JEE Main 2023 — Mathematics Algebra

Given,

$A=[\begin{matrix}1 & 5 \\ \lambda & 10\end{matrix}],{A}^{-1}=\alpha A+\beta I$A=[1λ​510​],A−1=αA+βI and $\alpha +\beta =-2$α+β=−2,

Now solving by using characteristic equation, we get

$$|A-k|=0\Rightarrow |\begin{matrix}1-k & 5 \\ \lambda & 10-k\end{matrix}|=0$$∣A−k∣=0⇒∣1−kλ​510−k​∣=0

$$\Rightarrow {k}^{2}-11k+10-5\lambda =0$$⇒k2−11k+10−5λ=0

$\Rightarrow {A}^{2}-11A+(10-5\lambda )I=0$⇒A2−11A+(10−5λ)I=0 {by putting $k=A$k=A}

$$\Rightarrow {A}^{-1}=\frac{1}{10-5\lambda }(-A+11I)$$⇒A−1=10−5λ1​(−A+11I)

Now on comparing with ${A}^{-1}=\alpha A+\beta I$A−1=αA+βI we get,

$$\Rightarrow \alpha =-\frac{1}{10-5\lambda },\beta =\frac{11}{10-5\lambda }$$⇒α=−10−5λ1​,β=10−5λ11​

And given,

$$\alpha +\beta =-2\Rightarrow \lambda =3,\alpha =\frac{1}{5},\beta =\frac{-11}{5}$$α+β=−2⇒λ=3,α=51​,β=5−11​

Hence, the value of $4{\alpha }^{2}+{\beta }^{2}+{\lambda }^{2}=14$4α2+β2+λ2=14