Animated Solution for Mathematics - Matrices and Determinants: Let $\lambda ,\mu \in \mathbb{R}$. If the system of equations $3x+5y+\lambda z=3$, $7x+11y-9z=2$, $97x+155y-189z=\mu$ has infinitely many solutions, then $\mu +2\lambda$ is equal to :

Unlocking the System

Welcome, my dear student. Today, we are going to peel back the layers of a classic JEE Advanced problem. It is not just about solving for $\lambda$ and $\mu$; it is about understanding the beautiful, rigid structure of linear algebra.

When we look at a system of equations, we are not just looking at numbers; we are looking at planes in three-dimensional space. Imagine three sheets of paper in space. Usually, they might meet at a single point.

But when we are told there are infinitely many solutions, we are being told that these planes are dancing in perfect harmony, intersecting along a single, shared line. This is the geometric soul of linear dependence.

The Linear Combination

Since the system has infinitely many solutions, the third equation, $97x+155y-189z=\mu$, cannot be independent of the first two. It must be a shadow, a reflection, a linear combination of the first two equations: $3x+5y+\lambda z=3$ and $7x+11y-9z=2$.

We express this mathematically as $L_3=a{L}_1+b{L}_2$. This is our master key. By equating the coefficients of $x$, $y$, and $z$, and the constant term, we create a bridge between the knowns and the unknowns.

Let us focus on the coefficients of $x$ and $y$. For $x$, we have $97=3a+7b$. For $y$, we have $155=5a+11b$.

This is a simple system of two linear equations with two variables, $a$ and $b$. Multiply the first by $5$ and the second by $3$ to eliminate $a$:

Subtracting these, we find $2b=20$, which gives us $b=10$. Substituting this back, $3a+7(10)=97$ leads us to $3a=27$, so $a=9$. We have found the weights of our linear combination!

Now that we have $a=9$ and $b=10$, the rest is a victory lap. For the $z$ coefficient, we have $-189=a(\lambda )+b(-9)$.

Substituting our values:

Solving for $\lambda$, we get $9\lambda =-99$, so $\lambda =-11$. Finally, for the constant term, $\mu =a(3)+b(2)$.

Substituting, we find:

The final step is to calculate $\mu +2\lambda$. Plugging in our values:

Look at what we have achieved. We did not just crunch numbers; we used the geometric property of linear dependence to force the system to reveal its secrets.

The beauty of this method is its elegance—it bypasses the messiness of determinants and goes straight to the heart of the relationship between the equations. The final result is 25. Keep this mindset, and no system of equations will ever intimidate you again.