A) \[0\]
B) \[1\]
C) \[3\]
D) \[-3\]
Correct Answer: B
Solution :
Given system of equation is \[(k+1)x+8y=4k\] and \[kx+(k+3)y=3k-1.\] \[\therefore \] For in fined many solution, \[\left| \begin{matrix} k+1 & 8 \\ k & k+3 \\ \end{matrix} \right|=0\] \[(k+1)\,(k+3)-8k=0\] \[\Rightarrow \] \[{{k}^{2}}+4k+3-8k=0\] \[\Rightarrow \] \[{{k}^{2}}-4k+3=0\] \[\Rightarrow \] \[k=\frac{4\pm \sqrt{16-12}}{2(1)}\] \[=\frac{4\pm 2}{2}=3,1\] Now, \[adj\,(A)=\left[ \begin{matrix} k+3 & -8 \\ -k & k+1 \\ \end{matrix} \right]\] Now, \[(adj\,A)\,B=\left[ \begin{matrix} k+3 & -8 \\ -k & k+1 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 4k \\ 3k-1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4{{k}^{2}}+12k-24k+8 \\ -4{{k}^{2}}+3{{k}^{2}}+2k-1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4{{k}^{2}}-12k+8 \\ -{{k}^{2}}+2k-1 \\ \end{matrix} \right]\] when \[k=1,\] \[(adj\,A)\,\,B=\left[ \begin{matrix} 4-12+8 \\ -1+2-1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\ \end{matrix} \right]\] satisfies when \[k=3,\] \[(adj\,\,A)\,\,B=\left[ \begin{matrix} 36-36+8 \\ -9+6-1 \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ -4 \\ \end{matrix} \right]\ne 0\] not satisfies Hence, \[k=1\] is the required solution.
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