Correct Answer is : (012121201212120)≤ft({array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0{array})021212102121210

≤ft({array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0{array})

≤ft({array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0{array})

cos 2θ≤ft({array}{ccc}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0{array})

cos 2θ≤ft({array}{ccc}0 & -1/2 & 1/2 \\ 1/2 & 0 & -1/2 \\ -1/2 & 1/2 & 0{array})

Correct Answer is : (012121201212120)≤ft({array}{ccc}0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0{array})021212102121210

Test Index

UPSC NDA 2 2022 Mathematics Solved Paper

Question Numbers: 37-39

Let A =

\left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \\ \cos^2 \theta & 0 & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right)

and A = P + Q where P is symmetric matrix and Q is skew-symmetric matrix.

Question : 37 of 120

Marks: +1, -0

What is P equal to ?

$\left(\begin{array}{ccc}0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0\end{array}\right)$
$\left(\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right)$
cos 2θ $\left(\begin{array}{ccc}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0\end{array}\right)$
cos 2θ $\left(\begin{array}{ccc}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0\end{array}\right)$

Solution:

Concept:

Every square matrix can be uniquely expressed as the sum of a symmetric matrix

P

and a skew-symmetric matrix

Q

, given by

P = \frac{1}{2}(A + A^T)

where

A^T

is the transpose.

Explanation:

Given

A = \begin{pmatrix}0 & \sin^2\theta & \cos^2\theta \\ \cos^2\theta & 0 & \sin^2\theta \\ \sin^2\theta & \cos^2\theta & 0\end{pmatrix}

Compute

A^T

by swapping rows and columns:

A^T = \begin{pmatrix}0 & \cos^2\theta & \sin^2\theta \\ \sin^2\theta & 0 & \cos^2\theta \\ \cos^2\theta & \sin^2\theta & 0\end{pmatrix}

Add

A

and

A^T

A + A^T = \begin{pmatrix}0 & \sin^2\theta+\cos^2\theta & \cos^2\theta+\sin^2\theta \\ \cos^2\theta+\sin^2\theta & 0 & \sin^2\theta+\cos^2\theta \\ \sin^2\theta+\cos^2\theta & \cos^2\theta+\sin^2\theta & 0\end{pmatrix}

Since

\sin^2\theta + \cos^2\theta = 1

A + A^T = \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}

Thus

P = \frac12(A + A^T) = \frac12 \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} = \begin{pmatrix}0 & \frac12 & \frac12 \\ \frac12 & 0 & \frac12 \\ \frac12 & \frac12 & 0\end{pmatrix}

Answer:

P = \begin{pmatrix}0 & \frac12 & \frac12 \\ \frac12 & 0 & \frac12 \\ \frac12 & \frac12 & 0\end{pmatrix}

(Option A).

Go to Question:

Prev Question

Next Question