Question

Find the value of $k$ for which each of the following system of equations have infinitely many solution:
$
  kx - 2y + 6 = 0 \\
  4x - 3y + 9 = 0 \\
 $

Answer 1

Hint – We will start solving this question by converting the given equations in matrix form and by using the given condition, i.e., equations have infinitely many solution, we can apply a formula, i.e., $\left( {adj.A} \right)B = 0$, from an application of matrix and determinants, i.e., linear system of equations, and obtain the required result.

Complete Step-by-Step solution:
The given equations are,
$
  kx - 2y + 6 = 0 \\
  4x - 3y + 9 = 0 \\
 $
We can write these equations as,
$
  kx - 2y = - 6 \\
  4x - 3y = - 9 \\
 $
Converting the given equations in matrix form, we get
$
  \left( {\begin{array}{*{20}{c}}
  k&{ - 2} \\
  4&{ - 3}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 6} \\
  { - 9}
\end{array}} \right) \\
   \Rightarrow AX = B \\
 $
where, $A = \left( {\begin{array}{*{20}{c}}
  k&{ - 2} \\
  4&{ - 3}
\end{array}} \right),X = \left( {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  { - 6} \\
  { - 9}
\end{array}} \right)$.
Now, it is given in the question that there are infinitely many solutions of the equations, therefore the equation is consistent.
Therefore, the formula which we have to use is,
$\left( {adj.A} \right)B = 0$…….. (1)
Let us find $adj.A$.
Let C be the matrix of the cofactors of the elements of A,
$C = \left( {\begin{array}{*{20}{c}}
  { - 3}&{ - 4} \\
  2&k
\end{array}} \right)$
As we know that $adj.A$ is equal to the transpose of the matrix of the cofactor, therefore
$adj.A$$ = \left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  { - 4}&k
\end{array}} \right)$
Now, for using the formula we need $\left( {adj.A} \right)B$, therefore by multiplying $adj.A$ $ = \left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  { - 4}&k
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  { - 6} \\
  { - 9}
\end{array}} \right)$, we obtain
$\left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  { - 4}&k
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  { - 6} \\
  { - 9}
\end{array}} \right)$
By using the method of multiplication of two matrices, we get
$
  \left( {\begin{array}{*{20}{c}}
  {18 - 18} \\
  {24 - 9k}
\end{array}} \right) \\
   = \left( {\begin{array}{*{20}{c}}
  0 \\
  {24 - 9k}
\end{array}} \right) \\
 $
Therefore, $\left( {adj.A} \right)B$ $ = \left( {\begin{array}{*{20}{c}}
  0 \\
  {24 - 9k}
\end{array}} \right)$
Putting this in (1), we obtain
$\left( {\begin{array}{*{20}{c}}
  0 \\
  {24 - 9k}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  0
\end{array}} \right)$
On comparing these two matrices, we get
$
  24 - 9k = 0 \\
   \Rightarrow - 9k = - 24 \\
   \Rightarrow 9k = 24 \\
   \Rightarrow k = \dfrac{{24}}{9} \\
   \Rightarrow k = \dfrac{8}{3} \\
 $
Hence, the value of $k$ is $\dfrac{8}{3}$.

Note – A matrix is an ordered rectangular array of numbers or functions. The system of equations is said to be consistent if it has one or more than one solutions. These types of questions become complex when we solve them using matrix formulas, to solve it perfectly, without making a mistake, all the basic formulas must be remembered.