JEE Main 2024 — Mathematics Algebra

Given: $|A-\mathrm{xI}|=0$∣A−xI∣=0 have roots $-1$−1 and $3$3.

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

$$\Rightarrow A-\mathrm{xI}=[\begin{matrix}a & b \\ c & d\end{matrix}]-[\begin{matrix}x & 0 \\ 0 & x\end{matrix}]$$⇒A−xI=[ac​bd​]−[x0​0x​]

$$\Rightarrow A-\mathrm{xI}=[\begin{matrix}a-x & b \\ c & d-x\end{matrix}]$$⇒A−xI=[a−xc​bd−x​]

$$\Rightarrow |A-\mathrm{xI}|=(a-x)(d-x)-\mathrm{bc}$$⇒∣A−xI∣=(a−x)(d−x)−bc

$$\Rightarrow |A-\mathrm{xI}|=\mathrm{ad}-\mathrm{ax}-\mathrm{dx}+{x}^{2}-\mathrm{bc}$$⇒∣A−xI∣=ad−ax−dx+x2−bc

$$\Rightarrow |A-\mathrm{xI}|={x}^{2}-(a+d)x+(\mathrm{ad}-\mathrm{bc})$$⇒∣A−xI∣=x2−(a+d)x+(ad−bc)

Sum of roots, $a+d=2$a+d=2

Product of roots, $\mathrm{ad}-\mathrm{bc}=-3$ad−bc=−3

Now finding matrix ${A}^{2}$A2,

$$\Rightarrow {A}^{2}=[\begin{matrix}a & b \\ c & d\end{matrix}]\times [\begin{matrix}a & b \\ c & d\end{matrix}]$$⇒A2=[ac​bd​]×[ac​bd​]

$$\Rightarrow {A}^{2}=[\begin{matrix}{a}^{2}+\mathrm{bc} & \mathrm{ab}+\mathrm{bd} \\ \mathrm{ac}+\mathrm{cd} & \mathrm{bc}+{d}^{2}\end{matrix}]$$⇒A2=[a2+bcac+cd​ab+bdbc+d2​]

Sum of diagonal elements, $S={a}^{2}+\mathrm{bc}+\mathrm{bc}+{d}^{2}$S=a2+bc+bc+d2

$$\Rightarrow S={a}^{2}+2\mathrm{bc}+{d}^{2}$$⇒S=a2+2bc+d2

$$\Rightarrow S=(a+d{)}^{2}-2(\mathrm{ad}-\mathrm{bc})$$⇒S=(a+d)2−2(ad−bc)

$$\Rightarrow S=(2{)}^{2}-2(-3)$$⇒S=(2)2−2(−3)

$$\Rightarrow S=4+6$$⇒S=4+6

$$\Rightarrow S=10$$⇒S=10