Question

माना लीजिए A = `[(1,-2,1),(-2,3,1),(1,1,5)]` हो तो सत्यापित कीजिए कि

1. [adj A]^-1 = adj (A^-1)
2. (A^-1)^-1 = A

Answer 1

A = `[(1,-2,1),(-2,3,1),(1,1,5)]` 

∴ |A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)

= 14 - 22 - 5

= -13

अब, `A_11 = 14, A_12 = 11, A_13 = -5`

`A_21 = 11, A_22 = 4, A_23 = -3`

`A_31 = -5, A_32 = -3, A_33 = -1`

∴ adj A = `[(14,11,-5),(11,4,-3),(-5,-3,-1)]`

∴ `A^-1 = 1/|A|`(adj A)

= `-1/13[(14,11,-5),(11,4,-3),(-5,-3,-1)]`

` = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)]`

(i) |adj A| = 14(-4 - 9) - 11(-11 - 15)-5(-33 + 20)

= 14(-13) - 11(-26) - 5(-13)

= -183 + 286 + 65 = 169

हमारे पास है,

adj(adj A) = `[(-13,26,-13),(26,-39,-13),(-13,-13,-65)]`

∴ `[adj A]^-1 = 1/|adj A|(adj(adjA))`

= `1/169[(-13,26,-13),(26,-39,-13),(-13,-13,-65)]`

= `1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`

अब, `A^-1 = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)] = [(-14/13, -11/13, 5/13),(-11/13,-4/13,3/13),(5/13,3/13,1/13)]`

∴ `adj(A^-1) = [(-4/169 - 9/169, -(-11/169 - 15/169),-33/169 + 20/169),(-(-11/169 - 15/169),(-14/169 - 25/169), -(42/169 + 55/169)),(-33/169 + 20/169, -(42/169 + 55/169), 56/169 - 121/169)]`

= `1/169[(-13,26,-13),(26,-39,-13),(-13,-13,-65)] = 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`

इसलिए, `[adjA]^-1 = adj(A^-1)`. 

(ii) हमने दिखाया है कि:

`A^-1 = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)]`

और, `adjA^-1 = 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`

अब,

`|A^-1| = (1/13)^3[-14 xx (-13) + 11 xx (-26) + 5 xx (-13)] `

`= (1/13)^3 xx (-169) `

`= -1/13`

∴ `(A^-1)^-1 = (adjA^-1)/|A^-1| `

=` 1/((-1/13)) xx 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)] `

`= [(1,-2,1),(-2,3,1),(1,1,5)]` = A

अत:, (A^-1)^-1 = A