Search any question & find its solution
Question:
Answered & Verified by Expert
Let $A$ be a $2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. $\operatorname{Tr}(A)$ denotes the sum of diagonal entries of $A$. Assume that $A^2=I$
Statement I If $A \neq I$ and $A \neq-1$, then $\operatorname{det} A=-1$
Statement II If $A \neq I$ and $A \neq-1$, then $\operatorname{Tr} A \neq 0$
Options:
Statement I If $A \neq I$ and $A \neq-1$, then $\operatorname{det} A=-1$
Statement II If $A \neq I$ and $A \neq-1$, then $\operatorname{Tr} A \neq 0$
Solution:
1868 Upvotes
Verified Answer
The correct answer is:
Statement I is true, statement II is false
Let $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$
Here, $A^2=I,|A|=-1$ and $A \neq \pm I$
$\operatorname{Tr}(A)=0$
$\therefore$ Statement I is true.
Statement II is false.
$A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$
Here, $A^2=I,|A|=-1$ and $A \neq \pm I$
$\operatorname{Tr}(A)=0$
$\therefore$ Statement I is true.
Statement II is false.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.