Q.Prove that : $\begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} = (a-b)(b-c)(c-a)(ab+bc+ca)$

Determinant Evaluation Using Identities

Expanding a $4\times4$ or $5\times5$ determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.

The geometric intuition

A determinant measures the signed "volume" of the box spanned by the rows in $n$-dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.

The three row (or column) operations

The identical rules hold for columns. There is also row-wise linearity: if a row is a sum $R_i = R_i' + R_i''$, the determinant splits into the sum of two determinants with all other rows fixed.

The strategy

1. Use operation 3 to create zeros in a row or column (value unchanged).
2. Factor out common factors with operation 2.
3. Swap rows if needed to reach upper-triangular form (track the sign change).
4. The determinant is then the product of the diagonal entries.

Worked example

Apply $R_2\to R_2-4R_1$ and $R_3\to R_3-7R_1$ (no change), then $R_3\to R_3-2R_2$: …