1 = \frac{1}{\pi} \int_0^{\pi} (A \sin t) \sin t dt = \frac{A}{\pi} \int_0^{\pi} \sin^2 t dt\] Using the identity \(\sin^2 t = \frac{1 - \cos(2t)}{2}\): \[b_1 = \frac{A}{2\pi} \int_0^{\pi} (1 - \cos(2t)) dt = \frac{A}{2\pi} \left[ t - \frac{\sin(2t)}{2} \right]_0^{\pi}\] \[b_1 = \frac{A}{2\pi} [ (\pi - 0) - (0 - 0) ] = \frac{A}{2\pi} \pi = \frac{A}{2}\]
The coefficients are \(a_1=0\) and \(b_1=\frac{A}{2}\). Ans- D
If \(\mathbf{A} = 2x\hat{i} + 3y\hat{j} + 4z\hat{k}\) and \(u = x^2 + y^2 + z^2\), then \(\text{div}(u\mathbf{A})\) at \((1, 1, 1)\) is .
We use the vector identity: \(\nabla \cdot (u\mathbf{A}) = u(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla u)\).
1. Calculate components at \((1, 1, 1)\):
\(u(1, 1, 1) = 1^2 + 1^2 + 1^2 = 3\)
\(\mathbf{A}(1, 1, 1) = 2\hat{i} + 3\hat{j} + 4\hat{k}\)
2. Calculate \(\nabla \cdot \mathbf{A}\) (Divergence of \(\mathbf{A}\)): \[\nabla \cdot \mathbf{A} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(4z) = 2 + 3 + 4 = 9\]
3. Calculate \(\nabla u\) (Gradient of \(u\)): \[\nabla u = \frac{\partial u}{\partial x}\hat{i} + \frac{\partial u}{\partial y}\hat{j} + \frac{\partial u}{\partial z}\hat{k} = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}\] At \((1, 1, 1)\): \[\nabla u(1, 1, 1) = 2\hat{i} + 2\hat{j} + 2\hat{k}\]
4. Calculate \(\text{div}(u\mathbf{A})\): \[\text{div}(u\mathbf{A}) = u(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla u)\] \[\text{div}(u\mathbf{A}) = (3)(9) + (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 2\hat{k})\] \[\text{div}(u\mathbf{A}) = 27 + (2\cdot 2 + 3\cdot 2 + 4\cdot 2)\] \[\text{div}(u\mathbf{A}) = 27 + (4 + 6 + 8) = 27 + 18 = 45\] Ans- 45
The probability of a resistor being defective is \(0.02\). There are \(50\) such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places) is .
This binomial distribution (\(n=50, p=0.02\)) is approximated by the Poisson Distribution with mean \(\lambda\): \[\lambda = np = 50 \times 0.02 = 1\]
We seek \(P(X \ge 2)\), which is calculated as: \[P(X \ge 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]\] Using the Poisson formula \(P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\) with \(\lambda = 1\): \[P(X=0) = \frac{e^{-1} 1^0}{0!} = e^{-1}\] \[P(X=1) = \frac{e^{-1} 1^1}{1!} = e^{-1}\]
\[P(X \ge 2) = 1 - [e^{-1} + e^{-1}] = 1 - 2e^{-1}\] Using the approximation \(e^{-1} \approx 0.367879\): \[P(X \ge 2) \approx 1 - 2(0.367879) = 1 - 0.735758 \approx 0.2642\]
Rounding off to two decimal places: \[P(X \ge 2) \approx 0.26\] Ans- 0.26