Animated Solution for Mathematics - Matrices and Determinants: The number of values of k for which the system of linear equations, $(k+2)x+10y=k$, $kx+(k+3)y=k-1$ has no solution is :

We are given two linear equations: $L_1:(k+2)x+10y=k$ $L_2:kx+(k+3)y=k-1$

Our objective is to find the value of $k$ such that the system has no solution. Geometrically, this occurs when the two lines are parallel but not coincident.

For a system of linear equations $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$, the lines are parallel and distinct (resulting in no solution) if and only if the ratios of the coefficients satisfy the following condition:

If the ratios were all equal, the lines would be coincident, leading to infinitely many solutions.

To find the potential values for $k$, we equate the ratios of the coefficients of $x$ and $y$:

Cross-multiplying these terms yields:

Expanding the left side, we obtain $k^2+5k+6=10k$. Rearranging this into a standard quadratic form gives:

Factoring the quadratic equation, we get $(k-2)(k-3)=0$. This provides two candidate values: $k=2$ and $k=3$.

Testing the Candidates

We must now verify these candidates against the constant ratio $\frac{c_1}{c_2}=\frac{k}{k-1}$ to ensure the lines are not coincident.

For $k=2$: The ratio of the constants is $\frac{2}{2-1}=2$. The ratio of the coefficients is $\frac{2+2}{2}=2$. Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the lines are coincident. We must reject $k=2$.

For $k=3$: The ratio of the coefficients is $\frac{3+2}{3}=\frac{5}{3}$. The ratio of the constants is $\frac{3}{3-1}=\frac{3}{2}$. Since $\frac{5}{3} eq \frac{3}{2}$, the lines are strictly parallel and distinct.

By rigorously testing our candidates, we determine that only $k=3$ satisfies the condition for no solution. Therefore, there is exactly $1$ value of $k$ that fulfills the requirement.