1
GATE CSE 2025 Set 1
MCQ (More than One Correct Answer)
+1
-0

Consider the given system of linear equations for variables $x$ and $y$, where $k$ is a realvalued constant. Which of the following option(s) is/are CORRECT?

$$\begin{aligned} & x+k y=1 \\ & k x+y=-1 \end{aligned}$$

A
There is exactly one value of $k$ for which the above system of equations has no solution.
B
There exist an infinite number of values of $k$ for which the system of equations has no solution.
C
There exists exactly one value of $k$ for which the system of equations has exactly one solution.
D
There exists exactly one value of $k$ for which the system of equations has an infinite number of solutions.
2
GATE CSE 2024 Set 1
MCQ (Single Correct Answer)
+1
-0.33

The product of all eigenvalues of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ is

A

-1

B

0

C

1

D

2

3
GATE CSE 2023
MCQ (Single Correct Answer)
+1
-0.33

The Lucas sequence $$L_n$$ is defined by the recurrence relation:

$${L_n} = {L_{n - 1}} + {L_{n - 2}}$$, for $$n \ge 3$$,

with $${L_1} = 1$$ and $${L_2} = 3$$.

Which one of the options given is TRUE?

A
$${L_n} = {\left( {{{1 + \sqrt 5 } \over 2}} \right)^n} + {\left( {{{1 - \sqrt 5 } \over 2}} \right)^n}$$
B
$${L_n} = {\left( {{{1 + \sqrt 5 } \over 2}} \right)^n} - {\left( {{{1 - \sqrt 5 } \over 3}} \right)^n}$$
C
$${L_n} = {\left( {{{1 + \sqrt 5 } \over 2}} \right)^n} + {\left( {{{1 - \sqrt 5 } \over 3}} \right)^n}$$
D
$${L_n} = {\left( {{{1 + \sqrt 5 } \over 2}} \right)^n} - {\left( {{{1 - \sqrt 5 } \over 2}} \right)^n}$$
4
GATE CSE 2023
MCQ (Single Correct Answer)
+1
-0.33

Let $$A = \left[ {\matrix{ 1 & 2 & 3 & 4 \cr 4 & 1 & 2 & 3 \cr 3 & 4 & 1 & 2 \cr 2 & 3 & 4 & 1 \cr } } \right]$$ and $$B = \left[ {\matrix{ 3 & 4 & 1 & 2 \cr 4 & 1 & 2 & 3 \cr 1 & 2 & 3 & 4 \cr 2 & 3 & 4 & 1 \cr } } \right]$$.

Let $$\mathrm{det}(A)$$ and $$\mathrm{det}(B)$$ denote the determinates of the matrices A and B, respectively.

Which one of the options given below is TRUE?

A
$$\mathrm{det}(A)=\mathrm{det}(B)$$
B
$$\mathrm{det}(B)=-\mathrm{det}(A)$$
C
$$\mathrm{det}(A)=0$$
D
$$\mathrm{det}(AB)=\mathrm{det}(A)+\mathrm{det}(B)$$
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