Let f lalpha - begin bmatrix coslalpha - -sin-alpha - 0 sin lalpha - costalpha - 011 0 - 0 - 1lendbmatrix - then f -1 is equal toA- noneB- f -C- f -D- f -1

The correct option is C f( α )(f(α ))^ 1=[ cosα amp; sinα amp; 0; sinα amp; cosα amp; 0; 0 amp; 0 amp; 1 ]=f( α) ...

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Byju's Answer

Standard IX

Mathematics

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

f (α)

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f (- α)

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f (α - 1)

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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

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} ⎤ ⎥ ⎦

, then

(f (α))^{- 1}

is equal to

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

Q. 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?

Q. If

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} ⎤ ⎥ ⎦

then

f (α + β) =

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s (α - β) & c o s α c o s (α - β) & 1 & c o s β c o s α & c o s β & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

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Let f(α)=⎡⎢⎣cosα−sinα0sinαcosα0001⎤⎥⎦, then (f(α))−1 is equal to

The correct option is B f(−α)(f(α))−1=⎡⎢⎣cosαsinα0−sinαcosα0001⎤⎥⎦=f(−α)

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

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