JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 10)
If $$A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$$ and $$\alpha+\beta=-2$$, then $$4 \alpha^{2}+\beta^{2}+\lambda^{2}$$ is equal to :
12
10
19
14
Explanation
$$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{cc}
1 & 5 \\
\lambda & 10
\end{array}\right] \\\\
& \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\
& \Rightarrow\left|\begin{array}{cc}
1-x & 5 \\
\lambda & 10-x
\end{array}\right|=0 \\\\
& \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\
& \Rightarrow 10-11 x+x^2-5 \lambda=0
\end{aligned}
$$
Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$
$$ \Rightarrow \alpha A^2+\beta A-I=0 $$
and $A^2-11 A+(10-5 \lambda) I=0$
On solving, we get
$$ \alpha=\frac{1}{5}, \beta=-\frac{11}{5} $$
$$ \begin{aligned} & \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned} $$
Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$
$$ \Rightarrow \alpha A^2+\beta A-I=0 $$
and $A^2-11 A+(10-5 \lambda) I=0$
On solving, we get
$$ \alpha=\frac{1}{5}, \beta=-\frac{11}{5} $$
$$ \begin{aligned} & \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned} $$
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