What is the inverse of the matrix? A={pmatrix} cosθ & sinθ & 0 \\ -sinθ & cosθ & 0 \\ 0 & 0 & 1 {pmatrix}

Correct Answer is : (cosθ−sinθ0sinθcosθ0001){pmatrix} cosθ & -sinθ & 0 \\ sinθ & cosθ & 0 \\ 0 & 0 & 1 {pmatrix}cosθsinθ0−sinθcosθ0001

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

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

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

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

Correct Answer is : (cosθ−sinθ0sinθcosθ0001){pmatrix} cosθ & -sinθ & 0 \\ sinθ & cosθ & 0 \\ 0 & 0 & 1 {pmatrix}cosθsinθ0−sinθcosθ0001

Test Index

UPSC NDA I 2018 Mathematics Solved Paper

Question : 48 of 120

Marks: +1, -0

A=\begin{pmatrix} cos\theta & sin\theta & 0 \\ -sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

$\begin{pmatrix} cos\theta & -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$
$\begin{pmatrix} cos\theta & 0 & -sin\theta \\ 0 & 1 & 0 \\ sin\theta & 0 & cos\theta \end{pmatrix}$
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & cos\theta & -sin\theta \\ 0 & sin\theta & cos\theta \end{pmatrix}$
$\begin{pmatrix} cos\theta & sin\theta & 0 \\ -sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Solution:

Concept:

The inverse of a matrix

A

is given by

A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)

, provided

\det(A) \neq 0

Explanation:

First, find the determinant of

A = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

Compute

\det(A) = \cos\theta (\cos\theta \cdot 1 - 0) - \sin\theta ((-\sin\theta) \cdot 1 - 0) + 0 = \cos^2\theta + \sin^2\theta = 1

Since

\det(A) = 1 \neq 0

, the inverse exists.

Next, find the minors of each element.

m_{11} = \cos\theta

m_{12} = -\sin\theta

m_{13} = 0

m_{21} = \sin\theta

m_{22} = \cos\theta

m_{23} = 0

m_{31} = 0

m_{32} = 0

m_{33} = \cos^2\theta + \sin^2\theta = 1

The cofactor matrix

C

is obtained by

c_{ij} = (-1)^{i+j} m_{ij}

Thus,

C = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

The adjugate of

A

is the transpose of

C

\text{adj}(A) = C^T = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

Finally,

A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

Answer:

A^{-1} = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

(Option A).

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