Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

If $A=\begin{bmatrix} acos\theta & bsin\theta \\-bsin\theta & acos\theta \end{bmatrix}$ then $A^{-1}$ is equal to :

Options:

$\frac{1}{a^2cos^2\theta - b^2sin^2\theta }\begin{bmatrix} acos\theta & bsin\theta \\-bsin\theta & acos\theta \end{bmatrix}$

$\frac{1}{a^2cos^2\theta + b^2sin^2\theta }\begin{bmatrix} acos\theta & bsin\theta \\-bsin\theta & acos\theta \end{bmatrix}$

$\frac{1}{a^2cos^2\theta + b^2sin^2\theta }\begin{bmatrix} acos\theta & -bsin\theta \\bsin\theta & acos\theta \end{bmatrix}$

$\frac{1}{a^2cos^2\theta + b^2sin^2\theta }\begin{bmatrix} asin\theta & bcos\theta \\-bcos\theta & asin\theta \end{bmatrix}$

Correct Answer:

$\frac{1}{a^2cos^2\theta + b^2sin^2\theta }\begin{bmatrix} acos\theta & -bsin\theta \\bsin\theta & acos\theta \end{bmatrix}$

Explanation:

The correct answer is Option (3) → $\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta }\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$

$A=\begin{bmatrix} a\cos\theta & b\sin\theta \\-b\sin\theta & a\cos\theta \end{bmatrix}$

$|A|=a^2\cos^2θ+b^2\sin^2θ$

$Adj\,A=\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$

$A^{-1}=\frac{Adj\,A}{|A|}=\frac{1}{a^2\cos^2θ+b^2\sin^2θ}\begin{bmatrix} a\cos\theta & -b\sin\theta \\b\sin\theta & a\cos\theta \end{bmatrix}$