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KVPY Sample Paper KVPY Stream-SX Model Paper-17

  • question_answer
    Let \[P=\left[ \begin{matrix}    3 & -1 & -2  \\    2 & 0 & \alpha   \\    3 & -5 & 0  \\ \end{matrix} \right]\] , where \[\alpha \in \mathbb{R}.\]suppose \[Q=[{{q}_{ij}}]\] is a matrix such that PQ=kI, where \[\operatorname{k}\in \mathbb{R}.\],\[\operatorname{k}\ne 0\]and I is the identity matrix of order 3. If \[{{\operatorname{q}}_{23}}=-\frac{k}{8}\] and \[\det (Q)=\frac{{{k}^{2}}}{2},\]then which of the following is incorrect

    A) \[a=0,k=8\]

    B) \[4a-k+8=0\]

    C) \[\det (P\,adj(Q))={{2}^{9}}\]

    D) \[\det (Q\,\,adj(P))={{2}^{9}}\]

    Correct Answer: A

    Solution :

    [a] \[PQ=kI\Rightarrow \frac{P.Q}{k}=I\Rightarrow {{P}^{-1}}=\frac{Q}{K}\]
    Also \[\left| P \right|=12\alpha +20\]
    Comparing the third elements of 2nd row on both sides, we get \[-\left( \frac{3\alpha +4}{12\alpha +20} \right)=\frac{1}{k}\times \frac{-k}{8}\]
    \[\Rightarrow \]\[24\alpha +32=12\alpha +20\Rightarrow \alpha =-1\]
    \[\therefore \]\[\left| P \right|=8\]
    Also \[PQ=kI\Rightarrow \left| P \right|\left| Q \right|={{k}^{3}}\Rightarrow 8\times \frac{{{k}^{2}}}{2}={{k}^{3}}\] \[\Rightarrow \]\[k=4\Rightarrow \left| Q \right|=\frac{{{k}^{2}}}{2}=8\]
    [b] \[4\alpha -k+8=4\times (-1)-4+8=0\]
    [c] Now \[\det (P\,adj\,\,Q)=\left| P \right|adjQ|\]\[=\left| P \right|{{\left| Q \right|}^{2}}=8\times {{8}^{2}}={{2}^{9}}\]
    [d] \[\left| QadjP \right|=\left| Q \right|{{\left| P \right|}^{2}}={{2}^{9}}\]


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