Question

sounow Let the given determinant be ( Delta ). Then,
[
begin{array}{l}
Delta=left|begin{array}{lll}
a+x & a-x & a-x
a-x & a+x & a-x
a-x & a-x & a+x
end{array}right|
=left|begin{array}{lll}
3 a-x & a-x & a-x
3 a-x & a+x & a-x
3 a-x & a-x & a+x
end{array}right| quadleft[C_{1} rightarrowleft(C_{1}+C_{2}+C_{3}right)right]
end{array}
]
( =(3 a-x) cdotleft|begin{array}{ccc}1 & a-x & a-x 1 & a+x & a-x 1 & a-x & a+xend{array}right| quadleft[operatorname{taking}(3 a-x) text { common from } C_{1}right] )
( =(3 a-x) cdotleft|begin{array}{ccc}1 & a-x & a-x 0 & 2 x & 0 0 & 0 & 2 xend{array}right|left[R_{2} rightarrowleft(R_{2}-R_{1}right) text { and } R_{3} rightarrowleft(R_{3}-R_{1}right)right] )
( left.=(3 a-x) cdot 1 cdotleft|begin{array}{cc}2 x & 0 0 & 2 xend{array}right| text { [expanding by } C_{1}right] )
[
=4(3 a-x) x^{2}
]
( therefore quad Delta=0 Leftrightarrow 4(3 a-x) x^{2}=0 )
( Leftrightarrow x=0 quad ) or ( quad x=3 a )
Hence, solution set ( ={0,3 a} )

# EXAMPLE 43 Solve for x: a+x a-x a-x a+x a-x a -x a -x| a-x= 0. a+x|

Solution