Question

The number of matrices $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, where a,b,c,d $\in\{-1,0,1,2,3, \ldots \ldots, 10\}$, such that $A=$_______ $A ^{-1}$, is

Answer 1

$A= \begin{bmatrix} a & b \\c & d\end{bmatrix}$
Given $A=A^{-1}$
$ \therefore A ^2= A \cdot A ^{-1}= I $
$\begin{bmatrix} a & b \\c & d\end{bmatrix}\begin{bmatrix} a & b \\c & d\end{bmatrix}=\begin{bmatrix} 1 & 0 \\0 & 1\end{bmatrix} $
$ \Rightarrow \begin{bmatrix} a^2+b c & a b+b d \\a c+c d & b c+d^2\end{bmatrix}=\begin{bmatrix} 1 & 0 \\0 & 1\end{bmatrix} $
$ \therefore a^2+b c=1 ....$(1)
$ a b+b d=0 .....$(2)
$ ac + cd =0 ......$(3)
$ bc + d ^2=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)
$1+b c=1 \Rightarrow b c=0$
$b =0 C =12$ possibilities
$c =0 b =12$ possibilities
but $(0,0)$ is repeated
$\therefore 2 \times 12=24$
$24-1($ repeated $)=23$ pairs
(b) $( a , d )=(1,-1) \Rightarrow$ bc $=0 \rightarrow 23$ 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)
$a \neq 0$ then $b=c=0$
$a ^2=1$
$a=\pm 1=d$
$( a , d )=(1,1),(-1,-1) \rightarrow 2$ pairs
$\therefore$ Total $=23+23+2+2$
$=50$ pairs