Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $x+y+az=b$, $2x+5y+2z=6$, $x+2y+3z=3$ has infinitely many solutions, then $2a+3b$ is equal to

Imagine you are standing in a 3D room. You have three planes, each represented by one of our equations: $x+y+az=b$, $2x+5y+2z=6$, and $x+2y+3z=3$.

Usually, these three planes intersect at a single point, giving us a unique solution. But today, the problem gives us a special condition: the system has infinitely many solutions.

This means these planes are not just meeting at a point; they are dancing together along a common line or even overlapping entirely. This geometric reality is the key to our algebraic approach.

To unlock this, we turn to Cramer's Rule. In the world of linear algebra, the determinant of the coefficient matrix, which we call $\Delta$, acts as a gatekeeper.

If $\Delta eq 0$, the gate is open for a unique solution. But when $\Delta =0$, the gate closes on uniqueness, and we enter the realm of either 'no solution' or 'infinitely many solutions'.

To guarantee the latter, we must ensure that the auxiliary determinants—${\Delta }_1,{\Delta }_2,\text{ and }{\Delta }_3$—are also zero. Think of these as the 'consistency checks' for our system.

Solving for $a$

Let's start by setting the main determinant $\Delta$ to zero. We extract the coefficients of $x,y,\text{ and }z$ to build our matrix:

Expanding this along the first row, we get:

Let's simplify this carefully. The first term is $1(15-4)=11$. The second is $-1(6-2)=-4$. The third is $a(4-5)=-a$.

So, our equation becomes $11-4-a=0$, which simplifies beautifully to $7-a=0$. Thus, we find $a=7$. We have our first piece of the puzzle!

Solving for $b$

Now that we know $a=7$, we need to find $b$. We use the condition ${\Delta }_1=0$. We replace the first column of our original determinant with the constants from the right side of our equations ($b,6,3$):

Expanding this again along the first row:

Calculating the terms: $b(15-4)-1(18-6)+7(12-15)=0$. This gives us $11b-12+7(-3)=0$, which is $11b-12-21=0$.

So, $11b=33$, and finally, $b=3$.

We have arrived at the finish line. With $a=7$ and $b=3$, the question asks for the value of $2a+3b$.

Substituting our values, we get:

It is a satisfying conclusion to a problem that seemed daunting at first. Remember, in JEE Advanced, the complexity is often just a mask for fundamental principles. Keep visualizing, keep calculating, and trust the math! The final answer is $23$.