Animated Solution for Mathematics - Matrices and Determinants: If $f(x)=\left|\begin{array}{ccc}{x}^3& 2x^2+1& 1+3x\\ 3x^2+2& 2x& x^3+6\\ x^3-x& 4& x^2-2\end{array}\right|$ for all $x\in \mathbb{R}$, then $2f(0)+f^{\prime }(0)$ is equal to

Welcome, future engineers! Today, we are going to dismantle a problem that often intimidates students in the JEE Advanced examination. We are given a function $f(x)$ defined as a $3\times 3$ determinant, and we need to find the value of $2f(0)+f^{\prime }(0)$.

At first glance, you might be tempted to expand the determinant into a massive polynomial. Resist that urge! In the world of competitive mathematics, the path of least resistance is often the most elegant one.

The Snapshot at the Origin

Our first objective is to find $f(0)$. This is our "snapshot" of the function at the origin. By substituting $x=0$ directly into the determinant, the expression simplifies significantly:

Expanding this along the first column—which is a strategic choice because it contains two zeros—we get:

Just like that, we have our first piece of the puzzle: $f(0)=12$.

The Row-Wise Differentiation Rule

Now, for the main event: $f^{\prime }(0)$. Many students panic here, thinking they must differentiate the entire expanded polynomial.

But we have a secret weapon: the row-wise differentiation rule. This rule states that the derivative of a determinant is the sum of three determinants, where in each, we differentiate exactly one row while keeping the others constant:

Since we only need the value at $x=0$, we can differentiate the rows and substitute $x=0$ immediately. This is much faster than finding the general derivative $f^{\prime }(x)$.

The Execution

Let us calculate each component:

1. For ${\Delta }_1(0)$: We differentiate the first row: $[x^3{]}^{\prime }=3x^2$, $[2x^2+1{]}^{\prime }=4x$, $[1+3x{]}^{\prime }=3$. At $x=0$, this row becomes $[0,0,3]$.

2. For ${\Delta }_2(0)$: We differentiate the second row: $[3x^2+2{]}^{\prime }=6x$, $[2x{]}^{\prime }=2$, $[x^3+6{]}^{\prime }=3x^2$. At $x=0$, this becomes $[0,2,0]$.

Notice the first column is all zeros? The determinant is $0$.

3. For ${\Delta }_3(0)$: We differentiate the third row: $[x^3-x{]}^{\prime }=3x^2-1$, $[4{]}^{\prime }=0$, $[x^2-2{]}^{\prime }=2x$. At $x=0$, this becomes $[-1,0,0]$.

Summing these up, we find $f^{\prime }(0)=24+0-6=18$. Finally, we compute our target expression:

And there you have it! By using the row-wise differentiation rule, we avoided the algebraic nightmare of expanding the determinant. Always look for the structural properties of the math—they are your best friends in the exam hall.