Question

For a real number $\alpha$, if the system $\left[ \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\\ \alpha & 1 & \alpha \\\ {{\alpha }^{2}} & \alpha & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]$ of linear equations, has infinitely many solutions, then what will be the value of $1+\alpha +{{\alpha }^{2}}$?

Answer 1

We know that if the system of equations has infinitely many solutions then it’s $\Delta$ = 0, i.e. the determinant of the coefficient matrix will be zero. So we will equate the coefficient matrix to be zero. And then solve for the values of $\alpha$. Then thereafter we will verify that the obtained values of the $\alpha$ will give all three different equations and if not then we will reject those values of $\alpha$.

Complete step-by-step answer :
We are given the system of the linear equations as,

1 & \alpha & {{\alpha }^{2}} \\\ \alpha & 1 & \alpha \\\ {{\alpha }^{2}} & \alpha & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]$$ Now we know that if the given system of linear equations have infinitely many solutions then the determinant of the coefficient matrix is going to be zero. i.e. $\Delta $ = 0. So we get, $$\left| \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\\ \alpha & 1 & \alpha \\\ {{\alpha }^{2}} & \alpha & 1 \\\ \end{matrix} \right|$$ = 0, This is a simple determinant so, we will directly expand it along the row 1, $$1\left( 1-{{\alpha }^{2}} \right)+\alpha \left( {{\alpha }^{3}}-\alpha \right)+{{\alpha }^{2}}\left( {{\alpha }^{2}}-{{\alpha }^{2}} \right)=0$$ $1-{{\alpha }^{2}}+{{\alpha }^{4}}-{{\alpha }^{2}}+0=0$ ${{\alpha }^{4}}-2{{\alpha }^{2}}+1=0$ $$\begin{aligned} & {{\left( {{\alpha }^{2}}-1 \right)}^{2}}=0 \\\ & \left( {{\alpha }^{2}}-1 \right)=0 \\\ \end{aligned}$$ $\begin{aligned} & {{\alpha }^{2}}=1 \\\ & \alpha =\pm 1 \\\ \end{aligned}$ If we put $\alpha =1$, then we get $$\left[ \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\\ \alpha & 1 & \alpha \\\ {{\alpha }^{2}} & \alpha & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]$$ $$\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & 1 & 1 \\\ 1 & 1 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]$$ Hence in this case two equations will come out to be same i.e. equation $x+y+z=1$ will be repeated. And the other equation will be $x+y+z=-1$.And for those two we have no solution as their LHS is same but RHS is different, so value of $\alpha $ cannot be 1. Now if we put $\alpha =-1$, then we get $$\left[ \begin{matrix} 1 & -1 & 1 \\\ -1 & 1 & -1 \\\ 1 & -1 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]$$ In this case we get all the three equations to be the same i.e. $x-y+z=1$, so this equation will have infinitely many solutions. Hence value of $\alpha $ = -1. Now we have to find the value of $1+\alpha +{{\alpha }^{2}}$ hence it’s value is, Putting $\alpha $ = -1, we get $\begin{aligned} & 1+\alpha +{{\alpha }^{2}}=1-1+1 \\\ & 1+\alpha +{{\alpha }^{2}}=1 \\\ \end{aligned}$ **Hence our answer is 1.** **Note** : You may make mistakes while taking the values of $\alpha $ and forget to reject the wrong value and verify it’s value so you need to be careful. And remember that for the set of linear equations to have infinitely many solutions then the determinant of the coefficient matrix will be zero.