Browse Questions

Using properties of determinants,solve 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:
• Elementary transformation can be done by
• (i)$R_i \leftrightarrow R_j$ or $C_i\leftrightarrow C_j.$
• (ii)$R_i\leftrightarrow kR_i$ or $C_i\leftrightarrow kC_i$
• (iii)$R_i\rightarrow R_i+kR_j$ or $C_i\rightarrow C_i+kC_j$
• The value of the determinant can be obtained by expanding it along any of its row(or column) and multiplying by its cofactors.
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 $C_1\rightarrow C_1+C_2+C_3$,we get

$\Delta=\begin{vmatrix}3a-x & a-x & a-x\\3a-x & a+x &a-x\\3a-x & a-x & a+x\end{vmatrix}$

Now take (3a-x) as the common factor from $C_1$

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

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

$\Delta=(3a-x)\begin{vmatrix}0 & -2x & 0\\0 & 2x &-2x\\1 & a-x & a+x\end{vmatrix}$

Now expanding along $R_1$,we get

$\Delta=(3a-x)\begin{vmatrix}2x & 0\\2x &-2x\end{vmatrix}=(3a-x)(4x^2-0)=4x^2(3a-x)$

Now it is given $\Delta=0.$

$4x^2(3a-x)=0\Rightarrow x=0$ or x=3a.

Hence the value of x are 0 or 3a.