# Using properties of determinants, solve the following for x : $\begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0$

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 $\Delta=\begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix}$
Now apply $C_2 \to C_2-2C_1$ and $C_3 \to C_3-3C_1$
$\Delta=\begin{vmatrix} x-2 & 1 & 2 \\ x-4 & -1 & -4 \\ x-8 & -11 & -40 \end{vmatrix}$
Apply $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$
$\Delta=\begin{vmatrix} x-2 & 1 & 2 \\ -2 & -2 & 6 \\ -6 & -12 & -42 \end{vmatrix}$
Let us take -2 and -6 as common factors from $R_2$ and $R_3$
$\Delta=(-2)(-6)\begin{vmatrix} x-2 & 1 & 2 \\ +1 & +1 & 3 \\ 1 & 2 & 7 \end{vmatrix}$
Apply $R_3 \to R_3 -R_2$
$\Delta=(-2)(-6)\begin{vmatrix} x-2 & 1 & 2 \\ 1 & 1 & 3 \\ 0 & 1 & 4 \end{vmatrix}$
Step 2:
Now expanding along $R_1$ we get,
$\Delta=12 \bigg[(x-2)(4 \times1- 3 \times 1)-1(1 \times 4-3 \times 0)+2 (1 \times 1-1 \times 0) \bigg]$
$=12 \bigg[(x-2)(1)-(4)+2\bigg]$
But given $|\Delta|=0$
Therefore $12\bigg[(x-2)-4+2 \bigg]=0$
$=> x-4=0$
Therefore $x=4$
Solution:Hence the value of x =4