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

Key idea: every row of the determinant adds to the same value $a+x+y+z$, so fold all columns into the first and pull that factor out.

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

Step 2 — $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$ give a triangular determinant:

Intuition

Whenever every row of a determinant adds up to the same thing, that common sum is hiding as a factor. You reveal it by folding all the columns into one, then clear the rest to a triangle whose diagonal you read straight off.

Setting up

Working the steps

1. Fold the columns in: apply $C_1 \to C_1+C_2+C_3$. Row by row the first entry becomes $(a+x)+y+z$, $x+(a+y)+z$, $x+y+(a+z)$ — all equal to $a+x+y+z$:

2. Pull the factor out of the first column:

3. Make zeros: $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$:

4. Triangular determinant = product of the diagonal $= 1\cdot a\cdot a = a^2$:

Check with $a=1,\ x=1,\ y=2,\ z=3$: the formula gives $1\cdot 7 = 7$, which matches a direct expansion.

Method: Column-Sum Trick (C₁→C₁+C₂+C₃) for Determinants Where Every Row Sums Alike

This method is the standard shortcut whenever every row (or column) of a determinant adds up to the same expression — a strong visual signal that a common factor is hiding inside the matrix.

Steps

Step 1: Check whether every row's entries add to a common expression

Add across each row and see if the sums match. If they do (here, every row of $\begin{vmatrix}a+x&y&z\\x&a+y&z\\x&y&a+z\end{vmatrix}$ sums to $a+x+y+z$), that common sum is the factor this trick will expose.

Step 2: Fold all columns into one via C₁→C₁+C₂+C₃

This operation does not change the determinant's value, but it replaces every entry of the first column with the common row-sum, since $C_1+C_2+C_3$ is exactly what you summed in Step 1.

Step 3: Factor the common expression out of the column

Since every entry in the new first column is identical, pull that common factor outside the determinant (the "scaling a row/column" property, used in reverse):

Step 4: Zero out the first column and read off the triangular determinant

Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ to turn the remaining first-column entries into zeros, reaching an upper-triangular matrix. The determinant of a triangular matrix is just the product of its diagonal entries, so multiply that product by the factor pulled out in Step 3 for the final answer.

Common Mistakes

Mistake 1: Adding the wrong rows/columns together

Why it's wrong: the trick only works because every ROW happens to sum to the same value $a+x+y+z$ when the three COLUMNS are added into one — a student who adds rows instead of columns (or checks only one row's sum instead of all three) may pull out a factor that doesn't actually apply uniformly. Correct approach: verify each of the three rows gives the same sum $a+x+y+z$ after $C_1\to C_1+C_2+C_3$ before factoring it out.

Mistake 2: Sign slip while zeroing out rows 2 and 3

Why it's wrong: applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ to the reduced matrix $[1,y,z;1,a+y,z;1,y,a+z]$ requires subtracting the same first row from each — mixing up which row is subtracted from which turns the diagonal $a,a$ into the wrong values, giving a wrong power of $a$ in the final answer. Correct approach: always subtract $R_1$ from $R_2$ and $R_3$ (never the reverse), and check that the first column becomes all zeros below the top row.

Mistake 3: Forgetting to reattach the factored constant

Why it's wrong: after reducing to the triangular determinant $1\cdot a\cdot a=a^2$, a student can report just $a^2$ and forget the $(a+x+y+z)$ factor pulled out earlier. Correct approach: always carry the factored-out term through to the final line — the answer is $a^2(a+x+y+z)$, not $a^2$ alone.