If I - 1 0- 0 1 - B- 0 1- -1 0 -and C- cos - sin - -sin- cos- - then C-

The correct option is A Icosθ + B sinθ I=[ 1 amp; 0; 0 amp; 1 ], B=[ 0 amp; 1; 1 amp; 0 ] C=[ cosθ amp; sinθ; ...

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Finding Inverse Using Elementary Transformations

If I =[ 1 0...

Question

If I $= (\begin{matrix} 1 & 0 0 & 1 \end{matrix}), B = (\begin{matrix} 0 & 1 - 1 & 0 \end{matrix}) a n d C = (\begin{matrix} c o s θ & s i n θ - s i n θ & c o s θ \end{matrix}) t h e n C =$

I c o s θ + B s i n θ

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I s i n θ + B c o s θ

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I c o s θ - B s i n θ

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- I c o s θ + B s i n θ

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Solution

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Similar questions

I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], J = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] and B = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]

\frac{3 + i sin θ}{4 - i cos θ}, θ \in [0, 2 π]

sin θ + i cos θ

Express $\frac{{(c o s θ + i s i n θ)}^{4}}{{i (c o s θ + i s i n θ)}^{- 4}}$ in a + ib form.

\frac{cos θ - i sin θ}{sin θ - i cos θ}

(1 + \sqrt{3}) (1 + i) (cos θ + i sin θ)

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