Q.If $a + b + c \neq 0$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$, then prove that $a = b = c$.

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

Step 1: Apply $C_1 \to C_1 + C_2 + C_3$ to the given determinant:

Step 2: Factor $(a+b+c)$ from $C_1$ (since $a+b+c \neq 0$, this factor is non-zero):

Step 3: Subtract $R_1$ from $R_2$ and $R_3$:

Step 4: Expand along $C_1$:

Simplify: $(c-b)(b-c) = -(b-c)^2$ and $(a-c)(a-b) = a^2 - a(b+c) + bc$. …

We are given $a+b+c\neq0$ and

Step 1 — Create a common factor. Apply $C_1\to C_1+C_2+C_3$. Every entry of the first column becomes $a+b+c$:

Because $a+b+c\neq0$, the condition $\Delta=0$ forces

Step 2 — Reduce. Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, then expand along the first column: …

Method: Creating a Common Factor via a Row/Column Sum, Then Reducing

Use this method whenever a determinant's rows (or columns) are cyclic rearrangements of the same three quantities, you're given that the determinant equals $0$, and asked to derive an algebraic consequence (such as three quantities being equal).

Steps

Step 1: Add all rows (or columns) into one

Apply an operation like $C_1\to C_1+C_2+C_3$ (or the row equivalent). When the rows/columns are cyclic permutations of the same three quantities, this sum collapses to the same expression in every entry of that column — typically the sum of the three quantities.

Step 2: Factor the common expression out

Since every entry in that column is now identical, factor it out as a scalar multiplying a simpler determinant (with a column of $1$'s standing where the common expression used to be).

Step 3: Use the given nonzero condition to cancel that factor

If you're told the extracted factor (e.g. $a+b+c$) is nonzero, and the original determinant equals $0$, then the remaining determinant — after the factor is pulled out — must itself be $0$: a nonzero number times something equal to zero forces that something to be zero. …

Common Mistakes

Mistake 1: Sign error applying $C_1\to C_1+C_2+C_3$

Why it's wrong: this operation must add all three columns into column 1 (not subtract, and not add only two of them); getting it wrong means the entries of column 1 won't come out as the clean, uniform $a+b+c$, and the whole "factor it out" strategy collapses.

Mistake 2: Cancelling the factor $(a+b+c)$ without stating why it's legal

Why it's wrong: dividing an equation like $(a+b+c)\cdot X = 0$ down to $X=0$ is only valid because the question explicitly gives $a+b+c\neq0$. Skipping this justification (or worse, cancelling when $a+b+c$ could be $0$) is a logical gap examiners specifically look for in "prove that" questions. …