Verified Solution • 2025–26 Session moderate

QJEEM23Ma69f0: JEE Main 2023 Mathematics - Class 12 JEEM23Ma69f0 Solution: If and , then is equal to :

If A=[1amp;5λamp;10],A−1=αA+βIA=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I} and α+β=−2\alpha+\beta=-2 , then 4α2+β2+λ24 \alpha^{2}+\beta^{2}+\lambda^{2} is equal to :

Multiple Choice Options

  • A 12
  • B 10
  • C 19
  • D 14

Step-by-Step Verified Solution

amp;A=[1amp;5λamp;10]amp;⇒∣A−xI∣=0amp;⇒∣1−xamp;5λamp;10−x∣=0amp;⇒(1−x)(10−x)−5λ=0amp;⇒10−11x+x2−5λ=0\begin{aligned}& \mathrm{A}=\left[\begin{array}{cc}1 & 5 \\\lambda & 10\end{array}\right] \\\\& \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\& \Rightarrow\left|\begin{array}{cc}1-x & 5 \\\lambda & 10-x\end{array}\right|=0 \\\\& \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\& \Rightarrow 10-11 x+x^2-5 \lambda=0\end{aligned}

Also, ⇒A−1=αA+βI\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}

⇒αA2+βA−I=0\Rightarrow \alpha A^2+\beta A-I=0

and A2−11A+(10−5λ)I=0A^2-11 A+(10-5 \lambda) I=0

On solving, we get

α=15,β=−115\alpha=\frac{1}{5}, \beta=-\frac{11}{5}

amp; So, 5λ−10=5⇒λ=3amp;∴4α2+β2+λ2amp;=4(125)+(12125)+9amp;=12525+9=14\begin{aligned}& \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\& \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\& =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\& =\frac{125}{25}+9=14\end{aligned}

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