Answer

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Hint: Write in matrix form and see the determinant values.

Given system of equations is:

$

kx + 2y = 5 \\

3x + y = 1 \\

$

Convert these equations in matrix format

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

5 \\

1

\end{array}} \right]$

The system of equation has no solution if the value of determinant$\left( D \right) = 0$and at least one of the determinants $\left( {{D_1}{\text{ and }}{D_2}} \right)$is non-zero.

So, ${\text{D = }}\left| {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right|$

Now put this determinant to zero and calculate the value of $k$for which system has no solution.

$

\Rightarrow {\text{D = }}\left| {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right| = 0 \\

\Rightarrow \left( {k \times 1} \right) - \left( {2 \times 3} \right) = 0 \\

\Rightarrow k = 6 \\

$

So, for$k = 6$, the value of the determinant is zero.

Now, calculate the value of determinant ${D_1}$at this value of$k$.

In determinant ${D_1}$the first column is replaced with column $\left[ {\begin{array}{*{20}{c}}

5 \\

1

\end{array}} \right]$

$ \Rightarrow {D_1} = \left| {\begin{array}{*{20}{c}}

5&2 \\

1&1

\end{array}} \right| = \left( {5 \times 1} \right) - \left( {2 \times 1} \right) = 3 \ne 0$

Therefore, the system of equations has no solution for $k = 6$

Hence, option b is correct.

Note: - Whenever you face such type of question the key point we have to remember is that put determinant D = 0, then calculate the value of k, then at this value of k if the value of determinant ${D_1}{\text{ or }}{D_2}$ is non-zero then the system of equations has no solution.

Given system of equations is:

$

kx + 2y = 5 \\

3x + y = 1 \\

$

Convert these equations in matrix format

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

5 \\

1

\end{array}} \right]$

The system of equation has no solution if the value of determinant$\left( D \right) = 0$and at least one of the determinants $\left( {{D_1}{\text{ and }}{D_2}} \right)$is non-zero.

So, ${\text{D = }}\left| {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right|$

Now put this determinant to zero and calculate the value of $k$for which system has no solution.

$

\Rightarrow {\text{D = }}\left| {\begin{array}{*{20}{c}}

k&2 \\

3&1

\end{array}} \right| = 0 \\

\Rightarrow \left( {k \times 1} \right) - \left( {2 \times 3} \right) = 0 \\

\Rightarrow k = 6 \\

$

So, for$k = 6$, the value of the determinant is zero.

Now, calculate the value of determinant ${D_1}$at this value of$k$.

In determinant ${D_1}$the first column is replaced with column $\left[ {\begin{array}{*{20}{c}}

5 \\

1

\end{array}} \right]$

$ \Rightarrow {D_1} = \left| {\begin{array}{*{20}{c}}

5&2 \\

1&1

\end{array}} \right| = \left( {5 \times 1} \right) - \left( {2 \times 1} \right) = 3 \ne 0$

Therefore, the system of equations has no solution for $k = 6$

Hence, option b is correct.

Note: - Whenever you face such type of question the key point we have to remember is that put determinant D = 0, then calculate the value of k, then at this value of k if the value of determinant ${D_1}{\text{ or }}{D_2}$ is non-zero then the system of equations has no solution.

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