A) \[0\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: C
Solution :
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 \[k=2\]
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