Question

The number of matrices $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, where a,b,c,d $\in \{-1,0,1,2,3,\dots \dots ,10\}$, such that $A=$_______ $A^{-1}$, is

Answer 1

Answer: 50

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Given $A=A^{-1}$
$\therefore A^2=A\cdot A^{-1}=I$
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\therefore a^2+bc=\mathrm{1....}$(1)
$ab+bd=\mathrm{0.....}$(2)
$ac+cd=\mathrm{0......}$(3)
$bc+d^2=\mathrm{1......}$(4)
$\Rightarrow (a+d)=0$ or $a-d=0$
Case - I
$a+d=0\Rightarrow (a,d)=(-1,1),(0,0),(1,-1)$
(a) $(a,d)=(-1,1)$
$\therefore$ from equation (1)
$1+bc=1\Rightarrow bc=0$
$b=0C=12$ possibilities
$c=0b=12$ possibilities
but $(0,0)$ is repeated
$\therefore 2\times 12=24$
$24-1($ repeated $)=23$ pairs
(b) $(a,d)=(1,-1)\Rightarrow$ bc $=0\to 23$ pairs
(c) $(a,d)=(0,0)\Rightarrow bc=1$
$\Rightarrow (b,c)=(1,1)\&(-1,-1),2$ pairs
Case - II
$a=d$
from (2) and (3)
$a\ne 0$ then $b=c=0$
$a^2=1$
$a=\pm 1=d$
$(a,d)=(1,1),(-1,-1)\to 2$ pairs
$\therefore$ Total $=23+23+2+2$
$=50$ pairs