Question

Let the value of the given determinant be ( Delta ). Then, ( Delta=left|begin{array}{lll}1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1end{array}right| )
[
=left|begin{array}{lll}
1+x+x^{2} & x & x^{2}
1+x+x^{2} & 1 & x
1+x+x^{2} & x^{2} & 1
end{array}right| text { [applying } C_{1} rightarrowleft(C_{1}+C_{2}+C_{3}right)
]
( =left(1+x+x^{2}right)left|begin{array}{ccc}1 & x & x^{2} 1 & 1 & x 1 & x^{2} & 1end{array}right| ) [taking ( left(1+x+x^{2}right) ) common from ( C_{1} )
( =left(1+x+x^{2}right)left|begin{array}{ccc}1 & x & x^{2} 0 & 1-x & x(1-x) 0 & x(x-1) & (1-x)(1+x)end{array}right| )
[applying ( left.R_{2} rightarrowleft(R_{2}-R_{1}right) text { and } R_{3} rightarrowleft(R_{3}-R_{1}right)right] )
( =left(1+x+x^{2}right)(1-x)^{2} cdotleft|begin{array}{ccc}1 & x & x^{2} 0 & 1 & x 0 & -x & (1+x)end{array}right| )
( left[operatorname{taking}(1-x) text { common from each of } R_{2} text { and } R_{3}right] ) ( =left(1+x+x^{2}right)(1-x)^{2} cdot 1 cdotleft|begin{array}{cc}1 & x -x & (1+x)end{array}right| quadleft[text { expanded by } C_{1}right] )
( =left(1+x+x^{2}right)(1-x)^{2} cdotleft(1+x+x^{2}right) )
( =left{(1-x)left(1+x+x^{2}right)right}^{2}=left(1-x^{3}right)^{2} )
Hence, ( Delta=left(1-x^{3}right)^{2} )

# EXAMPLE 11 Using properties of determinants, prove that x² 1 x=(1– x3). x x? 1

Solution