Basic Maths Grand Test- Q22

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asked Nov 19 in Basic Maths by gbmentor (85,520 points)
reshown Nov 20 by getgatebook

1 Answer

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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 tsshibam-senapati (1,040 points)
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