The multiplication inverse of A- cos - -sin - sin - cos - - is

The correct option is C [ cos θ amp; sin θ; sin θ amp; cos θ ] |A| =cos^2θ + sin^2θ = 1 adj(A) = [ cos θ amp; #160;sin θ; ...

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Standard XII

Mathematics

Finding Inverse Using Elementary Transformations

The multiplic...

Question

The multiplication inverse of $A = [\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix}]$ is

[\begin{matrix} - cos θ & sin θ - sin θ & - cos θ \end{matrix}]

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[\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}]

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[\begin{matrix} - cos θ & - sin θ sin θ & - cos θ \end{matrix}]

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[\begin{matrix} cos θ & sin θ sin θ & - cos θ \end{matrix}]

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Solution

The correct option is C $[\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}]$
$| A | = {cos}^{2} θ + {sin}^{2} θ = 1$
$a d j (A) = [\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}]$
$A^{- 1} = \frac{a d j (A)}{| A |} = [\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}]$

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Similar questions

Q. Simplify:

cos θ [\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}] + sin θ [\begin{matrix} sin θ & - cos θ cos θ & sin θ \end{matrix}]

Prove that following identities:
(i) $\frac{sin θ - cos θ}{s i n θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} = \frac{2}{2 s i n^{2} θ - 1}$
(ii) $\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{2}{1 - 2 {cos}^{2} θ}$