Q.Show that $\begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix} = (x-a)^2(x+2a)$.

Let $|A| = \begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix}$.

Step 1. Put $x = a$. Every row becomes $(a,a,a)$, so all three rows are identical. By the note on repeated rows, when $r=3$ rows coincide at $x=a$, $(x-a)^{r-1} = (x-a)^2$ is a factor of $|A|$.

Step 2. $|A|$ is a polynomial in $x$ of degree 3 (the leading-diagonal product $x\cdot x\cdot x = x^3$ fixes the degree), so after removing the degree-2 factor $(x-a)^2$, the remaining factor must be linear in $x$.

Step 3. Apply $C_1 \to C_1+C_2+C_3$. Every row-sum equals $x+2a$ (row 1: $x+a+a$; row 2: $a+x+a$; row 3: $a+a+x$), so column 1 becomes $(x+2a, x+2a, x+2a)^T$. Take $(x+2a)$ out of $C_1$:

Step 4. Apply $R_2 \to R_2-R_1$, $R_3 \to R_3-R_1$ to the reduced determinant: $\begin{vmatrix} 1 & a & a \\ 0 & x-a & 0 \\ 0 & 0 & x-a \end{vmatrix} = (x-a)^2$ (upper-triangular expansion). So $|A| = (x+2a)(x-a)^2$, confirming Steps 1–2: the linear factor is exactly $(x+2a)$ with constant $k=1$.

Factor Theorem — repeated-row substitution + column-sum factor + degree matching

• Forgetting the multiplicity rule: 3 identical rows at $x=a$ give $(x-a)^2$, not just $(x-a)$.
• Missing the $(x+2a)$ factor by not trying the column-sum trick $C_1\to C_1+C_2+C_3$.
• Not checking the leading coefficient (must equal 1) when writing the final factored form.