# Using properties of determinants, solve the following for x : $\begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0$

Toolbox:
• (i) If two rows or columns are identical then the value of the determinant is zero
• (ii)Elementary transformation can be made by
• (a) interchanging two rows or columns
• (b) By adding or subtracting two or more rows or columns.
Step 1:
Let $\Delta\begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix}$
Apply $R_1 \to R_1+R_2+R_3$
$\Delta=\begin{vmatrix} 3a-x & 3a-x & 3a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix}$
Let us take $(3a-x)$ as the common from $R_1$,
$\Delta=(3a-x)\begin{vmatrix} 1 & 1 & 1 \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix}$
Now apply $C_1 \to C_1+C_2$ and $C_2 \to C_2+C_3$
Step 2:
$\Delta=(3a-x)\begin{vmatrix} 0 & 0 & 0 \\ 2a & 2a & a-x \\ 2a-2x & 2a & a+x \end{vmatrix}$
Take $2a$ as common factor from $C_2$ and 2 as common factor from $C_1$
$\Delta=2(2a)(3a-x)\begin{vmatrix} 0 & 0 & 1 \\ a & 1 & a-x \\ a-x & 1 & a+x \end{vmatrix}$
Now expanding along $R_1$ we get
$\Delta=4a (3a-x) \bigg[0-0+1(a-(a-x))\bigg]$
$=(4a)(3a-x)[x]$
But it is given $|\Delta|=0$
Therefore $4ax(3a-x)=0$
$=>x=0$ or
$3a-x=0$
$=>x=3a$
Hence $x=0,3a$
Solution: option c is correct