Animated Solution for Mathematics - Matrices and Determinants: If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and the vectors $\vec{A}=(1,a,a^2)$, $\vec{B}=(1,b,b^2)$, $\vec{C}=(1,c,c^2)$, are non-coplanar, then the product $abc=.........$

Unlocking the Determinant

Welcome, future engineer. Today, we are not just solving a math problem; we are peeling back the layers of a beautiful geometric structure.

When you look at a determinant like

your first instinct might be to expand it. Resist that urge. In the world of JEE Advanced, brute force is rarely the intended path. Instead, let us look at the anatomy of this matrix.

The Power of Linearity

Look at that third column. It is not a simple list of numbers; it is a sum: $1+a^3$, $1+b^3$, and $1+c^3$. This is a massive hint.

In linear algebra, the determinant is a linear function of its columns. This means if a column is a sum, we can split the determinant into two separate, cleaner pieces.

We keep the first two columns, $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$ and $\left(\begin{array}{c}{a}^2\\ b^2\\ c^2\end{array}\right)$, identical in both new determinants. In the first, we place the $1$s in the third column. In the second, we place the cubic terms $a^3,b^3,c^3$.

Suddenly, the problem feels much lighter.

The Dance of Symmetry

Now, let us examine our second determinant:

Do you see the pattern? Every row has a common factor. Row one has $a$, row two has $b$, and row three has $c$.

By factoring $a,b,$ and $c$ out of their respective rows, we are left with:

Now, look at our first determinant from the split:

It is almost identical to the one we just factored, but the columns are in the wrong order. We need the $1$s in the first column.

We perform a swap of column two and column three (which flips the sign to negative), and then a swap of column one and column two (which flips the sign back to positive). Two swaps, two sign changes—the determinant remains unchanged. We have achieved perfect symmetry.

The Geometric Bridge

We now have:

This is where the physics of the problem kicks in. We are given vectors $\vec{A}=(1,a,a^2)$, $\vec{B}=(1,b,b^2)$, and $\vec{C}=(1,c,c^2)$.

The determinant we are looking at is the Scalar Triple Product of these vectors. The problem tells us these vectors are non-coplanar.

Geometrically, this means they span a 3D volume; they do not lie flat on a plane. Therefore, the volume of the parallelepiped they form cannot be zero. This implies that our determinant $\Delta eq 0$.

Since we know the determinant is not zero, we can safely divide both sides of our equation by it. We are left with the simple, elegant truth:

Therefore, $abc=-1$.

You see? We didn't need to expand a single complex polynomial. We used the properties of the determinant to reveal the underlying geometry, and the answer emerged naturally. Keep this mindset—look for the structure, respect the geometry, and the math will always guide you home.