Q.Using the properties of determinants, evaluate: $\begin{vmatrix} 3x & -x + y & -x + z \\ x - y & 3y & z - y \\ x - z & y - z & 3z \end{vmatrix}$

Key idea: the entries of each row add to $x+y+z$, so fold the columns into the first and factor it out.

Step 1 — $C_1 \to C_1+C_2+C_3$. Each new first-column entry is $x+y+z$:

Step 2 — $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$:

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

Intuition

The three entries in each row of this determinant add to the same total, $x+y+z$. That is the signal to add all columns into one column: it turns the first column into a constant you can factor straight out.

Setting up

Working the steps

1. Fold the columns in: $C_1 \to C_1+C_2+C_3$. Row 1: $3x+(-x+y)+(-x+z)=x+y+z$; row 2: $(x-y)+3y+(z-y)=x+y+z$; row 3: $(x-z)+(y-z)+3z=x+y+z$:

2. Make zeros in the first column: $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$:

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

Method: Row-Sum Factoring by Column-Folding

This method applies whenever every row of a determinant adds up to the same expression — the fastest route is to fold all the columns into one to expose that expression as a common factor, then reduce what's left to a $2\times2$ block.

Steps

Step 1: Check whether every row's entries sum to the same expression

Add the entries across each row separately. If all three sums come out equal to some expression $S$ (often a symmetric combination like $x+y+z$), that is the signal to use column-folding rather than direct expansion — direct expansion of a $3\times3$ with mixed-sign linear entries is slow and error-prone.

Step 2: Fold the columns into the first column

Apply $C_1 \to C_1 + C_2 + C_3$. Because every row sums to $S$, the entire new first column becomes $S,S,S$ — a column of a repeated value, which factors straight out:

Step 3: Zero out the first column below the top row …

Common Mistakes

Mistake 1: Folding the wrong columns together

Applying $C_1 \to C_1+C_2+C_3$ but adding only two of the three columns (or adding rows instead of columns) means the first column never collapses to the common value $x+y+z$ — the factoring step that makes the whole problem tractable is lost right at the start.

Mistake 2: Only zeroing part of a row

After factoring $(x+y+z)$ out of the first column, the operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ must be applied across the entire row, not just the first entry. A student who zeros the first column but leaves the other two entries untouched ends up with a determinant that still doesn't reduce to a simple $2\times2$ block. …