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# Solve the equation $\begin{vmatrix} x+a & x & x \\ x & x+a & x\\ x & x & x+a \end{vmatrix}=0, \; a \neq 0.$

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Toolbox:
• Determinants of a matrix of order three can be determined by expressing it in terms of second order determinants .This is known as expression of a determinant along a row or a column.
Solve for $x=\begin{vmatrix}x+a & x & x\\x & x+a & x\\x & x & x+a\end{vmatrix}=0$
Now expanding along $R_1$ we get,
$(x+a)[(x+a)(x+a)-x^2]-x[x(x+a)-x^2]+x[x^2-x(x+a)]=0.$
$(x+a)[(x+a)^2-x^2]-x[x^2+ax-x^2]+x[x^2-x^2-ax]=0.$
On simplifying we get,
$(x+a)[x^2+2ax+a^2-x^2]-x[ax]+x[-ax]=0.$
On simplifying further
$(x+a)(2ax+a^2)-ax^2-ax^2=0.$
$2ax^2+xa^2+2a^2x+a^3-2ax^2=0.$
$\Rightarrow x(a^2+2a^2)+a^3=0.$
$x(3a^2)=-a^3$
x=$\frac{-a^3}{3a^2}=\frac{-a}{3}.$
Hence $x=\frac{-a}{3}.$
answered Mar 1, 2013