Question

Use the properties of determinant, show that
\left| {\begin{array}{*{20}{c}} 1&x&{{x^2}} \\\ {{x^2}}&1&x \\\ x&{{x^2}}&1 \end{array}} \right| = {\left( {1 - {x^3}} \right)^2}

Answer 1

In this question apply determinant rule without opening the determinant doing this we can easily solve the determinant so, first of all add second and third row in first row.
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Complete step by step answer:
Given determinant is
\left| {\begin{array}{*{20}{c}} 1&x&{{x^2}} \\\ {{x^2}}&1&x \\\ x&{{x^2}}&1 \end{array}} \right| = {\left( {1 - {x^3}} \right)^2}
Consider L.H.S
= \left| {\begin{array}{*{20}{c}} 1&x&{{x^2}} \\\ {{x^2}}&1&x \\\ x&{{x^2}}&1 \end{array}} \right|
Now simplify the determinant using determinant properties and apply,
${R_1} \to {R_1} + {R_2} + {R_3}$, we have
= \left| {\begin{array}{*{20}{c}} {1 + x + {x^2}}&{1 + x + {x^2}}&{1 + x + {x^2}} \\\ {{x^2}}&1&x \\\ x&{{x^2}}&1 \end{array}} \right|
Noe take $\left( {1 + x + {x^2}} \right)$ outside the determinant from first row we have,
= \left( {1 + x + {x^2}} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ {{x^2}}&1&x \\\ x&{{x^2}}&1 \end{array}} \right|
Now again simplify the determinant using determinant properties and apply,
$\left( {{R_2} \to {R_2} - {R_1}} \right),{\text{ }}\left( {{R_3} \to {R_3} - {R_1}} \right)$, we have
= \left( {1 + x + {x^2}} \right)\left| {\begin{array}{*{20}{c}} 1&0&0 \\\ {{x^2}}&{1 - {x^2}}&{x - {x^2}} \\\ x&{{x^2} - x}&{1 - x} \end{array}} \right|
Now expand the determinant we have,
= \left( {1 + x + {x^2}} \right)\left[ {1\left| {\begin{array}{*{20}{c}} {1 - {x^2}}&{x - {x^2}} \\\ {{x^2} - x}&{1 - x} \end{array}} \right| - 0 + 0} \right]
Now again expand the mini determinant we have,
$= \left( {1 + x + {x^2}} \right)\left[ {\left( {1 - {x^2}} \right)\left( {1 - x} \right) - \left( {{x^2} - x} \right)\left( {x - {x^2}} \right)} \right]$
Now as we know that $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$, so use this property in above equation we have,
Here $\left( {a = 1,{\text{ }}b = x} \right)$
$= \left( {1 + x + {x^2}} \right)\left[ {\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 - x} \right) - {x^2}\left( {x - 1} \right)\left( {1 - x} \right)} \right] \\\ = \left( {1 + x + {x^2}} \right)\left[ {\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 - x} \right) + {x^2}\left( {1 - x} \right)\left( {1 - x} \right)} \right] \\\$
Now take ${\left( {1 - x} \right)^2}$ common we have
$= \left( {1 + x + {x^2}} \right){\left( {1 - x} \right)^2}\left[ {\left( {1 + x} \right) + {x^2}} \right] \\\ = {\left( {1 + x + {x^2}} \right)^2}{\left( {1 - x} \right)^2} \\\ = {\left( {\left( {1 + x + {x^2}} \right)\left( {1 - x} \right)} \right)^2} \\\$
Now as we know that $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, so use this property in above equation we have,
Here $\left( {a = 1,{\text{ }}b = x} \right)$
$\Rightarrow {\left( {\left( {1 + x + {x^2}} \right)\left( {1 - x} \right)} \right)^2} = {\left( {1 - {x^3}} \right)^2}$
= R.H.S
Hence proved

Note: In such types of questions solve the determinant without opening the determinant if we direct open the determinant it will lead us to a very complex situation that will not help us so, first simplify the determinant using determinant rules as above, then expand the determinant as above and simplify, we will get the required answer.