1 votes 1 votes Let $A$ be an $n\times n$ matrix such that $\mid\: A^{2}\mid=1.\:\: \mid A\:\mid$ stands for determinant of matrix $A.$ Then $\mid\:(A)\mid=1$ $\mid\:(A)\mid=0\:\text{or}\:1$ $\mid\:(A)\mid=-1,0\:\text{or}\:1$ $\mid\:(A)\mid=-1\:\text{or}\:1$ Linear Algebra isi2016-dcg linear-algebra matrix determinant + – gatecse asked Sep 18, 2019 recategorized Nov 18, 2019 by Lakshman Bhaiya gatecse 373 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $\mid A^{2} \mid = 1$ $\implies \mid A \mid ^{2}= 1\rightarrow (1)$ Let $\mid A \mid = x,$ put the value in equation$(1),$ then we get $x^{2} = 1\implies x = \pm 1$ Therefore $,\mid A \mid = \pm 1\implies \mid A \mid = -1\:\: \text{or}\:\: 1$ So, the correct answer is $(D)$. Reference: https://math.stackexchange.com/questions/900897/how-matrix-a-is-called-a2-i Lakshman Bhaiya answered Nov 18, 2019 edited Nov 18, 2019 by Lakshman Bhaiya Lakshman Bhaiya comment Share Follow See all 2 Comments See all 2 2 Comments reply Verma Ashish commented Nov 18, 2019 reply Follow Share It will be option D..right? 1 votes 1 votes Lakshman Bhaiya commented Nov 18, 2019 reply Follow Share Yes 0 votes 0 votes Please log in or register to add a comment.