Use row/column operations to simplify a 3×3 determinant.

Properties of Determinants — Row Operations

Question

Evaluate the determinant using row/column operations:

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

This is the Vandermonde determinant — one of the most frequently tested determinant forms in CBSE 12 and JEE Main. It appeared in CBSE 2024 Board Exam and similar forms show up almost every year.

Solution — Step by Step

We subtract Row 1 from Rows 2 and 3. The goal is to create zeros in the first column — this makes expansion much cleaner.

\Delta = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}

Notice that $b^2 - a^2 = (b-a)(b+a)$ and $c^2 - a^2 = (c-a)(c+a)$ . We factor out $(b-a)$ from R₂ and $(c-a)$ from R₃.

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

Pulling constants out of rows is one of the most powerful tools we have — it simplifies the determinant without changing its structure.

Now subtract the new Row 2 from Row 3:

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

The matrix is now upper triangular.

For an upper triangular matrix, the determinant is simply the product of diagonal elements.

\Delta = (b-a)(c-a) \times (1 \times 1 \times (c-b))

\boxed{\Delta = (b-a)(c-a)(c-b)}

Why This Works

The key property we used: row operations of the type $R_i \to R_i - R_j$ do not change the value of the determinant. This is because we’re adding a scalar multiple of one row to another — a fundamental property of determinants.

When we factored $(b-a)$ from R₂ and $(c-a)$ from R₃, we used the property that a scalar can be taken outside from any single row. These two operations together are what reduce an ugly $3 \times 3$ into a clean triangular form.

Upper triangular determinants are the endgame we’re always aiming for. Once you have zeros below the diagonal, the answer is just the diagonal product — no cofactor expansion needed.

Alternative Method

You can also expand directly using cofactor expansion along Column 1, since after Step 1 we already have two zeros there.

\Delta = 1 \cdot \begin{vmatrix} b-a & b^2-a^2 \\ c-a & c^2-a^2 \end{vmatrix}

= (b-a)(c^2-a^2) - (c-a)(b^2-a^2)

= (b-a)(c-a)(c+a) - (c-a)(b-a)(b+a)

= (b-a)(c-a)\left[(c+a)-(b+a)\right] = (b-a)(c-a)(c-b)

Same answer, slightly more algebra. The triangularization method in the main solution is faster under exam pressure.

The result $(b-a)(c-a)(c-b)$ can also be written as $(a-b)(b-c)(c-a)$ after factoring out $-1$ three times (net sign: $(-1)^3 = -1$ , so flip one pair). Both forms are correct — just verify signs if the question asks for a specific form.

Common Mistake

Most students forget to factor correctly in Step 2. After R₂ → R₂ − R₁, Row 2 becomes $[0,\ b-a,\ b^2-a^2]$ . When factoring out $(b-a)$ , the last entry becomes $\frac{b^2-a^2}{b-a} = b+a$ , not $b-a$ . Writing $b-a$ instead of $b+a$ here is the single most common error on this problem in boards — it gives a wrong final answer even though all the steps look right.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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