If A - - - cos - sin - 0- -sin - cos - 0- 0 0 e- - then A - - -1 is equal to

We have, (A (α, nbsp;β) )^ 1=adj A|A|=1e^β[ e^βcos α nbsp; amp; e^βsin α nbsp; amp;0; nbsp; e^βsin α nbsp; amp; e^ ...

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Inverse of a Matrix

If A α, β =[ ...

Question

If $A (α, β) = ⎡ ⎢ ⎣ \begin{matrix} cos α & sin α & 0 - sin α & cos α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦$ , then $(A (α, β))^{- 1}$ is equal to

A (- α, - β)

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A (- α, β)

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A (α, - β)

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A (α, β)

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Solution

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Similar questions

A (α, β) = ⎡ ⎢ ⎣ \begin{matrix} cos α & sin α & 0 - sin α & cos α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦,

A (α, β)^{- 1}

α, β

A (α, β) = ⎡ ⎢ ⎣ \begin{matrix} cos α & sin α & 0 - sin α & cos α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦

A (α, β)^{- 1}

If A $(α, β) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦, t h e n$

A (α, β) = ⎡ ⎢ ⎣ \begin{matrix} cos α & sin α & 0 - sin α & cos α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦

$A (α, β) = ⎛ ⎜ ⎝ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & e^{β} \end{matrix} ⎞ ⎟ ⎠ \Rightarrow {[A (α, β)]}^{- 1} =$

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