A) 0
B) 4
C) 5
D) 11
Correct Answer: D
Solution :
\[P=adj.Q\] \[\Rightarrow |P|=|adj.Q|=Q{{|}^{2}}\] \[\therefore \]\[|P|\,\,={{(4)}^{2}}\Rightarrow |P|\,\,=16\] \[\Rightarrow \]\[\left| \begin{matrix} 1 & c & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \\ \end{matrix} \right|=16\] \[\Rightarrow 1(0)-c(-2)+3(-2)=16\Rightarrow 2c=22\Rightarrow c=11\]You need to login to perform this action.
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