Animated Solution for Mathematics - Matrices and Determinants: The number of values of $k$, for which the system of equations : $(k+1)x+8y=4k;kx+(k+3)y=3k-1$ has no solution, is

A Journey into Linear Systems

Welcome, future engineers! Today, we are going to peel back the layers of a seemingly simple problem that hides a classic, treacherous trap. We are dealing with a system of linear equations, but with a twist: a parameter $k$.

Our goal is to find the number of values of $k$ for which the system has no solution. This isn't just about algebra; it is about understanding the geometric soul of linear equations.

The Geometric Reality

Imagine you are standing on a 2D coordinate plane. You have two lines, defined by the equations $(k+1)x+8y=4k$ and $kx+(k+3)y=3k-1$.

In the world of linear algebra, a system of two equations represents two lines. If these lines intersect at a single point, we have a unique solution. If they never meet, they are parallel and distinct, meaning we have no solution. If they lie exactly on top of each other, they are coincident, meaning they have infinite solutions.

To have 'no solution', our lines must be parallel but distinct. Mathematically, this requires the slopes to be equal, but the y-intercepts to be different. For the general form $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$, this translates to the elegant condition:

This is our compass for the journey ahead.

The Algebraic Hunt

Let us extract our coefficients. For the first equation, we have $a_1=k+1$, $b_1=8$, and $c_1=4k$. For the second, $a_2=k$, $b_2=k+3$, and $c_2=3k-1$.

Now, we set the ratio of the coefficients of $x$ and $y$ equal to each other:

This is where the algebra begins to dance. We cross-multiply to clear the denominators: $(k+1)(k+3)=8k$.

Expanding the left side, we get $k^2+3k+k+3=8k$, which simplifies beautifully to $k^2+4k+3=8k$. Bringing everything to one side, we arrive at the quadratic equation:

The Trap of the Quadratic

Factoring this quadratic is straightforward: $(k-1)(k-3)=0$. This gives us two potential candidates for $k$: $k=1$ and $k=3$.

Many students would stop here and answer '2'. But wait! Remember our condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}$. We have satisfied the first part, but we must verify the inequality.

Let us test $k=1$. Substituting $k=1$ into our ratios, we get:

Oh no! All three ratios are equal to 2. This means the lines are coincident; they are the same line! Thus, $k=1$ gives us infinite solutions, not zero. We must reject $k=1$.

Now, let us test $k=3$. Substituting $k=3$, we get:

Here, $\frac{4}{3} eq \frac{3}{2}$. The first two ratios are equal, but the third is different. This is exactly what we need for parallel, distinct lines!

The Elegance of Precision

By carefully testing our candidates, we discovered that only $k=3$ satisfies the condition for 'no solution'. The value $k=1$ was a siren song, leading us toward infinite solutions.

This problem teaches us that in JEE, the answer isn't just about solving the equation; it's about verifying the constraints of the physical reality the equation represents. There is only 1 value of $k$ that works. Keep this rigor in your toolkit, and you will conquer any system that comes your way!