Correct Answer is : cos 2θ(0−1212120−12−12120)≤ft({array}{ccc}0 & -1/2 & 1/2 \\ 1/2 & 0 & -1/2 \\ -1/2 & 1/2 & 0{array})021−21−2102121−210

≤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 : cos 2θ(0−1212120−12−12120)≤ft({array}{ccc}0 & -1/2 & 1/2 \\ 1/2 & 0 & -1/2 \\ -1/2 & 1/2 & 0{array})021−21−2102121−210

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 : 38 of 120

Marks: +1, -0

What is Q 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

A

can be written uniquely as

A = P + Q

, where

P

is symmetric and

Q

is skew‑symmetric.

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

and

Q = \frac{1}{2}(A - A^T)

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}

Here

Q

is the skew‑symmetric part, so

Q = \frac{1}{2}(A - A^T)

Compute

A - 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}

Using

\cos^2\theta-\sin^2\theta = \cos 2\theta

and

\sin^2\theta-\cos^2\theta = -\cos 2\theta

, we obtain:

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

Thus

Q = \frac{1}{2}(A - A^T) = \cos 2\theta \begin{pmatrix}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0\end{pmatrix}

Answer:

Option D.

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