Matrices A and B Satisfy `AB = B^(-1)`, where B `=[{:(2,-2),(-1,0):}]`, find the value of `lambda` for which `lambdaA - 2B^(-1) + 1=O`, Without finding `B^(-1)`.

Explanation

` because " " AB =B^(-1) or AB^(2) = I`
Now, `lambdaA-2B^(-1)+I=0`
`rArr" " lambdaAB-2B^(-1)B+IB=O`
`rArr" "lambdaAb-2I + B =O`
` rArr" "lambdaAB^(2) -2IB + B^(2) = O`
`rArr" "lambdaAB^(2) -2b + B^(2) =O`
`rArr " " lambdaI - 2B + B^(2) =O " " [because AB^(2) =I]`
`rArr lambda[{:(2,-2),(0,1):}]-2[{:(2,-2),(-1,0):}]+[{:(2,-2),(-1,0):}][{:(2,-2),(-1,0):}]=[{:(0,0),(0,0):}]`
`rArr " "[{:(lambda,0),(0,lambda):}]-[{:(4,-4),(-2,0):}]+[{:(6,-4),(-2,2):}]=[{:(0,0),(0,0):}]`
`rArr2 [{:(lambda+2,0),(0,lambda+2):}]=[{:(0,0),(0,0):}] rArr " " lamda+2=0`
`therefore " " lamda=-2`