For the matrix `A=[[3, 2],[ 1, 1]]` , find the numbers `a` and `b` such that `A^2+a A+b I=O` . Hence, find `A^(-1)` .

Explanation

we have, `A=[(3,1),(2,0)]` then `|A|=|(3,1),(2,1)|=3-2=1!=0`
`therefore" " A^(-1)`. Exists.
Now `A^(2)=A.A=[(3,1),(2,1)][(3,1),(2,1)]=[(11,4),(8,3)]`
`rArr" " [(11,4),(8,3)]+a[(3,1),(2,1)]+b[(1,0),(0,1)]=[(0,0),(0,0)]`
`rArr" " [(11+3a+b,4+a),(8+2a,3+a+b)]+[(0,0),(0,0)]`
Equating the corresponding elements, we get
`11a+3a+b=0`
`4+a=0`
` 3+a+b=0`
from Eqs. (ii) and (iv), we get `a=-4` and `b=1`
`therefore" " |a|+|b|=|-4|+|1|=4+1=5`
As`" " A^(2)+aA+bI=O`
` rArr " " A^(2)-4A+I+O rArr I=4A-A^(2)`
`rArr " " IA^(-1) = 4(A A^(-1))-A(A A^(-1))`
`=4I-AI=4I-A`
`= 4[(1,0),(0,1)]-[(3,1),(2,1)]=[(4,0),(0,4)]-[(3,1),(2,1)]`
`therefore " " A(-1)=[(1,-1),(-2,3)]`