यदि (If) `A= [(3,1),(7,-5)]` x तथा y निकाले ताकि
`A^(2) + xI = yA "इससे " A^(-1)` निकाले ।

Explanation

`A[(3,1),(7,5)]rArr A^(2) = [(3,1),(7,5)][(3,1),(7,5)]= [(16,8),(56,32)]`
अब ` A^(2) + I = yA " "`…(I)
`rArr [(16,8),(56,32)] + x [(1,0),(0,1)]=y[(3,1),(7,5)]`
` rArr [(16+x,8),(56 , 32 + x)]= y [(3,1),(7,5)] = [(3y,y),(7y,5y)]`
` therefore 16 + x = 3y, 8 = y`
` 56 = 7y, 32 + x = 5y`
इन समीकरणों को हल करने पर हमे मिलता है ,
` y = 8, x= 8 `
क्योकि ` |A| = 15 - 7 = 8 ne 0 rArr A^(-1) ` का अस्तित्व है ।
(1) के दोनों पक्षों को `A^(-1)` से गुणा करने पर ,
` (A^(2) + xI ) A^(-1) = yAA^(-1) `
` rArr A + 8A^(-1) = 8 I " "[because x = y = 8 "तथा " AA^(-1) = I]`
`8A^(-1) = 8I - A = 8 [(1,0),(0,1)]-[(3,1),(7,5)]= [(5,-1),(-7,3)]`
` therefore A^(-1) = (1)/(8) [(5,-1),(-7,3)]` .