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

For a given vector $${[\matrix{ 1 & 2 & 3 \cr } ]^T}$$, the vector normal to the plane defined by $${w^T}x = 1$$ is

A
$${[\matrix{ { - 2} & { - 2} & 2 \cr } ]^T}$$
B
$${[\matrix{ 3 & 0 & { - 1} \cr } ]^T}$$
C
$${[\matrix{ 3 & 2 & 1 \cr } ]^T}$$
D
$${[\matrix{ 1 & 2 & 3 \cr } ]^T}$$
2
GATE EE 2023
MCQ (Single Correct Answer)
+1
-0.33

In the figure, the vectors u and v are related as : Au = v by a transformation matrix A. The correct choice of A is

GATE EE 2023 Engineering Mathematics - Linear Algebra Question 3 English

A
$$\left[ {\matrix{ {{4 \over 5}} & {{3 \over 5}} \cr { - {3 \over 5}} & {{4 \over 5}} \cr } } \right]$$
B
$$\left[ {\matrix{ {{4 \over 5}} & { - {3 \over 5}} \cr {{3 \over 5}} & {{4 \over 5}} \cr } } \right]$$
C
$$\left[ {\matrix{ {{4 \over 5}} & {{3 \over 5}} \cr {{3 \over 5}} & {{4 \over 5}} \cr } } \right]$$
D
$$\left[ {\matrix{ {{4 \over 5}} & { - {3 \over 5}} \cr {{3 \over 5}} & { - {4 \over 5}} \cr } } \right]$$
3
GATE EE 2023
MCQ (Single Correct Answer)
+1
-0.33

One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation $$\sigma_0$$. The $$\sigma_0$$ is estimated by randomly drawing out 10,000 numbers of samples ($$x_n$$). The estimates $${\widehat \sigma _1}$$, $${\widehat \sigma _2}$$ are computed in the following two ways.

$$\matrix{ {\widehat \sigma _1^2 = {1 \over {100000}}\sum\nolimits\limits_{n = 1}^{10000} {x_n^2} } & {\widehat \sigma _2^2 = {1 \over {9999}}\sum\nolimits\limits_{n = 1}^{10000} {x_n^2} } \cr } $$

Which of the following statements is true?

A
$$E(\widehat \sigma _2^2) = \sigma _0^2$$
B
$$E(\widehat \sigma _2^{}) = {\sigma _0}$$
C
$$E(\widehat \sigma _1^2) = \sigma _0^2$$
D
$$E(\widehat \sigma _1^{}) = E({\widehat \sigma _2})$$
4
GATE EE 2023
Numerical
+1
-0.33

In the following differential equation, the numerically obtained value of $$y(t)$$, at $$t=1$$ is ___________ (Round off to 2 decimal places).

$${{dy} \over {dt}} = {{{e^{ - \alpha t}}} \over {2 + \alpha t}},\alpha = 0.01$$ and $$y(0) = 0$$

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