Q.Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.

Concept: Determinant Evaluation Using Cofactors

We expand along the third column because its entries are $yz$, $zx$, $xy$ — this will later reveal a factorisation.

Step 1: Cofactor expansion along column 3

where $C_{ij} = (-1)^{i+j} M_{ij}$.

Step 2: Compute each cofactor

• $C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = (1)(z - y)$
• $C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = (-1)(z - x) = x - z$
• $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = (1)(y - x)$

Step 3: Substitute and simplify

Expand each term: …

Why expand along the third column?

When a determinant has a column (or row) with entries that are products of variables, expanding along that column often reveals a common factor structure. Here, the third column contains $yz$, $zx$, $xy$ — each is the product of the two variables not appearing in that row's first column entry. This pattern suggests the determinant will factor nicely into differences of the variables.

Expanding along the third column means we compute:

where $C_{ij}$ is the cofactor of the element in row $i$, column $j$.

Step-by-step expansion

1. Find the cofactor $C_{13}$ (for element $yz$ in row 1, column 3)

The minor $M_{13}$ is the determinant of the matrix obtained by deleting row 1 and column 3:

The cofactor $C_{13} = (-1)^{1+3} M_{13} = (+1)(z - y) = z - y$.

2. Find the cofactor $C_{23}$ (for element $zx$ in row 2, column 3)

Delete row 2 and column 3:

Cofactor $C_{23} = (-1)^{2+3} M_{23} = (-1)(z - x) = x - z$.

3. Find the cofactor $C_{33}$ (for element $xy$ in row 3, column 3)

Delete row 3 and column 3:

Cofactor $C_{33} = (-1)^{3+3} M_{33} = (+1)(y - x) = y - x$.

4. Assemble the expansion

Now expand each term:

• $yz(z - y) = yz^2 - y^2z$
• $zx(x - z) = zx^2 - z^2x$
• $xy(y - x) = xy^2 - x^2y$

So:

5. Factor the expression

Group terms by common factors. Notice the expression is antisymmetric — swapping any two variables changes the sign. This suggests a factor of $(x-y)(y-z)(z-x)$.

Let's verify by expanding $(x-y)(y-z)(z-x)$: …

Method: Expanding Along a Column Chosen for Its Factorable Structure

Use this when the question specifies a row/column whose entries hint at a hidden factorisation (e.g. products of "the other" variables).

Steps

Step 1: Expand along the specified column using cofactors

Step 2: Compute each $2\times2$ minor and attach its sign carefully

The middle term (row $2$) always carries a negative sign, since $(-1)^{2+3}=-1$ regardless of which column is being expanded.

Step 3: Assemble the raw polynomial expression

Substitute the cofactors into $\Delta$ and expand each product fully. …

Common Mistakes

Mistake 1: Forgetting the alternating sign on the middle cofactor

Why it's wrong: for column $3$, the middle term's cofactor always carries a factor of $(-1)^{2+3}=-1$ — dropping this sign flips that term and corrupts the whole assembled expression. Correct approach: write $(-1)^{i+j}$ explicitly for every cofactor before substituting its minor, especially the middle one.

Mistake 2: Leaving the final answer as an unfactored sum of six terms instead of recognising the compact factored form …