If A - - cos- sin- 0- -sin- cos- 0- 0 0 e - - then- A - - T - A - - A -1- A -C- Adj A - - e - A -D- A - - T - A -

The correct options are A A(α, β)^T=A( α, β) B A(α β)^ 1=A( α, β) C Adj(A(α, β))=e^ β A( α, β)We have A(α, β)^T=[ cosα amp; sinα amp;0; sinα amp ...

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Adjoint 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)^{T} = A (- α, β)

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A (α - β)^{- 1} = A (- α, - β)

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A d j (A (α, β)) = e^{- β} A (- α, - β)

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A (α, β)^{T} = A (α, - β)

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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)^{- 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}

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} ⎤ ⎥ ⎦

|) A_{α} B_{β} = A_{α + β}

| |) A_{α} B_{β} = A_{α β}

| | |) (A_{α})^{- 1} = - A_{α}

) (A_{α})^{- 1} = A_{- α}

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

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