Let $\mathrm{A}$ be a $2 \times 2$ matrix with real entries. Let I be the $2 \times 2$ identity matrix.…
Let $\mathrm{A}$ be a $2 \times 2$ matrix with real entries. Let I be the $2 \times 2$ identity matrix. Denote by $\operatorname{tr}(\mathrm{A})$, the sum of diagonal entries of $A$. Assume that $A^2=1$.
Statement -1: If $A \neq 1$ and $A \neq-1$, then $\operatorname{det} A=-1$.
Statement $-2$ : If $A \neq 1$ and $A \neq-1$, then $\operatorname{tr}(A) \neq 0$.
Statement $-1$ is false, Statement $-2$ is true
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is false.
Solution
Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ so that $A^2=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & b c+d^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow a^2+b c=1=b c+d^2$ and $(a+d) c=0=(a+d) b$.
Since $A \neq I, A \neq 1, a=-d$ and hence $\operatorname{det} A=\left|\begin{array}{cc}\sqrt{1-b c} & b \\ c & -\sqrt{1-b c}\end{array}\right|=-1+b c-b c=-1$
Statement 1 is true.
But tr. $A=0$ and hence statement 2 is false.