Question

Find the value of $k$ for which each of the following systems of equations have infinitely many solutions:
$
  kx - 3y = 2k + 1 \\
  2\left( {k + 1} \right)x + 9y = 7k + 1 \\
 $

Answer 1

Hint - Here we will proceed by using the matrix method of linear equations for $2 \times 2$ matrix. Then by applying the conditions given in the question we will get our answer.

Complete Step-by-Step solution:
Let,
$A = \left( {\begin{array}{*{20}{c}}
  k&3 \\
  {2\left( {k + 1} \right)}&9
\end{array}} \right)$
$B = \left( \begin{gathered}
  2k + 1 \\
  7k + 1 \\
\end{gathered} \right)$
And, $x = \left( \begin{gathered}
  x \\
  y \\
\end{gathered} \right)$
Since, it is given that the following system of equations have infinitely many solutions. And, when there are infinitely many solutions of the given system of equations.
Then the value of $\left( {AdjA} \right)B = 0$, where, $\left( {AdjA} \right)$= Ad-joint of A.
In order to find the Ad-joint of $2 \times 2$ matrix, we interchange the value of the diagonal matrix, and the sign of the rest of the elements changes.
Therefore, $adj\left( A \right) = \left( {\begin{array}{*{20}{c}}
  9&{ - 3} \\
  { - 2k - 2}&k
\end{array}} \right)$
And $B = \left( \begin{gathered}
  2k + 1 \\
  7k + 1 \\
\end{gathered} \right)$ (It is already given)
$\left( {AdjA} \right) \times B = \left( {\begin{array}{*{20}{c}}
  9&{ - 3} \\
  { - 2k - 2}&k
\end{array}} \right) \times \left( \begin{gathered}
  2k + 1 \\
  7k + 1 \\
\end{gathered} \right)$
By applying the method of multiplication of two matrix, we get
$
  \left( {\begin{array}{*{20}{c}}
  {18k + 9 - 21k - 3} \\
  { - 4{k^2} - 2k - 4k + 7{k^2} + k}
\end{array}} \right) \\
   = \left( \begin{gathered}
   - 3k + 6 \\
  3{k^2} - 5k - 2 \\
\end{gathered} \right) \\
 $
According, to the given condition $\left( {AdjA} \right) \times B = 0$
Therefore, $\left( {\begin{array}{*{20}{c}}
  { - 3k + 6} \\
  {3{k^2} - 5k - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  0
\end{array}} \right)$
On comparing, we get
$
   \Rightarrow - 3k + 6 = 0 \\
   \Rightarrow - 3k = - 6 \\
   \Rightarrow k = \dfrac{{ - 6}}{{ - 3}} \\
   \Rightarrow k = 2 \\
 $
Then,
$
   \Rightarrow 3{k^2} - 5k - 2 = 0 \\
   \Rightarrow 3{k^2} - 5k = 2 \\
 $
On splitting the middle term we will get,
$
  3{k^2} - 6k + k = 0 \\
  3k\left( {k - 2} \right) + k\left( {k - 2} \right) = 0 \\
 $
So, $k - 2 = 0$
$ \Rightarrow k = 2$
Thus, the value of $k = 2$.
So, we can say that if $k = 2$, then infinite solutions exist for this system.

Note – Whenever we come up with this type of question, one must know that a $n \times n$ homogeneous system of linear equations has a unique solution if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions. It is a consistent system of equations.