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Question

If $A (α, β) = ⎣ ⎡ cos α - s in α 0 s in α cos α 0 00 e^{β} ⎦ ⎤$ , then $A (α, β)^{- 1}$ is equal to

A (- α, - β)

A (- α, β)

A (α, - β)

A (α, β)

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A (α, β) = ⎣ ⎡ cos α - s in α 0 s in α cos α 0 00 e^{β} ⎦ ⎤

, then

A (α, β)^{- 1}

is equal to

5 mins ago

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5 mins ago

Text solutionVerified

Given,

A (α, β) = ⎣ ⎡ cos α - s in α 0 s in α cos α 0 00 e^{β} ⎦ ⎤

∣ A (α, β) ∣ = e^{β} cos^{2} α + e^{β} sin^{2} α

∣ A (α, β) ∣ = e^{β}

Now,

a d j A (α, β) = C^{T} = ⎣ ⎡ e^{β} cos α - e^{β} s in α 0 e^{β} s in α e^{β} cos α 0 001 ⎦ ⎤^{T}

\Rightarrow a d j A (α, β) = ⎣ ⎡ e^{β} cos α e^{β} s in α 0 - e^{β} s in α e^{β} cos α 0 001 ⎦ ⎤

So,

A (α, β)^{- 1} = \frac{1}{∣ A ∣} . A d j A = ⎣ ⎡ cos α s in α 0 - s in α cos α 0 00 e^{- β} ⎦ ⎤

or,

A (α, β)^{- 1} = ⎣ ⎡ cos (- α) - s in (- α) 0 s in (- α) cos (- α) 0 00 e^{- β} ⎦ ⎤

Hence,

A (α, β)^{- 1} = A (- α, - β)

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Question Text	If $A (α, β) = ⎣ ⎡ cos α - s in α 0 s in α cos α 0 00 e^{β} ⎦ ⎤$ , then $A (α, β)^{- 1}$ is equal to
Answer Type	Text solution:1
Upvotes	150