Let F alpha - begin bmatrix cos lalpha - - sin alpha - 0sin lalpha - cos lalpha - 0 1 0 - 0 - 1 1lend bmatrix - where - R- Then - F -1 is equal toA- None of theseB- F -1C- F -D- F 2 -

The correct option is C F( α)We have F(α)F( α) = [ cos α amp; sin α amp;0; sin α amp; cos α amp; 0; 0 amp; 0 ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

Let F alpha =...

Question

Let $F (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , where $α ϵ R .$
Then $[F (α)]^{- 1}$ is equal to

F (- α)

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F (- α^{- 1})

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F (2 α)

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N o n e o f t h e s e

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Solution

The correct option is A $F (- α)$
We have
$F (α) F (- α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = I ∴ F (- α) = [F (α)]^{- 1}$

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Similar questions

Q. Let

f (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

, then

(f (α))^{- 1}

is equal to

Q. Let

f (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

, then

(f (α))^{- 1}

is equal to

Q. Let

A_{α} = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

, then which of the following option(s) is/are correct?

$I f A = [\begin{matrix} c o s^{2} α & c o s α s i n α c o s α s i n α & s i n^{2} α \end{matrix}] a n d B = [\begin{matrix} c o s^{2} β & c o s β s i n β c o s β s i n β & s i n^{2} β \end{matrix}]$ are two matrices such that AB is the null matrix, then

Q. Let

F (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

, where

α ϵ R

. Then

(F {(α))}^{- 1}

is equal to

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Let F(α)=⎡⎢⎣cosα−sinα0sinαcosα0001⎤⎥⎦, where αϵR. Then [F(α)]−1 is equal to

The correct option is A F(−α)We have F(α)F(−α)=⎡⎢⎣cosα−sinα0sinαcosα0001⎤⎥⎦⎡⎢⎣cosαsinα0−sinαcosα0001⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦=I∴F(−α)=[F(α)]−1

Let $F (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , where $α ϵ R .$
Then $[F (α)]^{- 1}$ is equal to