यदि (If) ` A = [(3,2),(1,1)] ` a और b का मान निकाले ताकि `A^(2) + aA = bI= O " इससे " A^(-1) ` निकाले ।

Explanation

जहाँ `A = [(3,2),(1,1)]rArr ।A। = ।(3,2),(1,1)।= 3 xx 1 - 1 xx 2 = 1 ne 0 `
` rArr A^(-1)` का अस्तित्व है ।
साथ ही `A^(2) = A । A = [(3,2),(1,1)]।[(3,2),(1,1)]= [(9+2,6+2),(3 + 1, 2 + 1)] = [(11,8),(4,3)]`
अब `A^(2) + aA + bI = O `
`rArr [(11,8),(4,3)]+ a [(3,2),(1,1)] + b [(1,0),(0,1)]= [(0,0),(0,0)]`
` rArr [(11 + 3a + b,8 + 2a),(4 + a,3 + a + b)]=[(0,0),(0,0)]`
`therefore 11 + 3a + b = 0, 8 + 2a = 0 `
` 4 + a + , 3 + a + b = 0 `
` rArr a + - 4 , b = 1`
Second part : (I) से , ` A^(2) - 4A + I = 0`
` rArr I = 4A - A^(2) " " [ A^(-1) ` से गुणा करने पर ]
` rArr A^(-1) = 4I - A " " [ because AA^(-1)= I]`
`= 4 [(1,0),(0,1)]-[(3,2),(1,1)]= [(1,-2),(-1,3)]`.