If the system of equations (alpha + 1)^3 x + | vedclass

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(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)

Vedclass · Accepted Answer

$-2$

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(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)^3 y = (\alpha + 3)^3$$(\alpha + 1) x + (\alpha + 2) y = (\alpha + 3)$$x + y = 1$The determinant of the augmented matrix must be zero:$\begin{vmatrix} (\alpha + 1)^3 & (\alpha + 2)^3 & -(\alpha + 3)^3 \ \alpha + 1 & \alpha + 2 & -(\alpha + 3) \ 1 & 1 & -1 \end{vmatrix} = 0$Applying column operations $C_1 	o C_1 + C_2$ and $C_2 	o C_2 + C_3$ (or evaluating directly),we find:$6\alpha + 12 = 0$$6\alpha = -12$$\alpha = -2$Thus,the correct option is $D$.

If the system of equations $(\alpha + 1)^3 x + (\alpha + 2)^3 y - (\alpha + 3)^3 = 0$,$(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0$,and $x + y - 1 = 0$ is consistent,what is the value of $\alpha$?

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