Q.Using the properties of determinants, prove that: $\begin{vmatrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{vmatrix} = 4xyz$

Key idea: add all rows to the first to expose a factor of $2$, then clear that row down to $(0,-x,-x)$ and expand.

Step 1 — $R_1 \to R_1+R_2+R_3$ makes the first row $\big(2(y+z),\,2(x+z),\,2(x+y)\big)$; factor out $2$:

Step 2 — $R_1 \to R_1-R_2$ then $R_1 \to R_1-R_3$ turn the top row into $(0,-x,-x)$; factor out $-x$:

$\Delta = -2x\begin{vmatrix} 0 & 1 & 1 \\ z & z+x & x \\ y & x & x+y \end{vmatrix}.$ …

Intuition

The matrix is built symmetrically from $x,y,z$, so the trick is to combine rows to manufacture common factors. Adding the rows together gives a first row that is $2\times$ a simpler row, and subtracting the other rows back out strips it down to something with a single common factor $-x$.

Setting up

Working the steps

1. Add all rows to the first: $R_1 \to R_1+R_2+R_3$ gives first row $\big(2y+2z,\;2z+2x,\;2x+2y\big)$. Factor $2$:

2. Strip the first row down. Do $R_1 \to R_1-R_2$: it becomes $\big((y+z)-z,\ (x+z)-(z+x),\ (x+y)-x\big)=(y,0,y)$. Then $R_1 \to R_1-R_3$: $\big(y-y,\ 0-x,\ y-(x+y)\big)=(0,-x,-x)$:

3. Factor $-x$ from the first row:

$\Delta = -2x\begin{vmatrix} 0 & 1 & 1 \\ z & z+x & x \\ y & x & x+y \end{vmatrix}.$ …

Method: Proving a Determinant Identity via Row Reduction and Expansion

This method proves that a symmetric $3\times3$ determinant equals a specific product (like $4xyz$ or a similar closed form) by reducing it to a matrix with a zero entry and then expanding the resulting $2\times2$ block.

Steps

Step 1: Combine all rows to expose a common factor

Apply $R_1 \to R_1+R_2+R_3$ (or whichever row/column combination the symmetry of the matrix suggests). In a symmetric determinant built from variables like $x,y,z$, this usually doubles a simpler row, letting you factor out a constant like $2$.

Step 2: Subtract the other rows to strip the combined row down further

With the doubled row in place, subtract the remaining rows one at a time ($R_1 \to R_1-R_2$, then $R_1 \to R_1-R_3$) until that row contains mostly zeros and one or two clean terms.

Step 3: Factor out any remaining common term …

Common Mistakes

Mistake 1: Subtracting the original row instead of the updated one

The sequence $R_1\to R_1-R_2$ followed by $R_1\to R_1-R_3$ must use the already-updated first row for the second subtraction. Reusing the original row both times silently performs a different (wrong) operation.

Mistake 2: Dropping the sign when factoring $-x$

The reduced first row is $(0,-x,-x)$. Factoring out $x$ instead of $-x$ flips the sign of the whole determinant, turning the correct $4xyz$ into $-4xyz$. …