JEE Main 2022 — Mathematics Algebra

Given, $A=[\begin{matrix}a & b \\ c & d\end{matrix}]$A=[ac​bd​] and $A={A}^{-1}$A=A−1

So, ${A}^{2}=A\cdot {A}^{-1}=I$A2=A⋅A−1=I

$$\Rightarrow [\begin{matrix}a & b \\ c & d\end{matrix}][\begin{matrix}a & b \\ c & d\end{matrix}]=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$⇒[ac​bd​][ac​bd​]=[10​01​]

$$\Rightarrow [\begin{matrix}{a}^{2}+bc & ab+bd \\ ac+cd & bc+{d}^{2}\end{matrix}]=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$⇒[a2+bcac+cd​ab+bdbc+d2​]=[10​01​]

On comparing both side we get,

$$\therefore {a}^{2}+bc=1\cdots (1)$$∴a2+bc=1⋯(1)

$$ab+bd=0\cdots (2)$$ab+bd=0⋯(2)

$$ac+cd=0\cdots (3)$$ac+cd=0⋯(3)

$$bc+{d}^{2}=1\cdots (4)$$bc+d2=1⋯(4)

Now equation $(1)-(4)$(1)−(4) gives

$${a}^{2}-{d}^{2}=0$$a2−d2=0

$\Rightarrow (a+d)=0$⇒(a+d)=0 or $a-d=0$a−d=0

Case-I

$$a+d=0\Rightarrow (a,d)=(-1,1),(0,0),(1,-1)$$a+d=0⇒(a,d)=(−1,1),(0,0),(1,−1)

Assuming case $(a)$(a)$\Rightarrow (a,d)=(-1,1)$⇒(a,d)=(−1,1)

Now from equation $(1)$(1)

$$1+bc=1\Rightarrow bc=0$$1+bc=1⇒bc=0

When $b=0,c=12$b=0,c=12 possibilities

When $c=0,b=12$c=0,b=12 possibilities

But $(0,0)$(0,0) is repeated

$$\therefore 2\times 12=24$$∴2×12=24

So, total case will be $24-1$24−1 (repeated) $=23$=23 pairs.

case $(b)$(b) $\Rightarrow (a,d)=(1,-1)\Rightarrow bc=0\rightarrow 23$⇒(a,d)=(1,−1)⇒bc=0→23 pairs

case $(c)\Rightarrow (a,d)=(0,0)\Rightarrow bc=1$(c)⇒(a,d)=(0,0)⇒bc=1

$\Rightarrow (b,c)=(1,1)$⇒(b,c)=(1,1) and $(-1,-1)\rightarrow 2$(−1,−1)→2 pairs

Case-II

When $a=d$a=d

from $(2)$(2) and $(3)$(3)

$a\neq 0$a=0 then $b=c=0$b=c=0

$${a}^{2}=1$$a2=1

$$\Rightarrow a=\pm 1=d$$⇒a=±1=d

$(a,d)=(1,1),(-1,-1)\rightarrow 2$(a,d)=(1,1),(−1,−1)→2 pairs

$\therefore$∴ Total $=23+23+2+2$=23+23+2+2

$=50$=50 pairs.