Question

Let $\alpha $ and $\beta $ be the roots of equation ${{x}^{2}}+x+1=0$ , then for $y\ne 0$ in R, $\left| \begin{matrix}
   1+y & \alpha & \beta \\
   \alpha & y+\beta & 1 \\
   \beta & 1 & y+\alpha \\
\end{matrix} \right|$ is equals to?
(a) ${{y}^{3}}$
(b) ${{y}^{3}}-1$
(c) $y({{y}^{2}}-1)$
(d) $y({{y}^{2}}-3)$

Answer 1

Hint: To solve this determinant, what we will do is firstly, we will find out the roots of the quadratic equation ${{x}^{2}}+x+1=0$. Then, we will use row and column elementary transformation and property of the cube root of unity which is $1+\omega +{{\omega }^{2}}=0$ to solve the determinant.

Complete step by step answer:
Now, before we start solving the questions, let us see how we calculate determinant and what are its various properties
Now , if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
Some of the properties of determinant are as follows,
( a ) Determinant evaluated across any row or column is the same.
( b ) If an element of a row or a column are zeros, then the value of the determinant is equal to zero.
( c ) If rows and columns are interchanged then the value of the determinant remains the same.
( d ) Determinant of an identity matrix is 1.
Firstly, finding roots of quadratic equation ${{x}^{2}}+x+1=0$, by quadratic formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ by comparing with $a{{x}^{2}}+bx+c=0$ , we get
a = 1, b = 1 c =1
so, $x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4(1)(1)}}{2(1)}$
On solving, we get
$x=\dfrac{-1\pm i\sqrt{3}}{2}$
Let, $\alpha =\omega =\dfrac{-1+i\sqrt{3}}{2}$ and $\beta ={{\omega }^{2}}=\dfrac{-1-i\sqrt{3}}{2}$
Now, we have to evaluate $\left| \begin{matrix}
   1+y & \alpha & \beta \\
   \alpha & y+\beta & 1 \\
   \beta & 1 & y+\alpha \\
\end{matrix} \right|$
Or, $\left| \begin{matrix}
   1+y & \omega & {{\omega }^{2}} \\
   \omega & y+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & y+\omega \\
\end{matrix} \right|$
Now, using elementary row operation ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ , we get
$=\left| \begin{matrix}
   1+\omega +{{\omega }^{2}}+y & 1+\omega +{{\omega }^{2}}+y & 1+\omega +{{\omega }^{2}}+y \\
   \omega & y+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & y+\omega \\
\end{matrix} \right|$
We know that, $1+\omega +{{\omega }^{2}}=0$
So, $=\left| \begin{matrix}
   y & y & y \\
   \omega & y+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & y+\omega \\
\end{matrix} \right|$
Taking y common from first row, we get
$=y\left| \begin{matrix}
   1 & 1 & 1 \\
   \omega & y+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & y+\omega \\
\end{matrix} \right|$
Using elementary row operation ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$ and elementary row operation ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$, we get
\[=y\left| \begin{matrix}
   0 & 0 & 1 \\
   \omega -y-{{\omega }^{2}} & y+{{\omega }^{2}}-1 & 1 \\
   {{\omega }^{2}}-1 & 1-y-\omega & y+\omega \\
\end{matrix} \right|\]
Expanding determinant along row ${{R}_{1}}$, we get
\[=y\left[ 0-0+1\left( \left( \omega -y-{{\omega }^{2}} \right)\cdot \left( 1-y-\omega \right)-(y+{{\omega }^{2}}-1)\cdot ({{\omega }^{2}}-1) \right) \right]\]
On simplifying by solving the brackets, we get
\[=y\left[ 0-0+{{y}^{2}} \right]\]
\[=y\left[ {{y}^{2}} \right]\]

On solving, we get
$={{y}^{3}}$
Hence, determinant $\left| \begin{matrix}
   1+y & \alpha & \beta \\
   \alpha & y+\beta & 1 \\
   \beta & 1 & y+\alpha \\
\end{matrix} \right|={{y}^{3}}$

So, the correct answer is “Option A”.

Note: It is very important to know how to solve determinant using it’s properties so knowledge of properties of determinant should be a priority. In determinant we can use both column and row elementary transformation. Also remember that finding roots of quadratic equation $a{{x}^{2}}+bx+c=0$, can be done by quadratic formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and if $\omega $ is complex cube root of infinity, then $1+\omega +{{\omega }^{2}}=0$ . Calculation should be done carefully while solving determinant problems.