If f(θ)={pmatrix} θ & θ \\ -θ & θ {pmatrix} then what is \{f(π)\}² equal to?

Correct Answer is : (1001){pmatrix} 1 & 0 \\ 0 & 1 {pmatrix}(1001)

{pmatrix} -1 & 0 \\ 0 & -1 {pmatrix}

{pmatrix} 1 & 1 \\ 1 & 1 {pmatrix}

{pmatrix} -1 & 0 \\ 0 & 1 {pmatrix}

{pmatrix} 1 & 0 \\ 0 & 1 {pmatrix}

Correct Answer is : (1001){pmatrix} 1 & 0 \\ 0 & 1 {pmatrix}(1001)

Test Index

UPSC NDA I 2025 Mathematics Paper

Question : 26 of 120

Marks: +1, -0

f(\theta)=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}

\{f(\pi)\}^2

equal to?

Solution:

Rotation matrix at

\theta=\pi

f(\pi)=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

To find

(f(\pi))^2

, multiply the matrix by itself:

f(\pi) \times f(\pi)=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \times \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

Using matrix multiplication:

Top-left element:

(-1)(-1)+(0)(0)=1

Top-right element:

(-1)(0)+(0)(-1)=0

Bottom-left element:

(0)(-1)+(-1)(0)=0

Bottom-right element:

(0)(0)+(-1)(-1)=1

Resulting matrix:

f(\pi)^2=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

\therefore (f(\pi))^2

is equal to the identity matrix, which is

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Go to Question:

Prev Question

Next Question