Animated Solution for Mathematics - Matrices and Determinants: If the system of equation $2x+\lambda y+3z=5$, $3x+2y-z=7$, $4x+5y+\mu z=9$ has infinitely many solutions, then $({\lambda }^2+{\mu }^2)$ is equal to :

Welcome, future engineer. Today, we are not just solving a system of equations; we are exploring the architecture of space itself.

When you look at the system $2x+\lambda y+3z=5$, $3x+2y-z=7$, and $4x+5y+\mu z=9$, do not see just a collection of variables and coefficients. See three planes in three-dimensional space.

Imagine you are standing in a room. Each equation represents a flat, infinite sheet—a plane.

Usually, three planes intersect at a single point, like the corner of a room where two walls meet the ceiling. That is a unique solution.

But the problem asks for infinitely many solutions. This means our three planes are not meeting at a point; they are behaving like the pages of an open book, all meeting along a single, common line.

This is the state of 'dependency.' Mathematically, this requires the system to be consistent and dependent, which brings us to our most powerful tool: Cramer's Rule.

To achieve this state of infinite intersection, we require the main determinant $\Delta$ to be zero, and crucially, all auxiliary determinants ${\Delta }_x$, ${\Delta }_y$, and ${\Delta }_z$ to be zero as well.

If $\Delta$ is zero but the auxiliary determinants are not, the planes would be parallel or inconsistent, leaving us with no solution at all.

Here is where we must be strategic. If we calculate $\Delta$ first, we are left with an equation involving both $\lambda$ and $\mu$. That is a dead end.

Instead, look at ${\Delta }_z$. By replacing the $z$-coefficients with the constants $(5,7,9)$, we effectively remove $\mu$ from the equation. This is our 'smart move.'

Let us set up the determinant ${\Delta }_z$:

Expanding along the first row, we carefully compute the minors:

Simplifying the arithmetic: $2(18-35)-\lambda (27-28)+5(15-8)=0$.

This reduces to $2(-17)-\lambda (-1)+5(7)=0$. With a sigh of relief, we see the equation simplify beautifully: $-34+\lambda +35=0$, which gives us $\lambda +1=0$, or $\lambda =-1$. We have our first key!

Now that we know $\lambda =-1$, the main determinant $\Delta$ is no longer a mystery. We substitute our value back into the matrix:

We expand this again along the first row:

This simplifies to $2(2\mu +5)+1(3\mu +4)+3(7)=0$.

Distributing the terms, we get $4\mu +10+3\mu +4+21=0$. Combining like terms, we arrive at $7\mu +35=0$, leading us directly to $\mu =-5$.

The problem asks for the value of ${\lambda }^2+{\mu }^2$.

With $\lambda =-1$ and $\mu =-5$, we calculate:

We have navigated the geometry, applied the algebraic constraints, and arrived at the solution. The final answer is $26$.

Remember, in JEE Advanced, the math is not just about calculation; it is about choosing the path of least resistance. You have mastered that today. Keep this confidence, and carry it into your next challenge.