Question

Find the value of $k$ for which each of the following systems of equations have infinitely many solutions:

$2x - 3y = 7$

$(k+2)x - (2k+1)y = 3(2k-1)$

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 answer:

Let,

$A = \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\ {k + 2}&{ - \left( {2k + 1} \right)} \end{array}} \right)$

$B = \left( {\begin{array}{*{20}{c}}  7 \\  3\left( {2k - 1} \right) \end{array}}  \right)$

And, $x = \left( {\begin{array}{*{20}{c}}  x \\  y  \end{array}}  \right)$

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 Adjoint of $2 \times 2$matrix, we interchange the value of the diagonal matrix, and the sign of the rest of the elements changes.

Therefore, $adj\left( A \right) = \left( {\begin{array}{*{20}{c}}  { - \left( {2k + 1} \right)}&3 \\  { - \left( {k + 2} \right)}&2 \end{array}} \right)$

And $B = \left( {\begin{array}{*{20}{c}}  7 \\  3\left( {2k - 1} \right) \end{array}}  \right)$   (It is already given)

$\left( {AdjA} \right)B = \left( {\begin{array}{*{20}{c}}  { - \left( {2k + 1} \right)}&3 \\   { - \left( {k + 2} \right)}&2 \end{array}} \right)\left( \begin{gathered}   7 \\   3\left( {2k - 1} \right) \\ \end{gathered} \right)$

By applying the method of multiplication of two matrix, we get

$\left( \begin{matrix} (-(2k+1) \times 7) + (3 \times 3(2k-1)) \\ (-(k+2) \times 7) + (2 \times 3(2k-1)) \\ \end{matrix} \right)$

$\left( \begin{matrix} (-7(2k+1)) + 9(2k-1) \\ (-7(k+2)) + ( 6 (2k-1)) \\ \end{matrix} \right)$

$\left( \begin{matrix} (-14k-7) + 18k-9) \\ (-7k-14) + ( 12k-6) \\ \end{matrix} \right)$

$\left( \begin{matrix} ( 4k-16) \\ (5k-20) \\ \end{matrix} \right)$

According, to the given condition $\left( {AdjA} \right) \times B = 0$

Therefore, $\left( {\begin{array}{*{20}{c}}  {4k - 16} \\   {5k - 20} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$

On comparing, we get

$   \Rightarrow 4k - 16 = 0 $

$   \Rightarrow 4k = 16 $

$   \Rightarrow k = \dfrac{{16}}{4} $

$   \Rightarrow k = 4  $

Then,

$   \Rightarrow 5k - 20 = 0 $

$   \Rightarrow 5k = 20 $

$   \Rightarrow k = \dfrac{{20}}{5} $

$   \Rightarrow k = 4 $

Thus, the value of $k = 4$.

So, we can say that if $k = 4$, then infinite solutions exist for this system.

Note: Whenever we come up with this type of question, one must know that there are various methods to solve such types of questions like simple method of linear equation, cross-multiplication method of linear equations. (Here we used the matrix method).