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How do you determine the value(s) of $k$ such that the system of linear equations has the indicated number of solutions: An infinite number of solutions for $4x + ky = 6\;and\;kx + y = - 3?$

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: First express the given equations in matrix form, then find the determinant of the matrix and then compare it to zero, because determinant of a matrix of linear equations equal zero proofs two lines are parallel. Then after getting the value(s) test for the consistency of the system, the value that satisfies the test will be the required solution.

Complete step by step solution:
In order to find the value(s) of $k$ such that the system of linear equation $4x + ky = 6\;and\;kx + y = - 3$ has infinite number of solutions, we will first express the equations in matrix form.
The given equations,
$4x + ky = 6\;and\;kx + y = - 3$
We can express it in matrix form as follows
$\left[ {\begin{array}{*{20}{c}}
4&k \\
k&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6 \\
{ - 3}
\end{array}} \right]$
Now, the given system of equation will have infinite solution if the determinant of the $2 \times 2$
matrix $\left[ {\begin{array}{*{20}{c}}
4&k \\
k&1
\end{array}} \right]$ is equal to zero.
$
\Rightarrow \left| {\begin{array}{*{20}{c}}
4&k \\
k&1
\end{array}} \right| = 0 \\
\Rightarrow (4 \times 1) - (k \times k) = 0 \\
\Rightarrow 4 - {k^2} = 0 \\
\Rightarrow {k^2} = 4 \\
\Rightarrow k = \pm 2 \\
$
So we get two values for $k,\;i.e.\; + 2\;{\text{and}}\; - 2$
Now we will test the consistency of the system of linear equations by substituting both the values in the original matrix equation one by one.
Testing for $k = + 2$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
4&2 \\
2&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6 \\
{ - 3}
\end{array}} \right]$
Performing row operation and taking row $1$ with $2$ we will get
$ \Rightarrow {}_{{R_1} - > \dfrac{1}{2}{R_1}}\left[ {\begin{array}{*{20}{c}}
2&1 \\
2&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 3}
\end{array}} \right]$
And now subtracting row $1\;{\text{from}}\;2$
$ \Rightarrow {}_{{R_2} - > {R_2} - {R_1}}\left[ {\begin{array}{*{20}{c}}
2&1 \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 3}
\end{array}} \right]$
Hence the $2nd$ row says $0x + 0y = - 3 \Rightarrow 0 + 0 = - 3 \Rightarrow 0 = - 3$ which is not possible, that is the system is inconsistent with $k = + 2$ so it is not the solution.
Testing for $k = - 2$
$
\Rightarrow \left[ {\begin{array}{*{20}{c}}
4&{ - 2} \\
{ - 2}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6 \\
{ - 3}
\end{array}} \right] \\
\Rightarrow {}_{{R_1} - > \dfrac{1}{2}{R_1}}\left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
{ - 2}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 3}
\end{array}} \right] \\
\Rightarrow {}_{{R_2} - > {R_1} + {R_2}}\left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
0
\end{array}} \right] \\
$
Since last line of matrix is all zero, hence it satisfies the condition
$\therefore k = - 2$ is the solution for the system of linear equations $4x + ky = 6\;and\;kx + y = - 3$ for infinite solutions.

Note: In a system of linear equation $a1x + b1y + c1 = 0\;{\text{and}}\;a2x + b2x + c2 = 0$ the system will have unique and consistent solution if $\dfrac{{a1}}{{a2}} \ne \dfrac{{b1}}{{b2}}$, will have infinitely many solution if $\dfrac{{a1}}{{a2}} = \dfrac{{b1}}{{b2}} = \dfrac{{c1}}{{c2}}$ and will be inconsistent and have no solution if $\dfrac{{a1}}{{a2}} = \dfrac{{b1}}{{b2}} \ne \dfrac{{c1}}{{c2}}$