If A = [ array rr 1 & 2 -1 & 4 array ] and A ⁻¹= I + A , R where I is the identity matrix of order 2 , then 4( + )=

Step-by-step solution

The inverse matrix A ⁻¹ is expressed as A ⁻¹ = I + A . One approach begins by computing A ⁻¹ directly for the given matrix. The determinant is ( A ) = (1)(4) - (2)(-1) = 6 . Constructing the adjugate yields [ array rr 4 & -2 1 & 1 array ] , so A ⁻¹ = 1 6 [ array rr 4 & -2 1 & 1 array ] = [ array rr 2 3 & - 1 3 1 6 & 1 6 array ] . Equating this to I + A = [ array rr + & 2 - & +4 array ] , the elements yield the system + = 2 3 2 = - 1 3 Solving gives = - 1 6 and thus = 2 3 + 1 6 = 5 6 . A more elegant method uses the