If the determinant of matrix with rows (a, a², 1+a³), (b, b², 1+b³), (c, c², 1+c³) is 0 and vectors

Q: Wait, why can we just split the determinant like that? Is that a standard property?

Yes, it is! The determinant is a linear function of each of its columns. If any column is a sum of two vectors, the determinant can be expressed as the sum of two determinants, where each determinant takes one part of the sum as its column. It is a powerful tool for simplifying complex expressions like this one.

Q: Why does swapping columns change the sign of the determinant?

This is a fundamental property of determinants. Swapping any two rows or any two columns effectively multiplies the determinant by $-1$. Think of it as a reflection in the geometry of the transformation; it reverses the orientation of the basis vectors, which is why the sign flips.

Q: What does 'non-coplanar' actually mean in terms of the determinant?

Geometrically, three vectors are non-coplanar if they do not lie in the same plane, meaning they span a 3D volume. The scalar triple product of three vectors, which is calculated as the determinant of the matrix formed by their components, gives the volume of the parallelepiped they span. If the volume is non-zero, the vectors are non-coplanar, and thus the determinant must be non-zero.

Question

If $\begin{vmatrix} a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3 \end{vmatrix} = 0$ and vectors $(1, a, a^2), (1, b, b^2)$ and $(1, c, c^2)$ are non-coplanar, then the product $abc$ equals

Options:

(A) 0

(B) 2

(C) -1

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring engineers! Today, we are going to unravel a problem that might look intimidating at first glance, but beneath the surface, it is a masterclass in the beauty of linear algebra.

We are dealing with a determinant equation involving variables $a$, $b$, and $c$, and a condition about non-coplanar vectors. Let us embark on this journey together.

### The Art of Splitting

We start with the given determinant equation:

$$
\begin{vmatrix} a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3 \end{vmatrix} = 0
$$

The third column is the key. It is a sum of two terms: $1$ and $a^3$, $1$ and $b^3$, and $1$ and $c^3$.

Using the linearity property of determinants, we can split this into two separate determinants:

$$
\begin{vmatrix} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{vmatrix} = 0
$$

This is the first step in our simplification. We have taken a complex structure and broken it down into two manageable pieces.

### The Transformation

Now, let us focus on the second determinant:

$$
\begin{vmatrix} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{vmatrix}
$$

Notice that the first row has a common factor of $a$, the second row has $b$, and the third row has $c$. By factoring these out, we get:

$$
abc \begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix}
$$

Now, let us turn to the first determinant:

$$
\begin{vmatrix} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{vmatrix}
$$

We want this to match the second one. By swapping column three with column two, and then column two with column one, we perform two swaps.

Each swap introduces a negative sign, so two swaps result in a positive sign. The determinant becomes:

$$
\begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix}
$$

### The Geometric Connection

We now have the following expression:

$$
(1 + abc) \begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix} = 0
$$

The problem states that the vectors $\vec{u} = (1, a, a^2)$, $\vec{v} = (1, b, b^2)$, and $\vec{w} = (1, c, c^2)$ are non-coplanar. Geometrically, this means they span a non-zero volume in 3D space.

The scalar triple product, which is exactly our determinant, must therefore be non-zero. Since the determinant is non-zero, the only way the equation holds is if $1 + abc = 0$.

Thus, we find that **$abc = -1$**. It is a beautiful result, isn't it? The geometry of the vectors and the algebra of the determinants come together perfectly to give us the answer.

Animated Solution for Mathematics - Matrices and Determinants: If $a b c a^{2} b^{2} c^{2} 1 + a^{3} 1 + b^{3} 1 + c^{3} = 0$ and vectors $(1, a, a^{2}), (1, b, b^{2})$ and $(1, c, c^{2})$ are non-coplanar, then the product $ab c$ equals

Select Answer:

Visualized Solution

Analyze the Given Determinant & Vectors

Splitting the Determinant

Factoring the Second Determinant

Transforming the First Determinant

Combining the Terms

Understanding the Vector Condition

Applying the Non-Coplanar Condition

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Art of Splitting

The Transformation

The Geometric Connection

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