JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 27)
Let $A=I_2-2 M M^T$, where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^T M=I_1$ holds. If $\lambda$ is a real number such that the relation $A X=\lambda X$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to __________.
Answer
2
Explanation
$\begin{aligned} & A=I_2-2 M^T \\\\ & A^2=\left(I_2-2 M M^T\right)\left(I_2-2 M^T\right) \\\\ & =I_2-2 M^T-2 M M^T+4 M^T M^T \\\\ & =I_2-4 M M^T+4 M M^T \\\\ & =I_2\end{aligned}$
$\begin{aligned} & \mathrm{AX}=\lambda \mathrm{X} \\\\ & \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX} \\\\ & \mathrm{X}=\lambda(\lambda \mathrm{X}) \\\\ & \mathrm{X}=\lambda^2 \mathrm{X} \\\\ & \mathrm{X}\left(\lambda^2-1\right)=0 \\\\ & \lambda^2=1 \\\\ & \lambda= \pm 1\end{aligned}$
Sum of square of all possible values $=2$
$\begin{aligned} & \mathrm{AX}=\lambda \mathrm{X} \\\\ & \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX} \\\\ & \mathrm{X}=\lambda(\lambda \mathrm{X}) \\\\ & \mathrm{X}=\lambda^2 \mathrm{X} \\\\ & \mathrm{X}\left(\lambda^2-1\right)=0 \\\\ & \lambda^2=1 \\\\ & \lambda= \pm 1\end{aligned}$
Sum of square of all possible values $=2$
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