What is the adjoint of the matrix {pmatrix} cos(-θ ) & -sin(-θ ) \\ -sin(-θ ) & cos(-θ ) {pmatrix} ?

Correct Answer is : (cosθ−sinθ−sinθcosθ){pmatrix} cosθ & -sinθ \\ -sinθ & cosθ {pmatrix}(cosθ−sinθ−sinθcosθ)

{pmatrix} cosθ & -sinθ \\ -sinθ & cosθ {pmatrix}

{pmatrix} cosθ & sinθ \\ sinθ & cosθ {pmatrix}

{pmatrix} cosθ & sinθ \\ -sinθ & cosθ {pmatrix}

{pmatrix} cosθ & -sinθ \\ sinθ & cosθ {pmatrix}

Correct Answer is : (cosθ−sinθ−sinθcosθ){pmatrix} cosθ & -sinθ \\ -sinθ & cosθ {pmatrix}(cosθ−sinθ−sinθcosθ)

Test Index

UPSC NDA II 2018 Mathematics Solved Paper

Question : 14 of 120

Marks: +1, -0

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

$\begin{pmatrix} cos\theta & -sin\theta \\ -sin\theta & cos\theta \end{pmatrix}$
$\begin{pmatrix} cos\theta & sin\theta \\ sin\theta & cos\theta \end{pmatrix}$
$\begin{pmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{pmatrix}$
$\begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{pmatrix}$

Solution:

Concept:

For a

2 \times 2

matrix

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the adjoint is the transpose of the cofactor matrix.

A shortcut:

\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Explanation:

First, simplify the given matrix using

\cos(-\theta) = \cos\theta

and

\sin(-\theta) = -\sin\theta

The matrix becomes:

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

Now apply the shortcut for the adjoint of a

2 \times 2

matrix.

Here

a = \cos\theta

b = \sin\theta

c = \sin\theta

d = \cos\theta

Interchange the diagonal elements:

d = \cos\theta

a = \cos\theta

Change the sign of the off-diagonal elements:

-b = -\sin\theta

-c = -\sin\theta

Thus the adjoint is:

\text{adj}(A) = \begin{pmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}.

Answer:

Option A:

\begin{pmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}

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