आव्यूह समीकरण
`[(2,1),(3,2)].A.[(-3,2),(5,-3)]=[(1,0),(0,1)]` को संतुष्ट करता हुआ आव्यूह A ज्ञात करे ।

Explanation

माना कि `B=[(2,1),(3,2)]" तथा "C=[(-3,2),(5,-3)]`
अब `।B।=।(2,1),(3,2)।= (4 - 3) = 1 ne 0`
` ।C।=।(-3,2),(5,-3)।= (9-10) = - 1ne 0`
अतः `B^(-1) "तथा " C^(-1)` का अस्तित्व है ।
`therefore` दिया गया आव्यूह समीकरण हो जाता है , `BAC = I_(2)`
अब ` BAC = I_(2) rArr B^(-1) (BAC)C^(-1) = B^(-1) I_(2) C^(-1)`
` rArr (B^(-1) B) A(CC^(-1)) = B^(-1) (I_(2) C^(-1))`
` rArr I_(2) Al_(2) = B^(-1) V^(-1)`
` rArr A= B^(-1) C^(-1)` ...(I)
B के अवयवो के सहखण्ड (cofactors ) है ,
` B_(1) = 2, B_(12) = - 3, B_(21) = - 1l B_(22) = 2`
` therefore " adiB " = [(2,-3),(-1,2)] "का transpose " = [(2,-1),(-3,2)]`
`therefore B^(-1) = (1)/(।B।)।"adi B"= [(2,-1),(-3,2)]" " [because ।B। = 1]`
पुनः C के अवयवों के सहखण्ड (cofactors ) है ,
`C_(11) = - 3, C_(12) = - 5, C_(21) = - 2, C_(22) = - 3 `
`therefore "adi C " = [(-3,-5),(-2,-3)]" का transpose "= [(-3,-3),(-5,-3)]`
` therefore C^(-1) = (1)/(।C।)। "adi C" = [(3,2),(5,3)]"" [ because ।C। = - 1]`
अब (1 ) से , `A = B^(-1)_C^(-1) = [(2,-1),(-3,2)][(3,2),(5,3)]=[(1,1),(1,0)]` .