Q.Using the properties of determinants, evaluate: $\begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}$

Concept: Determinant Evaluation Using Identities (Row/Column Operations)

Step 1: Apply $R_1 \to R_1 + R_2 + R_3$ to simplify the first row.

Step 2: Factor $(a+b+c)$ from $R_1$, then apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$.

$(a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix}$ …

Why This Approach Works

When a determinant has a symmetric pattern — each diagonal entry is a sum/difference of variables, and off-diagonals are multiples of the same variables — the trick is to combine rows to expose a common factor. Here, every row has the form:

• Diagonal: $(\text{sum of other two variables subtracted})$
• Off-diagonals: $2 \times$ the row's variable

Adding all three rows gives each entry in the first row a factor of $(a+b+c)$. That’s the golden moment: once you pull that factor out, the rest becomes much simpler.

Step-by-Step Solution

Let

1. Apply $R_1 \to R_1 + R_2 + R_3$

Add rows 2 and 3 to row 1. Compute each entry of the new $R_1$:

• First column: $(a - b - c) + 2b + 2c = a + b + c$
• Second column: $2a + (b - c - a) + 2c = a + b + c$
• Third column: $2a + 2b + (c - a - b) = a + b + c$

So the determinant becomes:

2. Factor out $(a+b+c)$ from $R_1$

Since every element in row 1 has the common factor $(a+b+c)$, we pull it out:

3. Simplify using column operations

Now apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$. This will create zeros in the first row.

• New $C_2$: $(1-1)=0$ in row 1; $(b-c-a) - 2b = -a - b - c$ in row 2; $2c - 2c = 0$ in row 3. …

Method: Row-Sum Folding + Column Clearing for Symmetric Triples

This method handles determinants built symmetrically from three variables, where adding all the rows together collapses every entry in one row to the same combined expression.

Steps

Step 1: Add all rows into one row

Try $R_1 \to R_1+R_2+R_3$ first. If the matrix is built symmetrically (each row uses the three variables in a rotated pattern), this typically makes every entry in the new row identical.

Step 2: Factor the repeated value out of that row

Once a row reads (constant, constant, constant), factor the constant outside the determinant, leaving a row of $1$'s in its place.

Step 3: Clear the rest of that column using column operations …

Common Mistakes

Mistake 1: Mishandling the double negative

After reducing to a triangular determinant with diagonal entries $1,\,-(a+b+c),\,-(a+b+c)$, a student who forgets that a negative times a negative is positive will conclude the product is $-(a+b+c)^2$ instead of $+(a+b+c)^2$ — flipping the overall sign of the final answer.

Mistake 2: Applying row operations to only one column …