Q.Using the properties of determinants, prove that: $\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \end{vmatrix} = 0$

Key idea: the messy powers clear if you multiply the rows by $x,y,z$; two columns then become proportional, forcing the determinant to $0$.

Step 1 — multiply $R_1,R_2,R_3$ by $x,y,z$ (this multiplies the determinant by $xyz$):

Step 2 — column 1 is $xyz\,(yz,\,zx,\,xy)$ and column 2 is $xyz\,(1,1,1)$; factor $xyz$ from each:

$xyz\,\Delta = x^2y^2z^2\begin{vmatrix} yz & 1 & x(y+z) \\ zx & 1 & y(z+x) \\ xy & 1 & z(x+y) \end{vmatrix}.$ …

Intuition

A determinant is zero exactly when its columns are linearly dependent. Here the first two columns look unrelated to the third — until you scale the rows by $x,y,z$. That scaling turns the entries into a symmetric pattern in which one column operation exposes two proportional columns.

Setting up

Working the steps

1. Scale the rows. Multiply $R_1,R_2,R_3$ by $x,y,z$ respectively. This multiplies the whole determinant by $xyz$:

2. Factor the columns. Column 1 is $xyz\,(yz,\,zx,\,xy)$ and column 2 is $xyz\,(1,1,1)$:

3. One column operation. Apply $C_3 \to C_3+C_1$:

So every entry of the new third column is $xy+yz+zx$, i.e. $C_3=(xy+yz+zx)\,C_2$: …

Method: Proving a Determinant is Zero via Linear Dependence

A determinant is exactly zero when its rows (or columns) are linearly dependent — two rows/columns are proportional, identical, or one is a combination of the others. When a "prove this determinant is zero" question has entries that are ratios or mixed powers, the strategy is to clear denominators by scaling, then manufacture that dependence.

Steps

Step 1: Scale rows (or columns) to clear awkward factors

If the entries mix different powers of the same variables, multiply each row by a chosen variable to make the entries homogeneous. Remember: multiplying a row by $k$ multiplies the whole determinant by $k$, so track that factor — you will divide it back out at the end.

Step 2: Factor out common terms from the resulting columns

After scaling, look for a common factor sitting in each column and pull it out using the column-scaling property. This usually simplifies the matrix into a much cleaner symmetric pattern. …

Common Mistakes

Mistake 1: Forgetting the scaling multiplies the whole determinant

Multiplying rows $R_1,R_2,R_3$ by $x,y,z$ respectively scales the entire determinant by $xyz$, not by $x+y+z$ or by nothing at all. Skipping this bookkeeping means the equation $xyz\,\Delta=0$ never gets set up correctly.

Mistake 2: Concluding $\Delta=0$ without addressing $xyz=0$

The identity $xyz\,\Delta=0$ only forces $\Delta=0$ directly when $xyz\neq0$. A fully honest proof needs to note the case where any of $x,y,z$ is $0$ separately (a column of $\Delta$ then has repeated/zero entries), otherwise the argument has a logical gap even though the final answer is right. …