21 votes 21 votes If matrix $X = \begin{bmatrix} a & 1 \\ -a^2+a-1 & 1-a \end{bmatrix}$ and $X^2 - X + I = O$ ($I$ is the identity matrix and $O$ is the zero matrix), then the inverse of $X$ is $\begin{bmatrix} 1-a &-1 \\ a^2& a \end{bmatrix}$ $\begin{bmatrix} 1-a &-1 \\ a^2-a+1& a \end{bmatrix}$ $\begin{bmatrix} -a &1 \\ -a^2+a-1& 1-a \end{bmatrix}$ $\begin{bmatrix} a^2-a+1 &a \\ 1& 1-a \end{bmatrix}$ Linear Algebra gateit-2004 linear-algebra matrix normal + – Ishrat Jahan asked Nov 2, 2014 Ishrat Jahan 4.4k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply HeadShot commented Oct 16, 2018 reply Follow Share Putting $a=0$ will make it more simpler to solve. 1 votes 1 votes N Prasanna kumar commented Jan 18, 2019 reply Follow Share For, who are interested to find the value of a ( i know it is not required to answer this question ) they given a eqn, $λ^2 - λ +1 = 0 $ ===> you can find λ values then We know that determinant = Product of eigen values ==> Find value of a 5 votes 5 votes AkashChandraGupta commented Aug 9, 2020 reply Follow Share (B) 0 votes 0 votes Please log in or register to add a comment.
Best answer 38 votes 38 votes Given, $X^2 - X + I = O$ $\quad \implies X^2 = X - I$ $\quad \implies (X^{-1}) (X^2 ) = (X^{-1})(X - I )$ (Multiplying $X^{-1}$ on both sides) $\quad \implies X = I - X^{-1} $ $\quad \implies X^{-1} = I - X$ Option (B) Himanshu1 answered Nov 16, 2015 • edited May 10, 2019 by Arjun Himanshu1 comment Share Follow See all 4 Comments See all 4 4 Comments reply Lakshman Bhaiya commented Nov 29, 2019 reply Follow Share @Satbir $X^{-1} = I - X$ When i subtract the matrix $'X'$ from the identity matrix, then i got the inverse $X^{-1} = \begin{bmatrix} 1-a &1 \\ -a^{2} + a -1 & a \end{bmatrix}$ which is not equal to the option $(B).$ Please correct me if i'm wrong? 0 votes 0 votes Satbir commented Nov 30, 2019 reply Follow Share $\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} - \begin{bmatrix} a & 1\\ -a^2+a-1&1-a \end{bmatrix} = \begin{bmatrix} 1-a &-1 \\ a^2-a+1& a \end{bmatrix}$ 6 votes 6 votes Lakshman Bhaiya commented Nov 30, 2019 reply Follow Share Ohh sorry, was in the night I just subtract the diagonal elements. Now I got it, it was just a silly mistake. Thank you so mcuh@Satbir 0 votes 0 votes harypotter0 commented Jul 20, 2020 reply Follow Share Lakshman sir I too faced the same issue thanks Satbir sir. 0 votes 0 votes Please log in or register to add a comment.
24 votes 24 votes It's a very simple question, We need to calculate the inverse of a 2x2 matrix, Inverse of a matrix $A = A^{-1} = \frac{Adjoint(A)}{|A|}$ Adjoint(A) = [cofactors of A]T, but for 2x2 matrix we have direct formula: A = a b c d is a 2x2 matrix then Adjoint of A = d -b -c a |A| = ad-bc So answer is (B)! Manu Thakur answered Nov 21, 2014 • edited Nov 21, 2017 by Manu Thakur Manu Thakur comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes You may not evaluate the given equation $X^{2} - X + I = O$. Rather start evaluating from the choices given. $X.X^{-1} = I$ and the first element should be 1. Option A. $\Rightarrow a\cdot (1-a)+1\cdot a^{2} = a$ Option B. $\Rightarrow a\cdot (1-a)+1\cdot (a^{2}-a+1) = 1$ Option C. $\Rightarrow a\cdot (-a)+1\cdot (-a^{2}+a-1) = -2a^{2}+a-1$ Option D. $\Rightarrow a\cdot (a^{2}-a+1)+1\cdot (1) = a^{3}-a^{2}+a+1$ You may verify the full multiplication for option B to see whether it indeed yields the identity matrix. dsomnath answered Jan 27, 2019 dsomnath comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Putting a=0 both in X and in the options (A),(B),(C) and (D) we see that X.X^-1= I hold goods for option (B). Hence option B is correct. DIBAKAR MAJEE answered Apr 21, 2020 DIBAKAR MAJEE comment Share Follow See 1 comment See all 1 1 comment reply Random_aspirant commented Jan 16 reply Follow Share I also did a = 0, then calculated determinant of the resultant matrix. Determinant of the matrix was 1 so I just took it as inverse of given matrix is equal to adjoint of the matrix. So option B. Was this correct method or did I simply got lucky? Thanks I am new to whole gate stuff. 0 votes 0 votes Please log in or register to add a comment.