Question

The number of values of $k4$ for which the following system of equations has at least three solutions
$ 8x+16y\,\,\,+8z=25, $ $ x+y+z=k $ and $ 3x+y+3z={{k}^{2}}, $ is

• A. $ 0 $
• B. $ 1 $
• C. $ 2 $
• D. $ 3 $

Answer 1

Answer: (C)

Given system of equation is $ 8x+16y+8z=25 $ $ x+y+z=k $ and $ 3x+y+3z={{k}^{2}} $ It can be written as, $ AX=B $ Where, $ A=\left[ \begin{matrix} 8 & 16 & 8 \\ 1 & 1 & 1 \\ 3 & 1 & 3 \\ \end{matrix} \right],\,\,B=\left[ \begin{matrix} 25 \\ k \\ {{k}^{2}} \\ \end{matrix} \right],\,\,\,X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right] $ Cofactors of A are $ {{A}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 1 & 1 \\ 1 & 3 \\ \end{matrix} \right|={{(-1)}^{2}}(3-1)=2 $ $ {{A}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} 1 & 1 \\ 3 & 3 \\ \end{matrix} \right|={{(-1)}^{3}}(3-3)=0 $ $ {{A}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} 1 & 1 \\ 3 & 1 \\ \end{matrix} \right|={{(-1)}^{4}}(1-3)=-2 $ $ {{A}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 16 & 8 \\ 1 & 3 \\ \end{matrix} \right|={{(-1)}^{3}}(48-8)=-40 $ $ {{A}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 8 & 8 \\ 3 & 3 \\ \end{matrix} \right|={{(-1)}^{4}}(24-24)=0 $ $ {{A}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 8 & 16 \\ 3 & 1 \\ \end{matrix} \right|={{(-1)}^{5}}(8-48)=40 $ $ {{A}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 16 & 8 \\ 1 & 1 \\ \end{matrix} \right|={{(-1)}^{4}}(16-8)=8 $ $ {{A}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 8 & 8 \\ 1 & 1 \\ \end{matrix} \right|={{(-1)}^{5}}(8-8)=0 $ $ {{A}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 8 & 16 \\ 1 & 1 \\ \end{matrix} \right|={{(-1)}^{6}}(8-16)=-8 $
$ \therefore $ $ adj\,(A)={{\left[ \begin{matrix} {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\ \end{matrix} \right]}^{T}} $
$ \Rightarrow $ $ adj\,(A)={{\left[ \begin{matrix} 2 & 0 & -2 \\ -40 & 0 & 40 \\ 8 & 0 & -8 \\ \end{matrix} \right]}^{T}} $
$ \Rightarrow $ $ adj\,(A)=\left[ \begin{matrix} 2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8 \\ \end{matrix} \right] $
Now, $ (adj\,A)\,B=\left[ \begin{matrix} 2 & -40 & 8 \\ 0 & 0 & 0 \\ -2 & 40 & -8 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 25 \\ k \\ {{k}^{2}} \\ \end{matrix} \right] $ $ (adj\,\,A)\,\,B=\left[ \begin{matrix} 50-40k+8{{k}^{2}} \\ 0+0+0 \\ -50+40k-8{{k}^{2}} \\ \end{matrix} \right] $ For atleast three solutions, $ (adj\,A)\,\,B=0 $
$ \Rightarrow $ $ \left[ \begin{matrix} 50-40k+8{{k}^{2}} \\ 0 \\ -50+40k-8{{k}^{2}} \\ \end{matrix} \right]=0 $
$ \Rightarrow $ $ \left[ \begin{matrix} 50-40k+8{{k}^{2}} \\ 0 \\ -50+40k-8{{k}^{2}} \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} \right] $
$ \Rightarrow $ $ 8{{k}^{2}}-40k+50=0 $
$ \Rightarrow $ $ 2({{k}^{2}}-20k+25)=0 $
$ \Rightarrow $ $ k=\frac{-(-20)\pm \sqrt{{{(-20)}^{2}}-4\times 1\times 25}}{2\times 1} $
$ \Rightarrow $ $ k=\frac{20\pm \sqrt{400-100}}{2} $
$ \Rightarrow $ $ k=\frac{20\pm \sqrt{300}}{2} $
$ \Rightarrow $ $ k=\frac{20\pm \sqrt{100\times 3}}{2} $
$ \Rightarrow $ $ k=\frac{20\pm 10\sqrt{3}}{2} $
$ \Rightarrow $ $ k=10\pm 5\sqrt{3} $
$ \Rightarrow $ $ k=10+5\sqrt{3},\,\,10-5\sqrt{3} $ The number of values of $ k=2 $