Animated Solution for Mathematics - Matrices and Determinants: Without expanding a determinant at any stage, show that $\left|\begin{array}{ccc}{x}^2+x& x+1& x-2\\ 2x^2+3x-1& 3x& 3x-3\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|=xA+B$, where $A$ and $B$ are determinants of order $3$ not involving $x$.

Mastering Determinants

Welcome, future engineers. Today, we are not just solving a determinant; we are performing a surgical operation on a mathematical structure.

When you see a problem like this in the JEE Advanced paper, your first instinct might be to panic at the sight of $x^2$ terms. But I want you to take a deep breath. We are going to dismantle this problem piece by piece, without ever resorting to the brute force of expansion.

The Quadratic Menace

Look at the initial determinant:

The enemy here is the $x^2$ term. It makes the determinant look bulky and unmanageable. Our goal is to simplify.

We look at the first column and realize that if we can eliminate these quadratic terms, the problem will collapse into something much simpler. We apply the row operations $R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$.

By doing this, we are not just changing numbers; we are stripping away the complexity. The first element of the second row becomes $(2x^2+3x-1)-2(x^2+x)=x-1$.

Suddenly, the quadratic term is gone. We have successfully reduced the degree of the polynomial. This is the first victory in our journey.

The Power of Zero

After our first pass, we are left with a new determinant:

Now, pause. Look at the second and third rows. Do you see it? The second column has $x-2$ in both rows, and the third column has $x+1$ in both rows.

This is not a coincidence; it is a gift. In the world of determinants, identical elements are an invitation to create zeros.

We apply $R_3\to R_3-R_2$. The second and third elements of the third row vanish into zero. The first element becomes $(x+3)-(x-1)=4$. We now have a row with a constant and two zeros. This is the most powerful position you can be in.

The Final Split

We are in the endgame. We have a constant $4$ in the third row. We use this to clear out the remaining $x$ terms in the first column.

By applying $R_1\to R_1-\frac{x^2+x}{4}{R}_3$ and $R_2\to R_2-\frac{x-1}{4}{R}_3$, we make the first column look like $(0,0,4{)}^T$. Now, we have a determinant where the first column is constant, but $x$ is still lurking in the second and third columns.

To reach our goal of $xA+B$, we need $x$ in only one column. We apply $C_2\to C_2-C_3$. This transforms the second column into $(3,-3,0{)}^T$.

Now, only the third column contains $x$. We can finally use the addition property of determinants to split the third column into two: one part containing $x$ and one part containing the constants.

We are left with:

This is exactly $xA+B$. We did not expand. We did not struggle with massive polynomials. We simply manipulated the structure until the answer revealed itself.

This is the elegance of mathematics. Keep practicing this mindset, and you will find that no determinant is too complex to conquer.