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Answer the following question: If = [cosαsinα-sinαcosα], show that Aα . Aβ = Aα+β - Mathematics and Statistics

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प्रश्न

Answer the following question:

If Aα = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`, show that Aα . Aβ = Aα+β 

योग

उत्तर

Aα = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`

∴ Aβ = `[(cosbeta, sinbeta),(-sinbeta, cosbeta)]`

∴ Aα.Aβ = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)] [(cosbeta, sinbeta),(-sinbeta, cosbeta)]`

`= [(cosalphacosbeta - sinalphasinbeta, cosalphasinbeta + sinalpha cosbeta),(-sinalpha cosbeta - cosalpha sinbeta, -sinalphasinbeta + cosalpha cosbeta)]`

`= [(cosalpha cosbeta - sinalpha sinbeta, cosalpha sinbeta + sinalpha cosbeta),(-[sinalpha cosbeta + cosalpha sinbeta], cosalphasinbeta - sinalpha cosbeta)]`

`= [(cos(alpha + beta), sin(alpha + beta)),(-sin(alpha + beta), cos(alpha + beta))]`

∴ Aα . Aβ = Aα+β

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Matrices - Properties of Matrix Multiplication
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Determinants and Matrices - Miscellaneous Exercise 4(B) [पृष्ठ १०१]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 4 Determinants and Matrices
Miscellaneous Exercise 4(B) | Q II. (10) | पृष्ठ १०१

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