A) -3
B) -1
C) 4
D) 2
Correct Answer: A
Solution :
Given \[[\begin{matrix} p & q & r \\ \end{matrix}]\left[ \begin{matrix} 3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2 \\ \end{matrix} \right]=[\begin{matrix} 3 & 0 & 1 \\ \end{matrix}]\] \[\Rightarrow \] \[[\begin{matrix} 3p+3q+2r & 4p+2q & p+3q+2r \\ \end{matrix}]\] \[=[\begin{matrix} 3 & 0 & 1 \\ \end{matrix}]\] \[\Rightarrow \] \[3p+3q+2r=3\] ?(i) \[4p+2q=0\Rightarrow q=-2p\] ?(ii) \[p+3q+2r=1\] On solving (i),(ii) and (iii), we get \[p=1,q=-2,r=3\] \[\therefore \] \[2p+q-r=2(1)+(-2)-(3)=-3.\]
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