Question

The equation \left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ {{x^2}}&1&x; \\\ x&{{x^2}}&1 \end{array}} \right| = 0 has
A.Exactly two distinct roots
B.One pair of equal real roots
C.Modulus of each root 1
D.Three pairs of equal roots

Answer 1

Here, we will apply the column transformation ${C_1} \to {C_1} + {C_2} + {C_3}$ in the left hand side of given determinant and then after taking factor $1 + x + {x^2}$ common out. Then we will apply row transformations, ${R_1} \to {R_1} - {R_2}$ and ${R_2} \to {R_2} - {R_3}$ in the obtained determinant and find the determinant to obtain the required equation.

Complete step-by-step answer:
We are given that the determinant is \left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ {{x^2}}&1&x; \\\ x&{{x^2}}&1 \end{array}} \right| = 0.
Applying column transformation ${C_1} \to {C_1} + {C_2} + {C_3}$ in the left hand side of above determinant, we get

{1 + x + {x^2}}&x;&{{x^2}} \\\ {{x^2} + 1 + x}&1&x; \\\ {x + {x^2} + 1}&{{x^2}}&1 \end{array}} \right| = 0$$ Taking the factor $$1 + x + {x^2}$$ common out from above, we get $$ \Rightarrow \left( {1 + x + {x^2}} \right)\left| {\begin{array}{*{20}{c}} 1&x;&{{x^2}} \\\ 1&1&x; \\\ 1&{{x^2}}&1 \end{array}} \right| = 0$$ Now, applying row transformations, $${R_1} \to {R_1} - {R_2}$$ and $${R_2} \to {R_2} - {R_3}$$ in the above determinant, we get

\Rightarrow \left( {1 + x + {x^2}} \right)\left| {\begin{array}{{20}{c}}
0&{x - 1}&{{x^2} - x} \\
0&{1 - {x^2}}&{x - 1} \\
1&{{x^2}}&1
\end{array}} \right| = 0 \\
\Rightarrow \left( {1 + x + {x^2}} \right)\left| {\begin{array}{{20}{c}}
{x - 1}&{{x^2} - x} \\
{1 - {x^2}}&{x - 1}
\end{array}} \right| = 0 \\

$$Finding the determinant of the above expression to simplify, we get$$

\Rightarrow \left( {1 + x + {x^2}} \right)\left( {\left( {x - 1} \right)\left( {x - 1} \right) - \left( {1 - {x^2}} \right)\left( {{x^2} - x} \right)} \right) = 0 \\
\Rightarrow \left( {1 + x + {x^2}} \right)\left( {{{\left( {x - 1} \right)}^2} - x\left( {1 - {x^2}} \right)\left( {x - 1} \right)} \right) = 0 \\
\Rightarrow \left( {1 + x + {x^2}} \right)\left( {x - 1} \right)\left( {\left( {x - 1} \right) - x\left( {1 - {x^2}} \right)} \right) = 0 \\
\Rightarrow \left( {1 + x + {x^2}} \right)\left( {x - 1} \right)\left( {x - 1 - x + {x^3}} \right) = 0 \\
\Rightarrow \left( {1 + x + {x^2}} \right)\left( {x - 1} \right)\left( {{x^3} - 1} \right) = 0 \\

Using the property, $${x^3} - 1 = \left( {1 + x + {x^2}} \right)\left( {x - 1} \right)$$ in the above expression, we get

\Rightarrow \left( {{x^3} - 1} \right)\left( {{x^3} - 1} \right) = 0 \\
\Rightarrow {\left( {{x^3} - 1} \right)^2} = 0 \\
\Rightarrow \left( {{x^3} - 1} \right)\left( {{x^3} - 1} \right) = 0 \\

$$ \Rightarrow {x^3} - 1 = 0$$ or $${x^3} - 1 = 0$$ Taking the square root of the above equation, we get $$ \Rightarrow {x^3} - 1 = 0$$ Adding the above equation by 1 on both sides, we get

\Rightarrow {x^3} - 1 + 1 = 0 + 1 \\
\Rightarrow {x^3} = 1 \\

$$Taking the cube root of the above equation on each side, we get$$

\Rightarrow x = \sqrt[3]{1} \\
\Rightarrow x = 1 \\

$$Thus, there is one pair of equal real roots. Hence, option B is correct. **Note:** In solving these types of questions, students can solve the given expression by simply solving the determinant. Here in this question the properties will only simplify the determinant, else it would be really hard for a student to solve it.$$