# 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}$

Last updated date: 22nd Mar 2023

•

Total views: 306k

•

Views today: 2.85k

Answer

Verified

306k+ 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.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?