Animated Solution for Mathematics - Matrices and Determinants: For the system of linear equations $2x-y+3z=5$, $3x+2y-z=7$, $4x+5y+az=\beta$, which of the following is NOT correct?

Welcome, future engineers and mathematicians. Today, we are not just solving a system of equations; we are performing a diagnostic on the very structure of three-dimensional space.

When you look at a system of linear equations like $2x-y+3z=5$, $3x+2y-z=7$, and $4x+5y+az=\beta$, I want you to stop seeing just numbers and variables. I want you to see three planes in space.

The question is: do these planes meet at a single point, do they form a prism, or do they all intersect along a single line? This is the heart of linear algebra.

In our journey, the first step is to identify the 'Gatekeeper'—the determinant of the coefficient matrix, which we denote as $D$. Why do we care about $D$? Because $D$ tells us if the system is 'well-behaved.'

If $D eq 0$, the system is non-singular, meaning the planes intersect at exactly one point. It is the safe zone. If $D=0$, the system is singular, and we enter the territory of infinite solutions or inconsistency.

Let us construct our matrix:

Expanding this along the first row, we perform the calculation with care. We take the first element, $2$, and multiply it by the minor $(2a-(-5))$. Then, we handle the middle term, $-(-1)$, which becomes $+1$, multiplied by $(3a-(-4))$. Finally, we add $3$ times $(15-8)$.

Simplifying this, we get $4a+10+3a+4+21$, which collapses beautifully into $7a+35$, or $7(a+5)$. This is our critical threshold. If $a=-5$, the system collapses. If $a eq -5$, the system is stable.

Now, let's look at Option D. It claims a unique solution for $a eq -5$. Since we found that $D=7(a+5)$, it is clear that as long as $a eq -5$, $D$ will never be zero. Therefore, a unique solution is guaranteed.

Option D is correct. But we are looking for the 'NOT correct' statement. We must dig deeper.

When $a=-5$, the system is on the edge of chaos. To see if it is consistent (infinitely many solutions) or inconsistent (no solution), we must look at the auxiliary determinant $D_3$. We replace the third column of our matrix with the constants $(5,7,\beta )$:

Expanding this, we get $2(2\beta -35)+1(3\beta -28)+5(15-8)$. Simplifying this yields $4\beta -70+3\beta -28+35$, which simplifies to $7\beta -63$, or $7(\beta -9)$.

Now, let's test the remaining options with our new knowledge.

For Option A, we test $a=-5$ and $\beta =9$. If $a=-5$, $D=0$. If $\beta =9$, $D_3=7(9-9)=0$. When $D=0$ and $D_3=0$, we have the condition for infinitely many solutions. Option A is correct.

For Option C, we test $a=-5$ and $\beta =8$. Again, $D=0$. But $D_3=7(8-9)=-7$. Since $D=0$ but $D_3 eq 0$, the system is inconsistent. Option C is correct.

Finally, we arrive at Option B. It claims infinitely many solutions for $a=-6$ and $\beta =9$. Let's check $D$ for $a=-6$:

Since $D=-7$, which is clearly not zero, the system must have a unique solution. It cannot have infinitely many solutions. The claim in Option B is mathematically impossible. We have found our culprit!

The Elegance of Logic

Do you see the beauty here? We didn't need to solve for $x,y,$ and $z$ individually.

By analyzing the 'DNA' of the system—the determinants—we could predict the behavior of the entire system without ever finding the specific coordinates. This is the power of linear algebra. It allows us to see the forest, not just the trees.

Keep practicing this logical dissection, and you will find that even the most intimidating JEE problems become clear, elegant, and solvable.