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(B) For the system of linear equations to be consistent,the determinant of the augmented matrix must be zero:$\begin{vmatrix} 2\alpha & 1 & 5 \ 1 & -

Vedclass · Accepted Answer

$23$

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(B) For the system of linear equations to be consistent,the determinant of the augmented matrix must be zero:$\begin{vmatrix} 2\alpha & 1 & 5 \ 1 & -6 & \alpha \ 1 & 1 & 2 \end{vmatrix} = 0$Expanding the determinant along the first row:$2\alpha(-12 - \alpha) - 1(2 - \alpha) + 5(1 + 6) = 0$$-24\alpha - 2\alpha^2 - 2 + \alpha + 35 = 0$$-2\alpha^2 - 23\alpha + 33 = 0$$2\alpha^2 + 23\alpha - 33 = 0$Wait,re-evaluating the expansion:$2\alpha(-12 - \alpha) - 1(2 - \alpha) + 5(1 - (-6)) = 0$$2\alpha(-12 - \alpha) - 2 + \alpha + 5(7) = 0$$-24\alpha - 2\alpha^2 - 2 + \alpha + 35 = 0$$-2\alpha^2 - 23\alpha + 33 = 0$$2\alpha^2 + 23\alpha - 33 = 0$Using the sum of roots formula $\alpha_1 + \alpha_2 = -b/a = -23/2$.Then $|2(\alpha_1 + \alpha_2)| = |2(-23/2)| = |-23| = 23$.

Let $\alpha_1, \alpha_2$ be two values of $\alpha$ for which the system $2\alpha x + y = 5$,$x - 6y = \alpha$,and $x + y = 2$ is consistent. Then $|2(\alpha_1 + \alpha_2)|$ is -

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