Answer

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Hint: To solve this question, first we will start with the condition for a non- trivial solution, converting these equations into a determinant of coefficients, find the value of k and then the solutions of the equations.

For a non-trivial solution D = 0 i.e. determinant = 0.

Determinant of the coefficients,

\[ \Rightarrow \left( {\begin{array}{*{20}{c}}

1&k&3 \\

3&k&{ - 2} \\

2&3&{ - 4}

\end{array}} \right) = 0\]

For making calculation easy we try to make the elements of determinant zero by applying \[{R_2} - 3{R_1},{R_3} - 2{R_1}\](here R means row)

∴ \[\Delta = \left( {\begin{array}{*{20}{c}}

1&k&3 \\

0&{ - 2k}&{ - 11} \\

0&{3 - 2k}&{ - 10}

\end{array}} \right) = 0\]

Or \[20k + 11(3 - 2k) = 0\]

Or 33-2k = 0

∴\[k = \dfrac{{33}}{2}\]

Putting t

the value of k, the equations are

\[x + \dfrac{{33}}{2}y + 3z = 0..........(1)\]

\[3x + \dfrac{{33}}{2}y - 2z = 0..........(2)\]

\[2x + 3y - 4z = 0............(3)\]

Multiply (1) by 3 and subtract from (2) and similarly multiply (1) by 2 and subtract from (3). Thus we get the equivalent system of equations as

\[x + \dfrac{{33}}{2}y + 3z = 0\]

\[ - 33y - 11z = 0\]

\[ - 30y - 10z = 0\]

From any of the last two, we get 3y = -z

Or\[\dfrac{y}{1} = \dfrac{z}{{ - 3}} = \lambda \], say

∴ y =\[\lambda \], z = -3\[\lambda \]

From (1), we get

$

x + \dfrac{{33}}{2}y + 3z = 0 \\

x + \dfrac{{33}}{2}y - 9\lambda = 0 \\

\therefore x = \dfrac{{ - 15}}{2}\lambda \\

\therefore x:y:z = - \dfrac{{15}}{2}:1: - 3 \\

$

Note: The system of equations in which the determinant of the coefficient is zero is called a non-trivial solution. And the system of equations in which the determinant of the coefficient matrix is not zero but the solutions are x=y=z=0 is called a trivial solution.

For a non-trivial solution D = 0 i.e. determinant = 0.

Determinant of the coefficients,

\[ \Rightarrow \left( {\begin{array}{*{20}{c}}

1&k&3 \\

3&k&{ - 2} \\

2&3&{ - 4}

\end{array}} \right) = 0\]

For making calculation easy we try to make the elements of determinant zero by applying \[{R_2} - 3{R_1},{R_3} - 2{R_1}\](here R means row)

∴ \[\Delta = \left( {\begin{array}{*{20}{c}}

1&k&3 \\

0&{ - 2k}&{ - 11} \\

0&{3 - 2k}&{ - 10}

\end{array}} \right) = 0\]

Or \[20k + 11(3 - 2k) = 0\]

Or 33-2k = 0

∴\[k = \dfrac{{33}}{2}\]

Putting t

the value of k, the equations are

\[x + \dfrac{{33}}{2}y + 3z = 0..........(1)\]

\[3x + \dfrac{{33}}{2}y - 2z = 0..........(2)\]

\[2x + 3y - 4z = 0............(3)\]

Multiply (1) by 3 and subtract from (2) and similarly multiply (1) by 2 and subtract from (3). Thus we get the equivalent system of equations as

\[x + \dfrac{{33}}{2}y + 3z = 0\]

\[ - 33y - 11z = 0\]

\[ - 30y - 10z = 0\]

From any of the last two, we get 3y = -z

Or\[\dfrac{y}{1} = \dfrac{z}{{ - 3}} = \lambda \], say

∴ y =\[\lambda \], z = -3\[\lambda \]

From (1), we get

$

x + \dfrac{{33}}{2}y + 3z = 0 \\

x + \dfrac{{33}}{2}y - 9\lambda = 0 \\

\therefore x = \dfrac{{ - 15}}{2}\lambda \\

\therefore x:y:z = - \dfrac{{15}}{2}:1: - 3 \\

$

Note: The system of equations in which the determinant of the coefficient is zero is called a non-trivial solution. And the system of equations in which the determinant of the coefficient matrix is not zero but the solutions are x=y=z=0 is called a trivial solution.

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