Without expanding prove that $\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)$.

> [!TLDR] > Pulling $x,y,z$ out of the columns leaves a Vandermonde determinant; two column operations then factor it as $xyz(x-y)(y-z)(z-x)$. > [!FORMULA] > A common factor of a column may be taken **outside** the determinant. > $C_i \to C_i - C_j$ leaves the determinant unchanged. 1. Start with $$\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}.$$ 2. Take $x$ common from $C_1$, $y$ from $C_2$, $z$ from $C_3$: $$\Delta = xyz\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}.$$ 3. Apply $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $$\Delta = xyz\begin{vmatrix} 0 & 0 & 1 \\ x-y & y-z & z \\ x^2-y^2 & y^2-z^2 & z^2 \end{vmatrix}.$$ 4. Factor $(x-y)$ from $C_1$ and $(y-z)$ from $C_2$, using $x^2-y^2=(x-y)(x+y)$ and $y^2-z^2=(y-z)(y+z)$: $$\Delta = xyz(x-y)(y-z)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & z \\ x+y & y+z & z^2 \end{vmatrix}.$$ …

Without expanding prove that Δ = x y z x² y² z² x³ y³ z³ = xyz(x-y)(y-z)(z-x).

Worked Examples · Example 30

\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)

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Concept understanding — Arithmetic Progression Determinant

What is an Arithmetic Progression Determinant?

You already know what an arithmetic progression (AP) is — a sequence where each term differs from the previous one by a fixed number called the common difference $d$ . For example, $2, 5, 8, 11, \dots$ is an AP with $d = 3$ .

Now imagine you take three terms from an AP — any three, not necessarily consecutive — and arrange them in a $3 \times 3$ determinant like this:

\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}

If the numbers in each row (or each column) form an arithmetic progression, something remarkable happens: the determinant is always zero. That is the core idea.

The Intuition

Why should that be true? Think about what a determinant measures. Geometrically, a $3 \times 3$ determinant gives the volume of a parallelepiped formed by three vectors. If those vectors are "linearly dependent" — meaning one can be written as a combination of the others — the volume collapses to zero.

When numbers in a row (or column) are in AP, that row is of the form:

a,\ a+d,\ a+2d

This is a linear function of the column index. The same linear pattern repeats across rows (or columns). That repetition creates a dependency: the second row is a linear combination of the first and third, or something similar. The determinant detects this dependency and returns zero.

Note

The determinant being zero does not mean the AP is "trivial" or the numbers are equal. It means the rows (or columns) are not independent — they are related by the AP structure.

The Precise Statement

3 \times 3

Let's write it clearly. Suppose we have a determinant:

\Delta = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

Case 1 — Rows in AP:

If $a_1, a_2, a_3$ are in AP, and $b_1, b_2, b_3$ are in AP, and $c_1, c_2, c_3$ are in AP, then $\Delta = 0$ .

Case 2 — Columns in AP:

If $a_1, b_1, c_1$ are in AP, and $a_2, b_2, c_2$ are in AP, and $a_3, b_3, c_3$ are in AP, then $\Delta = 0$ .

Watch out

The condition must hold for every row (or every column) simultaneously. If only one row is an AP and the others are not, the determinant is not necessarily zero.

A Quick Proof (for the curious)

Take the row-AP case. Let the first row be $a,\ a+d_1,\ a+2d_1$ , the second row $b,\ b+d_2,\ b+2d_2$ , and the third row $c,\ c+d_3,\ c+2d_3$ . Write the determinant:

\Delta = \begin{vmatrix} a & a+d_1 & a+2d_1 \\ b & b+d_2 & b+2d_2 \\ c & c+d_3 & c+2d_3 \end{vmatrix}

Now perform column operations: $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ . This gives:

\Delta = \begin{vmatrix} a & d_1 & 2d_1 \\ b & d_2 & 2d_2 \\ c & d_3 & 2d_3 \end{vmatrix}

Factor $2$ from the third column:

\Delta = 2 \begin{vmatrix} a & d_1 & d_1 \\ b & d_2 & d_2 \\ c & d_3 & d_3 \end{vmatrix}

Now the second and third columns are identical. A determinant with two equal columns is zero. Hence $\Delta = 0$ . …

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Pulling $x,y,z$ out of the columns leaves a Vandermonde determinant; two column operations then factor it as $xyz(x-y)(y-z)(z-x)$ .

C_i \to C_i - C_j

Start with

\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}.

Take $x$ common from $C_1$ , $y$ from $C_2$ , $z$ from $C_3$ :

\Delta = xyz\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}.

Apply $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$ :

\Delta = xyz\begin{vmatrix} 0 & 0 & 1 \\ x-y & y-z & z \\ x^2-y^2 & y^2-z^2 & z^2 \end{vmatrix}.

Factor $(x-y)$ from $C_1$ and $(y-z)$ from $C_2$ , using $x^2-y^2=(x-y)(x+y)$ and $y^2-z^2=(y-z)(y+z)$ : $\Delta = xyz(x-y)(y-z)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & z \\ x+y & y+z & z^2 \end{vmatrix}.$ …

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←Previous question · 2.6.5Example 29 — Given that a, b and c are in A.P., evaluate Δ = 2y+4 5y+7 8y+a \3y+5 6…→Next question · 2.6.5Example 31 — Without expanding, evaluate Δ = (b+c)² a² a² b² (c+a)² b² c² c² (a+b)²…

Q.Without expanding prove that Δ=∣xyzx2y2z2x3y3z3∣=xyz(x−y)(y−z)(z−x)\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)Δ=​xx2x3​yy2y3​zz2z3​​=xyz(x−y)(y−z)(z−x).

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Q.Without expanding prove that $\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)$ .

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