JEE Main 2018 — Mathematics Algebra

Let $A=[\begin{matrix}a & b \\ c & d\end{matrix}]$A=[ac​bd​]

Now, $A.[\begin{matrix}1 & 2 \\ 0 & 3\end{matrix}]=[\begin{matrix}a & b \\ c & d\end{matrix}][\begin{matrix}1 & 2 \\ 0 & 3\end{matrix}]=[\begin{matrix}a & 2a+3b \\ c & 2c+3d\end{matrix}]$A.[10​23​]=[ac​bd​][10​23​]=[ac​2a+3b2c+3d​] is a scalar.

$$\therefore c =0, 2a+3b=0, a=2c+3d\Rightarrow a=3d$$∴c=0,2a+3b=0,a=2c+3d⇒a=3d

Given, $|3A|=108$∣3A∣=108

$$\Rightarrow |A|=\frac{108}{{3}^{2}}=12\Rightarrow ad-bc=ad=12$$⇒∣A∣=32108​=12⇒ad−bc=ad=12

$$\Rightarrow (3d)d=12\Rightarrow {d}^{2}=4[\because a=3d]$$⇒(3d)d=12⇒d2=4[∵a=3d]

$\therefore {a}^{2}={(3d)}^{2}=9{d}^{2}==9\times 4=36$∴a2=(3d)2=9d2==9×4=36 and

$$\because 2a+3b=0\Rightarrow b=\frac{-2a}{3}=\frac{-2\times 6}{3}\Rightarrow b=-4$$∵2a+3b=0⇒b=3−2a​=3−2×6​⇒b=−4

Now, ${A}^{2}=[\begin{matrix}a & b \\ 0 & d\end{matrix}][\begin{matrix}a & b \\ 0 & d\end{matrix}]=[\begin{matrix}{a}^{2} & ab+bd \\ 0 & {d}^{2}\end{matrix}]=[\begin{matrix}36 & -32 \\ 0 & 4\end{matrix}]$A2=[a0​bd​][a0​bd​]=[a20​ab+bdd2​]=[360​−324​]