If A - begin pmatrix cos x -sin x - 0 1sin x -cos x - 0 0 - 0 - 1lendpmatrix - then adj A-1- A- - cos x sin x 0- -sin x cos x 0- 0 0 1 - cos x -sin x 0- sin x cos x 0- 0 0 1 -C- - 0 0 0- 0 0 0- 0 0 0 -D- - 1 0 0- 0 1 0- 0 0 1 -

The correct option is B |A|=1 ⇒ (adj A) =I ⇒ (A(adjA))^ 1 = I^ 1 ⇒ (adj A)^ 1 A^ 1 =1 ⇒ (adj A)^ 1 = A ...

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Determinant

If A = begin ...

Question

$I f A = ⎛ ⎜ ⎝ \begin{matrix} c o s x & - s i n x & 0 s i n x & c o s x & 0 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠, t h e n (a d j A)^{- 1} =$

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f (x) = ⎡ ⎢ ⎣ \begin{matrix} cos x & - sin x & 0 sin x & cos x & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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$sin x cos x 0$

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show that $F (x) F (y) = F (x + y)$

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} ⎤ ⎥ ⎦

g (y) = ⎡ ⎢ ⎣ \begin{matrix} c o s y & 0 & s i n y 0 & 1 & 0 - s i n y & 0 & c o s y \end{matrix} ⎤ ⎥ ⎦

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