Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations : $x+ay+z=3$, $x+2y+2z=6$, $x+5y+3z=b$ has no solution, then :-

Imagine you are standing in a room where three massive sheets of glass are suspended in the air. Each sheet represents one of our linear equations: $x+ay+z=3$, $x+2y+2z=6$, and $x+5y+3z=b$.

In the world of linear algebra, these are planes in three-dimensional space. When we ask for a 'solution' to the system, we are asking: 'Where do these three sheets of glass meet?'

Usually, they meet at a single point—a unique solution. But today, the problem gives us a fascinating constraint: the system has no solution.

Geometrically, this means there is no single point where all three planes intersect. They might be parallel, or they might form a triangular prism where each pair of planes intersects, but the three together never share a common ground. This is the essence of an inconsistent system.

Cramer's Rule

To translate this geometric reality into the language of algebra, we turn to the elegant machinery of Cramer's Rule. Cramer's Rule tells us that for a system of linear equations to have a unique solution, the determinant of the coefficient matrix, which we denote as $D$, must be non-zero.

If $D eq 0$, the planes intersect at exactly one point. But we are looking for the opposite. We need the system to be inconsistent.

This forces our hand: the main determinant $D$ must be zero. This is our first gatekeeper.

If $D=0$, the system is either inconsistent (no solution) or dependent (infinitely many solutions). To ensure we land in the 'no solution' territory, we need a second condition: at least one of the numerator determinants ($D_x,D_y,\text{or }{D}_z$) must be non-zero.

If all of them were zero, the planes would overlap in a way that creates infinite solutions. We must avoid that at all costs.

Unlocking the Parameters

Let us build our main determinant $D$ using the coefficients of $x,y,\text{and }z$:

Expanding this along the first row, we get:

Let's simplify this step-by-step. The first term is $1(6-10)=-4$. The second term is $-a(3-2)=-a$. The third term is $1(5-2)=3$.

Combining these, we find:

For the system to be inconsistent, we set $D=0$, which gives us $-1-a=0$, or simply $a=-1$. We have cracked the first part of the code!

Why $b$ Matters

Now, we must not get complacent. We have $a=-1$, but we still need to find the condition for $b$. We must ensure that the system does not collapse into 'infinitely many solutions.'

We construct $D_z$ by replacing the third column of our matrix with the constants from the right-hand side of our equations: $3,6,\text{and }b$. Substituting $a=-1$, we get:

Expanding this along the first row:

Let's simplify this carefully. The first term is $2b-30$. The second term is $+1(b-6)=b-6$. The third term is $3(3)=9$.

Adding them all together:

For the system to have no solution, we require $D_z eq 0$. Therefore:

We have arrived at our destination. For the system to be inconsistent, we must have $a=-1$ and $b eq 9$.

I want you to pause and appreciate what you just did. You didn't just solve an equation; you navigated the geometry of 3D space, you respected the rules of Cramer, and you successfully avoided the 'infinite solutions' trap.

This is the mindset of a JEE Advanced topper—not just calculating, but understanding the conditions that govern the system. Keep this rigor, keep this curiosity, and you will conquer any problem that comes your way.