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प्रश्न
Show that AB = BA where,
A = `[(costheta, - sintheta),(sintheta, costheta)], "B" = [(cosphi, -sinphi),(sinphi, cosphi)]`
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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