Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $x-2y+z=-4,2x+\alpha y+3z=5,3x-y+\beta z=3$ has infinitely many solutions, then $12\alpha +13\beta$ is equal to

Imagine standing in a 3D room. Each equation in our system, $x-2y+z=-4$, $2x+\alpha y+3z=5$, and $3x-y+\beta z=3$, represents a flat plane slicing through space.

Usually, three planes intersect at a single point—a unique solution. But today, we are dealing with something far more elegant: infinitely many solutions.

This means these three planes are not just meeting at a point; they are meeting along a common line, or perhaps they are even coincident. This is a state of perfect geometric harmony. To find the parameters $\alpha$ and $\beta$ that allow this harmony to exist, we must turn to the language of matrices and determinants.

Cramer's Rule

In the world of JEE Advanced, Cramer's Rule is your most trusted ally for $3\times 3$ systems. The condition for infinitely many solutions is strict.

It demands that the main determinant, $D$, which captures the essence of the coefficient matrix, must vanish: $D=0$. But that is only the first step.

If $D=0$, the system is either inconsistent or has infinite solutions. To ensure the latter, we must also ensure that the sub-determinants $D_1,D_2,$ and $D_3$—where we replace the columns of $x,y,$ and $z$ with the constant vector—are also zero. This is the gatekeeper condition.

Solving for $\beta$

Now, we must be tactical. We have two unknowns, $\alpha$ and $\beta$. If we jump straight into the main determinant $D$, we will be stuck with an equation involving both variables, which is a headache we do not need.

Instead, look at $D_2$. By replacing the $y$-coefficients with the constants, we isolate $\beta$. Let us set up the determinant:

Expanding this along the first row is straightforward. We calculate $1(5\beta -9)+4(2\beta -9)+1(6-15)=0$.

Simplifying this, we get $5\beta -9+8\beta -36-9=0$, which collapses beautifully into $13\beta -54=0$. Thus, we have our first victory: $\beta =\frac{54}{13}$.

Finding $\alpha$

With $\beta$ in our pocket, the main determinant $D$ is no longer a mystery. Let us write it out:

Expanding this, we get $1(\alpha \beta +3)+2(2\beta -9)+1(-2-3\alpha )=0$. This simplifies to $\alpha \beta +3+4\beta -18-2-3\alpha =0$, or more cleanly, $\alpha \beta -3\alpha +4\beta -17=0$.

Now, we substitute our known value of $\beta =\frac{54}{13}$ into this equation. It might look like a fraction-heavy nightmare, but stay calm. We have $\alpha (\frac{54}{13})-3\alpha +4(\frac{54}{13})-17=0$.

Grouping the $\alpha$ terms, we get $\alpha (\frac{54-39}{13})+\frac{216-221}{13}=0$. This simplifies to $\frac{15\alpha }{13}-\frac{5}{13}=0$, which leads us directly to $15\alpha =5$, or $\alpha =\frac{1}{3}$.

We have conquered the variables. We found $\alpha =\frac{1}{3}$ and $\beta =\frac{54}{13}$. The final step is simply to calculate the value requested: $12\alpha +13\beta$.

Substituting our values:

Look at that! The fractions vanish, the complexity dissolves, and we are left with a clean, integer answer. This is the beauty of mathematics.

No matter how intimidating the parameters look at the start, if you follow the logical path, the universe of the problem eventually aligns. You have successfully navigated the system, and that is the mark of a true problem solver. The final answer is 58.