Answer
453.3k+ views
Hint: 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.
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)