Animated Solution for Mathematics - Matrices and Determinants: If $a>0$ and discriminant of $a{x}^2+2bx+c$ is $-ve$, then $\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$ is equal to

Welcome, student. Today, we are going to peel back the layers of a problem that, at first glance, looks like a terrifying wall of algebra. We have a determinant, a quadratic expression, and a condition on the discriminant.

It is easy to feel overwhelmed, but I want you to take a deep breath. In JEE Advanced, the most complex-looking problems are often the ones that hide the most elegant, simple truths.

Let us start by visualizing the quadratic expression $f(x)=a{x}^2+2bx+c$. We are told that $a>0$, which tells us our parabola opens upwards.

The real magic lies in the discriminant, $D=(2b{)}^2-4ac<0$. Because $D<0$, the equation $a{x}^2+2bx+c=0$ has no real roots.

Geometrically, this means the parabola never touches or crosses the x-axis. It floats entirely above it! This is our first pillar of truth: for any real value of $x$, $a{x}^2+2bx+c>0$. Keep this in your pocket; we will need it for the final act.

Now, let us turn our attention to the determinant $\Delta$:

When you see a determinant like this, do not rush to expand it using the standard formula. That is a trap!

Look at the third column. The elements are $ax+b$ and $bx+c$. Do you see the pattern? They are linear combinations of the first two columns.

Specifically, if you take the first column and multiply it by $x$, then add the second column, you get exactly the third column. This is not a coincidence; it is an invitation to use row operations to simplify our lives.

Creating Zeros

Our goal is to create as many zeros as possible in the third row. We apply the row operation $R_3\to R_3-(x{R}_1+R_2)$.

For the first column, we take $(ax+b)$ and subtract $(x\cdot a+b)$, which leaves us with $0$. For the second column, we take $(bx+c)$ and subtract $(x\cdot b+c)$, which also leaves us with $0$.

We have successfully created two zeros in the third row! Now, for the final element in the third row, we take $0$ and subtract the combination $x(ax+b)+(bx+c)$.

Expanding this, we get $-(a{x}^2+bx+bx+c)$, which simplifies beautifully to $-(a{x}^2+2bx+c)$.

Our determinant now looks like this:

Expanding along the third row is now trivial. We get:

This simplifies to $-(a{x}^2+2bx+c)(ac-b^2)$. We can absorb that negative sign into the second bracket to get $(a{x}^2+2bx+c)(b^2-ac)$.

Now, recall our pillars of truth: $a{x}^2+2bx+c$ is always positive, and $b^2-ac$ is always negative (since $4(b^2-ac)<0$). A positive number multiplied by a negative number is always negative.

And there you have it! The determinant is negative. You have just navigated a complex problem by trusting the geometry and the symmetry of the matrix. Well done.