Mathematics - Matrix Question with Solution | TestHub

$\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$ $$\begin{array}{rl}& \left[\begin{array}{ll}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ & b(a+d)=0,b=0\text{ or }a=-d\\ & c(a+d)=0,c=0\text{ or }a=-d\\ & a^2+bc=1,bc+d^2=1\end{array}$$ ' $a$ ' and ' $d$ ' are diagonal elements $a+d=0$ statement- 1 is correct. Now, $\det (A)=ad-bc$ Now, from (3) $a^2+bc=1$ and $d^2+bc=1$ So, $a^2-d^2=0$ Adding $a^2+d^2+2bc=2$ $$=(a+d{)}^2-2ad+2bc=2$$ or $0-2(ad-bc)=2$ So, $ad-bc=1\Rightarrow \det (A)=-1$ So, statement $-2$ is also true. But statement $-2$ is not the correct explanation of statement-I