First slide

Applications of determinant

Question

If $(\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array}) {(\begin{array}{cc} 1 & \tan θ \\ - \tan θ & 1 \end{array})}^{- 1} = (\begin{array}{cc} a & - b \\ b & a \end{array})$ then

Moderate

A

$a = b = 1$
B

$a = \cos 2 θ, b = \sin 2 θ$
C

$a = \sin 2 θ, b = \cos 2 θ$
D

$a = 1, b = \sin 2 θ$

Solution

$\begin{array}{l} (\begin{array}{cc} a & - b \\ b & a \end{array}) = (\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array}) (\begin{array}{cc} \cos^{2} θ & - \sin θ \cos θ \\ \sin θ \cos θ & \cos^{2} θ \end{array}) \\ = (\begin{array}{cc} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & \cos 2 θ \end{array}) \\ \Rightarrow a = \cos 2 θ, b = \sin 2 θ \end{array}$

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