Question

If the system of equations $ (k+1)x+8y=4k $ and $ kx+(k+3)y=3k-1 $ has infinitely many solutions, then $k$ is equal to

• A. $ 0 $
• B. $ 1 $
• C. $ 3 $
• D. $ -3 $

Answer 1

Answer: (B)

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.