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Show that AB=BA where, A = [cosθsinθsinθcosθ],B[cosθ-sinθsinθcosθ]

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प्रश्न

Show that AB = BA where, 

A = `[(costheta, - sintheta),(sintheta, costheta)], "B" = [(cosphi, -sinphi),(sinphi, cosphi)]`

योग
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उत्तर

AB = `[(costheta, -sintheta),(sintheta, costheta)] [(cosphi, -sinphi),(sinphi, cosphi)]`

`= [(costhetacosphi - sintheta sinphi, -costhetasinphi - sinthetacosphi),(sinthetacosphi + costhetasinphi, -sinthetasinphi + costhetacosphi)]`

`= [(cos (theta + phi) - sin (theta + phi)), (sin (theta + phi)  cos (theta + phi))]`      ...(i)

BA = `[(cosphi, -sinphi),(sinphi, cosphi)] [(costheta, - sintheta),(sintheta, costheta)]`

`= [(costhetacosphi - sintheta sinphi, -sinthetacosphi - costhetasinphi),(costhetasinphi + sinthetacosphi, -sinthetasinphi + costhetacosphi)]`

`= [(cos (theta + phi) - sin (theta + phi)), (sin (theta + phi)  cos (theta + phi))]`      ...(ii)

From (i) and (ii), we get

AB = BA

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अध्याय 4: Determinants and Matrices - Exercise 4.6 [पृष्ठ ९४]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 4 Determinants and Matrices
Exercise 4.6 | Q 4. (ii) | पृष्ठ ९४

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