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JEE MAIN 2024
1-2-2024 S2
Question
Let $A=I_2-2 M^{\mathrm{T}}$, where M is real matrix of order $2 \times 1$ such that the relation $\mathrm{M}^{\mathrm{T}} \mathrm{M}=\mathrm{I}_1$ 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 :
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Solution
$$ \begin{aligned} & \mathrm{A}=\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}} \\ & \mathrm{~A}^2=\left(\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}\right)\left(\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}\right) \\ & =\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}-2 \mathrm{MM}^{\mathrm{T}}+4 \mathrm{MM}^{\mathrm{T}} \mathrm{MM}^{\mathrm{T}} \\ & =\mathrm{I}_2-4 \mathrm{MM}^{\mathrm{T}}+4 \mathrm{MM}^{\mathrm{T}} \\ & =\mathrm{I}_2 \\ & \mathrm{AX}=\lambda \mathrm{X} \\ & \mathrm{~A}^2 \mathrm{X}=\lambda \mathrm{AX} \\ & \mathrm{X}=\lambda(\lambda \mathrm{X}) \\ & \mathrm{X}=\lambda^2 \mathrm{X} \\ & \mathrm{X}\left(\lambda^2-1\right)=0 \\ & \lambda^2=1 \\ & \lambda= \pm 1 \end{aligned} $$ Sum of square of all possible values $=2$
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