यदि `(If)A = [(6,5),(7,6)]`, दिखाएँ कि ` A^(2) - 12 A + I = O` .
इससे `A^(-1)` निकाले ।

Explanation

यहाँ `A = [(6,5),(7,6)] therefore A^(2) [(6,5),(7,6)][(6,5),(7,6)]= [(71,60),(84,71)]`
अब `A^(2) - 1 3 A + I` ltबरgt`=[(71,60),(84,71)]- 12 [(6,5),(7,6)] + [(1,0),(0,1)]`
`[(71,60),(84,71)]-[(72,60),(84,72)]+ [(1,0),(0,1)]`
`=[(71-72+1,60-60+0),(84- 84 + 0 ,71-72+1)]= [(0,0),(0,0)] = O = RHS `
Second part :
`|A| = |(6,5),(7,6)| = 6xx6- 7 xx5 = 1 ne 0 rArrA^(-1)` का अस्तित्व है ।
अब `A^(2) - 12 A + I = 0 `
` rArr I = 12 A - A^(2)`
` rArr IA^(-1) = (12 A - A^(2)) A^(-1) " "[ A^(-1) ` से दोनों तरफ गुणा करने पर ]
` 12 AA^(-1) - A (AA^(-1))`
`12I - AI`
`12I - A `
`= 12 [(1,0),(0,1)]-[(6,5),(7,6)]=[(12,0),(0,12)]- [(6,5),(7,6)]= [(6,-5),(-7,6)]`