Let Alalpha - begin bmatrix cos alpha - -sin lalpha - 0 sin- alpha - cos lalpha - 0110 - 0 - 1lend bmatrix - then which of the following options is-are correct-

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Inverse of a Matrix

Let Alalpha =...

Question

Let $A_{α} = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then which of the following option(s) is/are correct?

A_{α + β} = A_{α} A_{β}

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(A_{α})^{- 1} = A_{- α}

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(A_{α})^{- 1} = - (A_{α})

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A_{α}^{2} = - I

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Solution

The correct options are
A $A_{α + β} = A_{α} A_{β}$
B $(A_{α})^{- 1} = A_{- α}$

$A_{α} A_{β} = A_{α + β}$

Also , $A_{0} = I because A_{α} A_{- α} = A_{α - α} \Rightarrow A_{0} = I$

$∴$ We get $(A_{α})^{- 1} = A_{- α}$

However, $(A_{α})^{- 1} = - (A_{α})$ and $A_{α}^{2} = - I do not hold true$

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Similar questions

Q. 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} ⎤ ⎥ ⎦

, where

α ϵ R .

Then

[F (α)]^{- 1}

is equal to

$I f A = [\begin{matrix} c o s^{2} α & c o s α s i n α c o s α s i n α & s i n^{2} α \end{matrix}] a n d B = [\begin{matrix} c o s^{2} β & c o s β s i n β c o s β s i n β & s i n^{2} β \end{matrix}]$ are two matrices such that AB is the null matrix, then

If $A = [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}]$ , then $A^{2} =$

If $A = [\begin{matrix} s i n α & c o s α - c o s α & s i n α \end{matrix}]$ , then verify that A'A=I.

If A = $[\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}]$ and $A^{- 1}$ = A', then find the value of $α$ .

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Let Aα=⎡⎢⎣cos α−sin α0sin αcos α0001⎤⎥⎦, then which of the following option(s) is/are correct?

The correct options are A Aα+β=AαAβ B (Aα)−1=A−α AαAβ=Aα+β Also , A0=I because AαA−α=Aα−α⇒A0=I ∴ We get (Aα)−1=A−α However, (Aα)−1=−(Aα) and A2α=−I do not hold true

Let $A_{α} = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then which of the following option(s) is/are correct?