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Inverse of a Matrix

Let f lalpha ...

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

A

f (α)

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B

f (- α)

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C

f (α - 1)

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D

n o n e

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Solution

The correct option is B $f (- α)$
$(f (α))^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = f (- α)$

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Similar questions

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) = ⎡ ⎢ ⎣ \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} ⎤ ⎥ ⎦

α ϵ R .

[F (α)]^{- 1}

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?

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

f (α + β) =

If $A = [\begin{matrix} α & 0 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} 1 & 0 3 & 1 \end{matrix}],$ Then the value of $α$ for which $A^{2} = B$ is

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