Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $x+4y-z=\lambda ,7x+9y+\mu z=-3,5x+y+2z=-1$ has infinitely many solutions, then $(2\mu +3\lambda )$ is equal to :

Welcome, future engineers. Today, we are going to peel back the layers of a classic JEE Advanced problem. We are dealing with a system of three linear equations in three variables: $x$, $y$, and $z$.

But there is a twist—two of our coefficients, $\mu$ and $\lambda$, are unknowns. The problem gives us a powerful hint: the system has 'infinitely many solutions.'

What does this mean? Imagine three planes in 3D space. Usually, they intersect at a single point—that is your unique solution.

But when we say 'infinitely many solutions,' we are describing a scenario where these planes are not independent. They might intersect along a common line, or perhaps they are all the same plane.

Algebraically, this means the system is dependent. This is where our toolkit—Cramer's Rule—becomes our best friend.

To handle this, we look at the determinant of the coefficient matrix, which we call $D$. If $D eq 0$, the system is independent and has a unique solution.

But we want infinity! So, we must force $D=0$.

Furthermore, for the system to be consistent (meaning it actually has solutions, rather than being impossible), we must also have $D_x=D_y=D_z=0$. This is the golden rule for dependency in linear systems.

Unlocking $\mu$ via the Main Determinant

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

Now, we expand this determinant along the first row. This is where precision is paramount. Do not rush!

Let us simplify this carefully. The first term is $1(18-\mu )$. The second term is $-4(14-5\mu )$. The third term is $-1(7-45)$.

Expanding these, we get:

Look at the constants: $18-56+38$. That is $56-56=0$. The constants vanish!

We are left with:

Thus, we find that $\mu =0$. We have successfully cracked the first part of the puzzle.

Finding $\lambda$ with $D_x$

Now that we know $\mu =0$, we turn our attention to $\lambda$. We use the condition $D_x=0$.

To form $D_x$, we replace the first column of our original determinant with the constants from the right-hand side of the equations: $\lambda$, $-3$, and $-1$.

Expanding this along the first row again:

Let us simplify the terms inside the brackets:

This leads us directly to $18\lambda =-18$, which means $\lambda =-1$.

We have our values: $\mu =0$ and $\lambda =-1$. The question asks us to calculate the value of $2\mu +3\lambda$.

Substituting our values:

And there we have it! The answer is $-3$.

This problem was not just about calculation; it was about understanding the geometric constraints of linear systems. When you see 'infinitely many solutions,' let your mind immediately jump to the determinant conditions.

Stay curious, keep practicing, and remember that every complex problem is just a series of simple steps waiting to be solved.