Animated Solution for Mathematics - Matrices and Determinants: For $\alpha ,\beta \in \mathbb{R}$, suppose the system of linear equations $x-y+z=5$, $2x+2y+\alpha z=8$, $3x-y+4z=\beta$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of

Welcome, future engineer! Today, we are not just solving a system of equations; we are exploring the beautiful, interconnected nature of linear algebra.

When you look at a system of three linear equations, I want you to stop seeing just numbers and variables. Instead, visualize three planes in three-dimensional space.

Usually, these planes intersect at a single, unique point—a specific coordinate $(x,y,z)$. But today, the problem tells us something different: the system has infinitely many solutions.

Geometrically, this means our three planes are not meeting at a single point; they are meeting along a common line. They are dancing together in perfect harmony.

To unlock this mystery, we turn to the elegant machinery of Cramer's Rule. We know that for a system of linear equations to have a unique solution, the determinant of the coefficient matrix, $\Delta$, must be non-zero.

But when we are dealing with infinite solutions, the system is 'dependent.' This forces the main determinant to collapse: $\Delta =0$.

Furthermore, to ensure the system is consistent (meaning the planes actually intersect and aren't just parallel and disjoint), we must also have ${\Delta }_x={\Delta }_y={\Delta }_z=0$. This is our roadmap.

Unmasking $\alpha$

Let us construct our main determinant $\Delta$ using the coefficients of $x,y,$ and $z$ from our equations:

Now, let us expand this along the first row. Take a deep breath and focus on the signs.

We have $1(2\cdot 4-(-1)\cdot \alpha )-(-1)(2\cdot 4-3\cdot \alpha )+1(2\cdot (-1)-3\cdot 2)=0$. Simplifying this, we get $1(8+\alpha )+1(8-3\alpha )+1(-2-6)=0$.

Combining these terms gives us $8+\alpha +8-3\alpha -8=0$. The constants $8$ and $-8$ cancel out, leaving us with $-2\alpha +8=0$. Solving this, we find $\alpha =4$. We have successfully cracked the first part of the code!

The Hunt for $\beta$

With $\alpha =4$ in our pocket, we move to the next phase. We need to find $\beta$. We use the condition ${\Delta }_x=0$.

We replace the first column of our determinant with the constant terms from the right side of our equations: $5,8,$ and $\beta$.

Expanding this along the first row again: $5(2\cdot 4-(-1)\cdot 4)-(-1)(8\cdot 4-\beta \cdot 4)+1(8\cdot (-1)-\beta \cdot 2)=0$.

This simplifies to $5(8+4)+1(32-4\beta )+1(-8-2\beta )=0$. Calculating the numbers: $5(12)+32-4\beta -8-2\beta =0$, which becomes $60+32-8-6\beta =0$.

This simplifies beautifully to $84-6\beta =0$, leading us to $\beta =14$.

Connecting to Quadratics

We have our roots: $\alpha =4$ and $\beta =14$. The problem asks for the quadratic equation that has these as roots.

Remember the standard form: $x^2-(\text{sum of roots})x+(\text{product of roots})=0$.

The sum is $4+14=18$, and the product is $4\times 14=56$. Thus, our equation is $x^2-18x+56=0$.

Look at that! We started with a system of linear equations and ended with a quadratic polynomial. This is the beauty of JEE Advanced mathematics—it tests not just your ability to calculate, but your ability to weave different concepts together into a single, coherent narrative. You did great today!