Prove that $\begin{vmatrix} (x-2)^2 & (x-1)^2 & x^2 \\ (x-1)^2 & x^2 & (x+1)^2 \\ x^2 & (x+1)^2 & (x+2)^2 \end{vmatrix}=-8$

Answer 1

Toolbox:

• (i) If two rows or columns are identical then the value of the determinant is zero
• (ii)Elementary transformation in a determinant can be made
• (a) by interchanging two rows or columns
• (b) by adding or subtracting two or more rows or columns.

Step 1:

Let $\Delta=\begin{vmatrix} (x-2)^2 & (x-1)^2 & x^2 \\ (x-2)^2 & x^2 & (x+1)^2 \\ x^2 & (x+1)^2 & (x+2)^2 \end{vmatrix}$

Apply $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$

$\Delta=\begin{vmatrix} (x-2)^2 & (x-1)^2 & x^2 \\ (2x-3) & (2x-1) & (2x+1) \\ 4(x-1) & 4x & 4(x+1) \end{vmatrix}$

Take 4 as common vector from $R_3$

$\Delta=4\begin{vmatrix} (x-2)^2 & (x-1)^2 & x^2 \\ (2x-3) & (2x-1) & (2x+1) \\ (x-1) & x & (x+1) \end{vmatrix}$

Apply $R_2 \to 2R_3-R_2$

$\Delta=4\begin{vmatrix} (x-2)^2 & (x-1)^2 & x^2 \\ 1 & 1 & 1 \\ (x-1) & x & (x+1) \end{vmatrix}$

Step 2:

Apply $C_2 \to R_2-C_3$ and $C_3 \to C_3-C_1$

$\Delta=4\begin{vmatrix} (x-2)^2 & (-2x+1) & 4x-4 \\ 1 & 0 & 0 \\ (x-1) & -1 & 2 \end{vmatrix}$

Expanding along $R_2$

$\Delta =4 [1(-4x+2+4x-4)]$

$\Delta=4 \times -2$

$\Delta =-8$

Hence proved