# Basic Maths Grand Test- Q22

reshown Nov 20

it is given $1 +\alpha + a^{2}$ but i suppose it is $1 +\alpha + \alpha^{2}$

So now to solve the linear equations let us first have our augmented matrix

$\begin{bmatrix} 1& \alpha & \alpha^2 & 1 \\ \alpha & 1 & \alpha & -1 \\ \alpha^2 & \alpha &1 &1 \end{bmatrix}$

applying $R2 := R2 - \alpha*R1$ and $R3 := R3 - \alpha^2*R1$

$\begin{bmatrix} 1 & \alpha & \alpha^2 & 1\\ 0 & 1-\alpha^2 &\alpha - \alpha^3 &-1-\alpha \\ 0&\alpha-\alpha^3 & 1-\alpha^4 &1-\alpha^2 \end{bmatrix}$

again applying $R3 := R3 - \alpha*R2$

$\begin{bmatrix} 1 & \alpha & \alpha^2 &1 \\ 0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha\\ 0& 0 & 1-\alpha^2 & 1+\alpha \end{bmatrix}$

Now that we have got here we see that in question it is given that this linear equations have infinitely many solutions so the value of $1-\alpha^2$ must be zero otherwise we would get 3 independent rows that would imply that the matrix is invertible and we will be left with only one solution for any right hand side.

as we make $1-\alpha^2$ as zero also we see that $1+\alpha$ also be zero as it will make the equations consistent

from above we conclude that the value of $\alpha$ = -1.

So by putting the value in rthe equation $1+\alpha+\alpha^2$ we get the answer as 1.

answered Nov 24 by (1,040 points)