दिखाएँ कि आव्यूह `A=[(1,2,2),(2,1,1),(2,2,1)]`
समीकरण `A^(2) - 4A- 5I_(3) = O ` को संतुष्ट करता है और इससे ` A^(-1)`

Explanation

`A^(2) = [(1,2,2),(2,1,2),(2,2,1)][(1,2,2),(2,1,2),(2,2,1)]`
`=[(1.1+2.2+2.2,1.2+ 2.1+ 2.2,1.2+ 2.2+ 2.1),(2.1+1.2+2.2,2.2+1.1 + 2.2,2.2 + 1.2+2.1),(2.1 + 2.2 + 1.2,2.2 + 2.1 + 1.2,2.2 + 2.2 + 1.1)]=[(9,8,8),(8,9,8),(8,8,9)]`
अब `A^(2) - 4A - 5I_(3) =[(9,8,8),(8,9,8),(8,8,9)]- [(4,8,8),(8,4,8),(8,8,4)]-[(5,0,0),(0,5,5),(0,0,5)]`
` [(0,0,0),(0,0,0),(0,0,0)]= O`
तथा `|A|[(1,2,2),(2,1,2),(2,2,1)]= 1 (1-4)-2(2-4)+ 2(4-2)`
` = - 3 + 4 + 4 = 5 ne 0 `
अतः `A^(-1)` अस्तित्व है ।
अब `A^(-1) - 4A - 5I_(3) = O rArr A^(2) - 4A = 5I_(3)`
` rArr A^(-1) , A^(2) - 4A^(-1) . A= 5A^(-1) , I_(3)`
[दोनों तरफ `A^(-1)` को बाएँ तरफ रखकर गुणा करने पर ]
` rArr (A^(-1) . A ) A- 4I_(3) = 5A^(-1)`
` rArr A- 4I_(3) = 5A^(-1) rArr A - 1 = (1)/(5) (A - 4I_(3))`
`= (1)/(5) .{[(1,2,2),(2,1,2),(2,2,1)]-[(4,0,0),(0,4,0),(0,0,4)]} = (1)/(5) . [(-3,2,2),(2,-3,2),(2,2,-3)]` .