Animated Solution for Mathematics - Matrices and Determinants: Let $\alpha ,\beta (\alpha \ne \beta )$ be the values of m, for which the equations $x+y+z=1$; $x+2y+4z=m$ and $x+4y+10z=m^2$ have infinitely many solutions. Then the value of ${\sum }_{n=1}^{10}(n^{\alpha }+n^{\beta })$ is equal to :

Welcome, future engineers! Today, we are not just solving a system of equations; we are embarking on a journey through three-dimensional space. When you look at the system of equations provided, I want you to stop seeing just variables like $x,y,$ and $z$. Instead, I want you to visualize three distinct planes in space.

Usually, when we solve a system of three linear equations, we are hunting for the unique point where these three planes meet. But this problem is different. It asks for the condition where there are infinitely many solutions.

Geometrically, this means these planes are not meeting at a single point. Instead, they are behaving like the spine of an open book, intersecting along an entire common line. This is the core geometric reality we must capture.

Cramer's Rule

To translate this geometric intuition into algebraic rigor, we turn to Cramer's Rule. For a system to have infinitely many solutions, the main determinant of the coefficient matrix, denoted as $D$, must vanish. That is, $D=0$.

However, as we discussed in our FAQs, $D=0$ is a necessary but not sufficient condition. We must also ensure that the determinants $D_x,D_y,$ and $D_z$ are all zero. If $D=0$ but one of the others is non-zero, the planes would be parallel and inconsistent, leaving us with no solution at all.

Let us construct our main determinant $D$ using the coefficients of $x,y,$ and $z$:

Expanding this along the first row, we calculate: $1(20-16)-1(10-4)+1(4-2)$. This simplifies beautifully to $4-6+2=0$. The condition is satisfied! Our planes are indeed positioned to allow for infinite solutions.

Now, we must find the values of $m$ that make this system dependent. We set $D_x=0$. We replace the first column of our matrix with the constants from the right-hand side of our equations: $1,m,$ and $m^2$.

This is where many students stumble. Take a deep breath and expand carefully. We get $1(20-16)-1(10m-4m^2)+1(4m-2m^2)=0$.

Simplifying this, we arrive at $4-10m+4m^2+4m-2m^2=0$. Combining like terms, we get the quadratic equation $2m^2-6m+4=0$.

Dividing by $2$, we find $m^2-3m+2=0$. Factoring this, we get $(m-1)(m-2)=0$. Thus, our values are $\alpha =1$ and $\beta =2$.

Summation

We have conquered the algebra. Now, we face the final challenge: evaluating the sum ${\sum }_{n=1}^{10}(n^{\alpha }+n^{\beta })$. Substituting our values, we need to compute ${\sum }_{n=1}^{10}(n^1+n^2)$.

We split this into two standard summations:

1. The sum of the first $n$ natural numbers:

2. The sum of the squares of the first $n$ natural numbers:

Adding these together, $55+385=440$.

Look at what we have achieved. We started with a system of equations, visualized the intersection of planes, applied the rigor of Cramer's Rule, solved a quadratic, and finished with a series summation.

This is the essence of JEE Advanced physics and mathematics—connecting disparate concepts into a single, elegant solution. Never fear the complexity; embrace the process, and the answer will always reveal itself. The final result is 440.