19 votes 19 votes Let the characteristic equation of matrix $M$ be $\lambda ^{2} - \lambda - 1 = 0$. Then. $M^{-1}$ does not exist. $M^{-1}$ exists but cannot be determined from the data. $M^{-1} = M + I$ $M^{-1} = M - I$ $M^{-1}$ exists and can be determined from the data but the choices (c) and (d) are incorrect. Linear Algebra tifr2010 linear-algebra matrix + – makhdoom ghaya asked Oct 4, 2015 • edited Aug 17, 2020 by soujanyareddy13 makhdoom ghaya 12.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 18 votes 18 votes We can solve using Cayley- Hamilton Theorem $λ^2−λ−1=0$ $\implies M^2 - M -I = 0$ $\implies I= M^2 - M$ Now premultiplying by $M^{-1}$ $M^{-1}I=M^{-1} M^2 - M^{-1}M$ $M^{-1} = M-I$ Correct Answer: $D$ Umang Raman answered Oct 4, 2015 • edited Apr 25, 2019 by Naveen Kumar 3 Umang Raman comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Lakshman Bhaiya commented Nov 19, 2018 i edited by Lakshman Bhaiya Jan 29, 2019 reply Follow Share Yeah that's right Given $\lambda^{2}-\lambda-1=0$ Compare with the equation $x^{2}-(\lambda_{1}+\lambda_{2})x+\lambda_{1}\lambda_{2}=0$ $\lambda_{1}\lambda_{2}=Det(M)=-1$ Eigen values are $\lambda_{1}=\frac{1+\sqrt{5}}{2}$ and $\lambda_{2}=\frac{1-\sqrt{5}}{2}$ $(1)M^{-1}$ exist becasue $|M|\neq0$ $M^{-1}=\frac{adj(M)}{|M|}$ if $|M|=0$ entire expreesion made $\infty.$ 1 votes 1 votes Lakshman Bhaiya commented Nov 19, 2018 i edited by Lakshman Bhaiya Nov 30, 2019 reply Follow Share what about option $(b)?$ We can't find the $M^{-1} $, because data is insufficient. 2 votes 2 votes ankitgupta.1729 commented Sep 7, 2020 reply Follow Share saying pre-multiplying by $M^{-1}$ is technically incorrect because we don’t know matrix M is invertible. If $AB=I$ then $A^{-1}=B$ and $B^{-1} = A$ for squared matrices A and B of same size. Since, Cayley- Hamilton theorem is for square matrices, so M should be a square matrix and I is already a square matrix. Hence, M-I will also be a square matrix and both M and M-I will be of the same size. So, here, $I= M^2-M = M(M-I)$ which means $M^{-1} = M- I $ and $(M-I)^{-1} = M$ Another way to say, M is invertible as : $I = M(M-I)$ $ \det(I) = \det(M(M-I))$ $ \det(I) = \det(M) \det(M-I)$ $ 1 = \det(M) \det(M-I)$ So, $ \det(M) \neq 0$ which implies M is invertible. 1 votes 1 votes Please log in or register to add a comment.