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 need to find the number and nature of roots of the given equation. First, we will use the properties of determinants to simplify the determinant and obtain the equation. Then, we will use the algebraic identity for the difference of cubes of two numbers to factorise the equation. Finally, we will equate the factored expressions to 0 to find the possible values of $x$.

Formula Used:
We will use the following formulas:
1.If a scalar multiple of the elements of a row (or column) are added or subtracted to another row (or column), the value of the determinant remains the same.
2.The difference of the cubes of two numbers is given by the algebraic identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$.
3.The discriminant $D$ of a quadratic equation $A{x^2} + Bx + C = 0$ is given by the formula $D = {B^2} - 4AC$.

Complete step by step solution:
First, we will simplify the determinant using the properties of determinants.
If a scalar multiple of the elements of a row (or column) are added or subtracted to another row (or column), the value of the determinant remains the same.
We will subtract the product of $x$ and the elements of column 3 from the corresponding elements of column 1.
Therefore, applying ${C_1} \to {C_1} - x{C_3}$, we get
\Rightarrow \left| {\begin{array}{*{20}{c}}{1 - x\left( {{x^2}} \right)}&x;&{{x^2}}\\\\{{x^2} - x\left( x \right)}&1&x;\\\\{x - x\left( 1 \right)}&{{x^2}}&1\end{array}} \right| = 0
Multiplying the terms of the expression, we get
\Rightarrow \left| {\begin{array}{*{20}{c}}{1 - {x^3}}&x;&{{x^2}}\\\\{{x^2} - {x^2}}&1&x;\\\\{x - x}&{{x^2}}&1\end{array}} \right| = 0
Subtracting the terms in the expression, we get
\Rightarrow \left| {\begin{array}{*{20}{c}}{1 - {x^3}}&x;&{{x^2}}\\\0&1&x;\\\0&{{x^2}}&1\end{array}} \right| = 0
Now, we will expand the determinant to find the value.
Expanding the determinant along column 1, we get
$\Rightarrow \left( {1 - {x^3}} \right)\left[ {1\left( 1 \right) - {x^2}\left( x \right)} \right] - 0\left[ {x\left( 1 \right) - {x^2}\left( {{x^2}} \right)} \right] + 0\left[ {x\left( x \right) - 1\left( {{x^2}} \right)} \right] = 0$
Simplifying the expression, we get
$\Rightarrow \left( {1 - {x^3}} \right)\left[ {1 - {x^3}} \right] - 0 + 0 = 0$
Therefore, we get
$\Rightarrow {\left( {1 - {x^3}} \right)^2} = 0$
The expression in the parentheses is a cubic polynomial.
We will use algebraic identities to simplify the cubic polynomial.
Rewriting the expression, we get
$\begin{array}{l} \Rightarrow {\left( {{1^3} - {x^3}} \right)^2} = 0\\\ \Rightarrow \left( {{1^3} - {x^3}} \right)\left( {{1^3} - {x^3}} \right) = 0\end{array}$
The difference of the cubes of two numbers is given by the algebraic identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$.
Substituting $a = 1$ and $b = x$ in the identity, we get
$\Rightarrow {1^3} - {x^3} = \left( {1 - x} \right)\left[ {{1^2} + \left( 1 \right)\left( x \right) + {x^2}} \right]$
Simplifying the expression, we get
$\Rightarrow {1^3} - {x^3} = \left( {1 - x} \right)\left( {1 + x + {x^2}} \right)$
Rewriting the equation ${\left( {1 - {x^3}} \right)^2} = 0$ using the algebraic identity, we get
$\Rightarrow \left( {1 - x} \right)\left( {1 + x + {x^2}} \right)\left( {1 - x} \right)\left( {1 + x + {x^2}} \right) = 0$
Thus, we get
$\Rightarrow 1 - x = 0$ or $1 + x + {x^2} = 0$ or $1 - x = 0$ or $1 + x + {x^2} = 0$
We will find the roots of the quadratic equation $1 + x + {x^2} = 0$ using the quadratic formula.
The quadratic formula states that the roots of a quadratic equation $A{x^2} + Bx + C = 0$ are given by $x = \dfrac{{ - B \pm \sqrt D }}{{2A}}$, where $D$ is the discriminant given by the formula $D = {B^2} - 4AC$.
First, let us find the value of the discriminant.
Comparing the equation $1 + x + {x^2} = 0$ with the standard form of a quadratic equation $A{x^2} + Bx + C = 0$, we get
$A = 1$, $B = 1$, and $C = 1$
Substituting $A = 1$, $B = 1$, and $C = 1$ in the formula for discriminant, we get
$\Rightarrow D = {\left( 1 \right)^2} - 4\left( 1 \right)\left( 1 \right)$
Simplifying the expression, we get
$\Rightarrow D = 1 - 4$
Subtracting the terms, we get
$\Rightarrow D = - 3$
If the discriminant of a quadratic equation is less than 0, then the quadratic equation has no real roots.
Therefore, since $D = - 3$, the quadratic equations $1 + x + {x^2} = 0$ have no real roots.
Thus, we get
$\Rightarrow 1 - x = 0$ or $1 - x = 0$
Simplifying the expression, we get
$\Rightarrow x = 1,1$
$\therefore$ The cubic equation $1 - {x^3} = 0$ has only one pair of equal real roots, that is 1 and 1.
Thus, the correct option is option (b).

Note: We used the term “cubic polynomial” in the solution. A cubic polynomial is a polynomial whose degree is 3. It is of the form $a{x^3} + b{x^2} + cx + d$, where $a$ is not equal to 0. There are other types of polynomials as well. Linear polynomial and quadratic polynomial are two commonly used polynomials. Linear polynomials have the highest degree of 1 and quadratic polynomials have the highest degree of 2.