Question

If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` then find A⁻¹ by the adjoint method.

Answer 1

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

∴ |A| = `|(2,-1,1),(-1,2,-1),(1,-1,2)|` 

= 2(4 − 1) + 1(−2 + 1) + 1(1 − 2)

= 6 − 1 − 1

= 4 ≠ 0

∴ A⁻¹ exists.

We have to find cofactor matrix = [A_ij]_3×3, where A_ij = (−1)^i+j M_ij

Now, `A_11 = (−1)^(1 + 1) M_11`

= `|(2,-1),(-1,2)|`

= 4 − 1

= 3

`A_12 = (−1)^(1+2) M_12`

= `-|(-1,-1),(1,2)|`

= −(−2 + 1)

= 1

`A_13 = (−1)^(1+3) M_13`

= `|(-1,2),(1,-1)|`

= 1 − 2

= −1

`A_21 = (−1)^(2+1) M_21`

= `-|(-1,1),(-1,2)|`

= −(−2 + 1)

= 1

`A_22 = (−1)^(2+2) M_22`

= `|(2,1),(1,2)|`

= 4 − 1

= 3

`A_23 = (−1)^(2+3) M_23`

= `-|(2,-1),(1,-1)|`

= −(−2 + 1)

= 1

`A_31 = (−1)^(3+1) M_31`

= `|(-1,1),(2,-1)|`

= 1 − 2

= −1

`A_32 = (−1)^(3+2) M_32`

= `-|(2,1),(-1,-1)|`

= −(−2 + 1)

= 1

`A_33 = (−1)^(3+3) M_33`

= `|(2,-1),(-1,2)|`

= 4 − 1

= 3

∴ The cofactor matrix = `[(A_11,A_12,A_13),(A_21,A_22,A_23),(A_31,A_32,A_33)]`

= `[(3,1,-1),(1,3,1),(-1,1,3)]`

Adj A = `[(3,1,-1),(1,3,1),(-1,1,3)]`

∴ `A^-1 = 1/|A| ("Adj A")`

∴ `A^-1 = 1/4[(3,1,-1),(1,3,1),(-1,1,3)]`