Question

Find values of k for which the system of equation (3k-8)x+3y+3z=0,3x+(3k-8)y+3z=0,3x+3y+(3k-8)z=0 has a non -trival solution

Answer 1

Answer:

K=2/3 for non trivial solution

Answer 2

Given,

$\[\begin{array}{l}\left( {3k - 8} \right)x + 3y + 3z = 0\\3x + \left( {3k - 8} \right)y + 3z = 0\\3x + 3y + \left( {3k - 8} \right)z = 0\end{array}\]$

has a non- trivial solution.

Solution,

$\[\begin{array}{*{20}{c}}\vline& {\left( {3k - 8} \right)}&3&3\vline& \\\vline& 3&{\left( {3k - 8} \right)}&3\vline& \\\vline& 3&3&{\left( {3k - 8} \right)}\vline& \end{array} = 0\]$

Add all rows and write it in first row.

$\[ \Rightarrow \begin{array}{*{20}{c}}\vline& {\left( {3k - 2} \right)}&{\left( {3k - 2} \right)}&{\left( {3k - 2} \right)}\vline& \\\vline& 3&{\left( {3k - 8} \right)}&3\vline& \\\vline& 3&3&{\left( {3k - 8} \right)}\vline& \end{array} = 0\]$

Taking $\[{\left( {3k - 2} \right)}\]$ as a common from first row.

$\[ \Rightarrow \left( {3k - 2} \right)\begin{array}{*{20}{c}}\vline& 1&1&1\vline& \\\vline& 3&{\left( {3k - 8} \right)}&3\vline& \\\vline& 3&3&{\left( {3k - 8} \right)}\vline& \end{array} = 0\]$

$\[\begin{array}{l} \Rightarrow 3k - 2 = 0\\ \Rightarrow 3k = 2\end{array}\]$

$\[ \Rightarrow k = \frac{2}{3}\]$

Hence the value of k is $\[\frac{2}{3}\]$ for non- trivial solution.