Question papers
GATE 2025 - Q1
Consider the matrix A below:$$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&\alpha&\beta \\0&0&0&\gamma \end{bmatrix}$$ For which of the following combinations of $\alpha$, $\beta$, and $\gamma$, is the rank of the matrix A atleast three?
(i) $\alpha=0$ and $\beta = \gamma \neq 0$
(ii) $\alpha=\beta = \gamma = 0$
(iii) $\alpha \neq 0$ and $\beta = \gamma = 0$
(iv) $\alpha=\beta = \gamma \neq 0$
a) only (i), (iii) and (iv)
b) only (iv)
c) only (ii)
d) only (i) and (iii)
Correct Answer: A
Rank of A has to be atleast three means rank can be $3$ or more ($4$).
(i) $\alpha = 0$, $\beta \neq 0$, $\gamma \neq 0$ and $\beta =\gamma$.$$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&0&\beta \\0&0&0&\beta \end{bmatrix}$$ $R_4\leftarrow R_4-R_3$, we will get $$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&0&\beta \\0&0&0&0 \end{bmatrix}$$$$\rho(A)=3$$
(ii) $\alpha = 0$, $\beta = 0$ and $\gamma = 0$.$$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&0&0 \\0&0&0&0 \end{bmatrix}$$ $$\rho(A)=2$$
(iii) $\alpha \neq 0$, $\beta = 0$, $\gamma = 0$.$$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&\alpha&0 \\0&0&0&0 \end{bmatrix}$$ Exchanging column $3$ and column $4$ , we will get $$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&0&\alpha \\0&0&0&0 \end{bmatrix}$$$$\rho(A)=3$$
(iv) $\alpha \neq 0$, $\beta \neq 0$, $\gamma \neq 0$ and $\alpha=\beta=\gamma$.$$\begin{bmatrix}2&3&4&5\\0&6&7&8\\0&0&\alpha &\alpha \\0&0&0&\alpha \end{bmatrix}$$$$\rho(A)=4$$
Hence (i), (iii) and (iv) are correct.
GATE 2025 - Q2
Consider a part of an electrical network as shown below. Some node voltages and the current flowing through the $3\Omega$ resistor are as indicated. The voltage (in volts) at node X is
a) $20/3$
b) $32/3$
c) $22/3$
d) $2/3$
Correct Answer: A
Step $1$: Voltage across $3\Omega$ resistor is $$V_{3\Omega}=1×3=3V$$
Step $2$: Voltage at node A, $$9-V_{A}=3V$$$$V_{A}=9-3=6V$$
Step $3$: Now finding the current flowing through $2\;\Omega$ and $1\;\Omega$ will be$$I=\frac{8-6}{3}=2/3$$
Step $4$: Voltage across $2\Omega$ resistor is $$V_{2\Omega}=2/3×2=4/3V$$
Step $5$: Voltage at node X, $$8-V_{X}=4/3$$$$V_{X}=8-4/3=20/3$$
GATE 2025 - Q3
Consider an additive white Gaussian noise (AWGN) channel with bandwidth W and noise spectral density $\frac{N_0}{2}$. Let $P_{av}$ denote the average transmit power constraint. Which one of the following plots illustrates the dependence of the channel capacity $C$ on the bandwidth $W$ (keeping $P_{av}$ and $N_0$ fixed)?
a) $103\;nJ$
b) $60\;nJ$
c) $40\;nJ$
d) $20\;nJ$
Correct Answer: A
Step $1$: Voltage across $3\Omega$ resistor is $$V_{3\Omega}=1×3=3V$$
Step $2$: Voltage at node A, $$9-V_{A}=3V$$$$V_{A}=9-3=6V$$
Step $3$: Now finding the current flowing through $2\;\Omega$ and $1\;\Omega$ will be$$I=\frac{8-6}{3}=2/3$$
Step $4$: Voltage across $2\Omega$ resistor is $$V_{2\Omega}=2/3×2=4/3V$$
Step $5$: Voltage at node X, $$8-V_{X}=4/3$$$$V_{X}=8-4/3=20/3$$
GATE 2025 - Q4
Consider a frequency modulated (FM) signal $$f(t)=A_c\;cos(2\pi ft+3sin(2\pi f_1 t)+4sin(2\pi f_1 t))$$
a) $103\;nJ$
b) $60\;nJ$
c) $40\;nJ$
d) $20\;nJ$
Correct Answer: A
Step $1$: Voltage across $3\Omega$ resistor is $$V_{3\Omega}=1×3=3V$$
Step $2$: Voltage at node A, $$9-V_{A}=3V$$$$V_{A}=9-3=6V$$
Step $3$: Now finding the current flowing through $2\;\Omega$ and $1\;\Omega$ will be$$I=\frac{8-6}{3}=2/3$$
Step $4$: Voltage across $2\Omega$ resistor is $$V_{2\Omega}=2/3×2=4/3V$$
Step $5$: Voltage at node X, $$8-V_{X}=4/3$$$$V_{X}=8-4/3=20/3$$