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If \mathrm{P} is a 3 \times 3 real matrix such that \mathrm{P}^{\mathrm{T}}=\mathrm{aP}+(\mathrm{a}-1) \mathrm{I},where \mathrm{a}>1,then

Option: 1

|\operatorname{AdjP}|=\frac{1}{2}


Option: 2

|\operatorname{AdjP}|=1


Option: 3

\mathrm{P}  is a singular matrix


Option: 4

|\operatorname{AdjP}|>1


Answers (1)

best_answer

\left(\mathrm{P}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{aP}^{\mathrm{T}}(\mathrm{a}-1) \mathrm{I}

\mathrm{P}=\mathrm{a}(\mathrm{aP}+(\mathrm{a}-1) \mathrm{I})+(\mathrm{a}-1) \mathrm{I}
=\mathrm{a}^{2} \mathrm{P}+\left(\mathrm{a}^{2}-\mathrm{a}\right) \mathrm{I}+(\mathrm{a}-1) \mathrm{I}
=\mathrm{a}^{2} \mathrm{P}+\left(\mathrm{a}^{2}-\mathrm{a}+\mathrm{a}-1\right) \mathrm{I}
\mathrm{P}=\mathrm{a}^{2} \mathrm{P}+\left(\mathrm{a}^{2}-1\right) \mathrm{I} \quad \Rightarrow \quad P=\left(1-\mathrm{a}^{2}\right)=\left(\mathrm{a}^{2}-1\right) \mathrm{I}
                                                              \mathrm{P}=-\mathrm{I}

|\operatorname{AdjP}|=|\mathrm{P}|^{3-1}=(-1)^{2}
=1

Posted by

Pankaj

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