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, $A=\left[\begin{array}{ccc}8& 16& 8\\ 1& 1& 1\\ 3& 1& 3\end{array}\right],B=\left[\begin{array}{c}25\\ k\\ k^2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ Cofactors of A are $A_{11}={(-1)}^{1+1}\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|={(-1)}^2(3-1)=2$ $A_{12}={(-1)}^{1+2}\left|\begin{array}{cc}1& 1\\ 3& 3\end{array}\right|={(-1)}^3(3-3)=0$ $A_{13}={(-1)}^{1+3}\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|={(-1)}^4(1-3)=-2$ $A_{21}={(-1)}^{2+1}\left|\begin{array}{cc}16& 8\\ 1& 3\end{array}\right|={(-1)}^3(48-8)=-40$ $A_{22}={(-1)}^{2+2}\left|\begin{array}{cc}8& 8\\ 3& 3\end{array}\right|={(-1)}^4(24-24)=0$ $A_{23}={(-1)}^{2+3}\left|\begin{array}{cc}8& 16\\ 3& 1\end{array}\right|={(-1)}^5(8-48)=40$ $A_{31}={(-1)}^{3+1}\left|\begin{array}{cc}16& 8\\ 1& 1\end{array}\right|={(-1)}^4(16-8)=8$ $A_{32}={(-1)}^{3+2}\left|\begin{array}{cc}8& 8\\ 1& 1\end{array}\right|={(-1)}^5(8-8)=0$ $A_{33}={(-1)}^{3+3}\left|\begin{array}{cc}8& 16\\ 1& 1\end{array}\right|={(-1)}^6(8-16)=-8$
$\therefore$ $adj(A)={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^T$
$\Rightarrow$ $adj(A)={\left[\begin{array}{ccc}2& 0& -2\\ -40& 0& 40\\ 8& 0& -8\end{array}\right]}^T$
$\Rightarrow$ $adj(A)=\left[\begin{array}{ccc}2& -40& 8\\ 0& 0& 0\\ -2& 40& -8\end{array}\right]$
Now, $(adjA)B=\left[\begin{array}{ccc}2& -40& 8\\ 0& 0& 0\\ -2& 40& -8\end{array}\right]\left[\begin{array}{c}25\\ k\\ k^2\end{array}\right]$ $(adjA)B=\left[\begin{array}{c}50-40k+8k^2\\ 0+0+0\\ -50+40k-8k^2\end{array}\right]$ For atleast three solutions, $(adjA)B=0$
$\Rightarrow$ $\left[\begin{array}{c}50-40k+8k^2\\ 0\\ -50+40k-8k^2\end{array}\right]=0$
$\Rightarrow$ $\left[\begin{array}{c}50-40k+8k^2\\ 0\\ -50+40k-8k^2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow$ $8k^2-40k+50=0$
$\Rightarrow$ $2(k^2-20k+25)=0$
$\Rightarrow$ $k=\frac{-(-20)\pm \sqrt{{(-20)}^2-4\times 1\times 25}}{2\times 1}$
$\Rightarrow$ $k=\frac{20\pm \sqrt{400-100}}{2}$
$\Rightarrow$ $k=\frac{20\pm \sqrt{300}}{2}$
$\Rightarrow$ $k=\frac{20\pm \sqrt{100\times 3}}{2}$
$\Rightarrow$ $k=\frac{20\pm 10\sqrt{3}}{2}$
$\Rightarrow$ $k=10\pm 5\sqrt{3}$
$\Rightarrow$ $k=10+5\sqrt{3},10-5\sqrt{3}$ The number of values of $k=2$