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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