Let f- - cos- -sin- 0- sin - cos- 0- 0 0 1 - then f- -1 is equal to

(f(α ))^ 1=[ cosα amp; sinα amp; 0; sinα amp; cosα amp; 0; 0 amp; 0 amp; 1 ]=f( α) ...

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Inverse of a Matrix

Let fα =[ cos...

Question

Let $f (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then $(f (α))^{- 1}$ is equal to

f (α)

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f (- α)

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f (α - 1)

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n o n e

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Solution

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Similar questions

F (α) = ⎛ ⎜ ⎝ \begin{matrix} cos α & - sin α & 0 sin α & cos α & 0 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠

{F (x)}^{- 1} = ?

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} ⎤ ⎥ ⎦

[F (α)]^{- 1} = F (- α)

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

[G (β)]^{- 1} = G (- β)

f (α) = ⎡ ⎢ ⎣ \begin{matrix} cos α & - sin α & 0 sin α & cos α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

f (- α)

f (- α) + f (α)

f (α) = ⎡ ⎢ ⎣ \begin{matrix} cos α & - sin α & 0 sin α & cos α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

f (- α) + f (α)

A = ⎡ ⎢ ⎣ \begin{matrix} cos α & - sin α & 0 sin α & cos α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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