The question is from JEE Advanced(2017), where it asks to identify matrices which are the square of a matrix with real entries:
I first found out the determinant of all the matrices. Options (A & B) both have negative determinant value and hence can't be expressed as square of a matrix with real entries. For explanation let any of the two be called as $A$ . Given they are square of another matrix(let $B$) so $B^2=A$ . Taking the determinant on both sides, I get $|B|^2=|A|$ and since $|A|$ is negative, I get $|B|^2<0$. So $B$ can't have all real entries. Option C is $I$ whose square is $I$ or vice versa. I am having trouble with option D . Since it's determinant is also positive and I can't find a simple matrix which when squared gives that option. I have talked to my teacher . He said there is a method to find square root of a matrix but that's far beyond our level. I am a High school student studying in grade 12. So please, if possible, give a simplified hint/answer.
Thanks in advance!
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2 Answers
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The determinant of a $3\times 3$ matrix being positive is a necessary and sufficient condition for it to have atleast one real root.
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$\begingroup$ I don't think so. Can you prove it or provide any reference $\endgroup$ Commented Feb 10, 2020 at 2:04
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Regarding only the $yz$ plane, $D$ rotates by 180°. What if you rotate by 90° twice?
answered Feb 9, 2020 at 15:30
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$\begingroup$ von I don't understand what you're trying to say . They represent planes if combined with x,y,z. But it's only numerical matrix $\endgroup$ Commented Feb 10, 2020 at 2:05