Question

Let A be a $2\times 2$ matrix with real entries. Let I be the $2\times 2$ identity matrix. Denote by tr(A), the sum of diagonal entries of $A$. Assume that $A^2=$ I.
Statement-l: If $A\ne I$ and $A\ne -I$, then $detA=-1$.
Statement-2: If $A\ne I$ and $A\ne -I$, then $t_1\left\{A\right)\ne 0$ .

• A. Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is false.
• C. Statement-1 is false, Statement-2 is true.
• D. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Answer 1

The correct option is B Statement-1 is true, Statement-2 is false.

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ where $a,b,c,d\in R$

$tr\left(A\right)=a+d$

Now, $A^2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]$
Given $A^2=I$

$\Rightarrow \left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow a^2+bc=1=bc+d^2$ and $ab+bd=0=ac+cd$

Since, $A\ne I$ and $A\ne -I$

$\Rightarrow a=-d$

So,$a=\sqrt{1-bc}$ and $d=-\sqrt{1-bc}$

Hence, $A=\left[\begin{array}{cc}\sqrt{1-bc}& b\\ c& -\sqrt{1-bc}\end{array}\right]$

$\Rightarrow |A|=\left|\begin{array}{cc}\sqrt{1-bc}& b\\ c& -\sqrt{1-bc}\end{array}\right|$

$\Rightarrow |A|=-1$

Statement 1 is true.

Here, $tr\left(A\right)=\sqrt{1-bc}-\sqrt{1-bc}=0$

So, statement 2 is false.