Question

The possible combination of values of \[k\], \[d \in R\] from the given set of options, for which the system of equations \[x + dy + 3z = 0\], \[kx + 2y + 2z = 0\], \[2x + 3y + 4z = 0\] admits of non-trivial solution is
This question has multiple correct options
A. \[k = 2\]
B. \[d = 2\]
C. \[k = 3\]
D. \[d = \dfrac{5}{4}\]

Answer 1

Hint: In this question first of all, consider the augmented matrix of the given system of equations. Then equate the determinant of the coefficient matrix to zero as the given system of equations have non-trivial solutions.

Complete Step-by-Step solution:
The given equations are \[x + dy + 3z = 0\], \[kx + 2y + 2z = 0\], \[2x + 3y + 4z = 0\]
Clearly the given system of equations are homogeneous equations.
Now consider the augmented matrix i.e., \[AX = O\] for the given system of homogeneous equations where \[A = \left[ {\begin{array}{*{20}{c}}
  1&d&3 \\
  k&2&2 \\
  2&3&4
\end{array}} \right]\], \[B = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right]\]
As the system of equations have non-trivial solution coefficient matrix i.e., \[\left| A \right|\] must be equal to zero.
So, we have
\[
   \Rightarrow \left| A \right| = 0 \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&d&3 \\
  k&2&2 \\
  2&3&4
\end{array}} \right| = 0 \\
\]
Expanding the determinant, we have
\[
   \Rightarrow 1\left[ {\left( {2 \times 4} \right) - \left( {3 \times 2} \right)} \right] - d\left[ {\left( {k \times 4} \right) - \left( {2 \times 2} \right)} \right] + 3\left[ {\left( {k \times 3} \right) - \left( {2 \times 2} \right)} \right] = 0 \\
   \Rightarrow 1\left( {8 - 6} \right) - d\left( {4k - 4} \right) + 3\left( {3k - 4} \right) = 0 \\
   \Rightarrow 2 - 4dk + 4d + 9k - 12 = 0 \\
   \Rightarrow - 4dk + 4d + 9k - 9 - 1 = 0 \\
\]
Grouping the terms, we have
\[
   \Rightarrow - 4d\left( {k - 1} \right) + 9\left( {k - 1} \right) = 1 \\
   \Rightarrow \left( {9 - 4d} \right)\left( {k - 1} \right) = 1 \\
\]
The above equation will be satisfied when
\[ \Rightarrow \left( {9 - 4d} \right) = \left( {k - 1} \right) = 1\]
So possible values are
\[
   \Rightarrow \left( {k - 1} \right) = 1 \\
   \Rightarrow k = 1 + 1 \\
  \therefore k = 2 \\
\]
And
\[
   \Rightarrow \left( {9 - 4d} \right) = 1 \\
   \Rightarrow 4d = 9 - 1 \\
   \Rightarrow 4d = 8 \\
  \therefore d = 2 \\
\]
Thus, the correct options are A. \[k = 2\] and B. \[d = 2\]

Note: Here the given system of equations are homogeneous equations. The system of equations in which the determinant of the coefficient is zero is called a non-trivial solution. For the answer we can substitute the obtained value in the coefficient matrix and see whether its determinant is equal zero or not.