Consider the system of equationsbegin{cases} | vedclass

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(B) The given system of equations is a homogeneous system $AX = 0$,where the coefficient matrix is $A = \begin{bmatrix} 1 & 1 & 1 \ \alpha & \beta &

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infinite number of solutions,if any two of $\alpha, \beta$ and $\gamma$ are equal.

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(B) The given system of equations is a homogeneous system $AX = 0$,where the coefficient matrix is $A = \begin{bmatrix} 1 & 1 & 1 \ \alpha & \beta & \gamma \ \alpha^2 & \beta^2 & \gamma^2 \end{bmatrix}$.The determinant of the coefficient matrix is $|A| = \begin{vmatrix} 1 & 1 & 1 \ \alpha & \beta & \gamma \ \alpha^2 & \beta^2 & \gamma^2 \end{vmatrix}$.This is a Vandermonde determinant,which evaluates to $|A| = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$.$(i)$ If $\alpha, \beta, \gamma$ are distinct,then $|A| 
eq 0$. In this case,the system has only the trivial solution $(x, y, z) = (0, 0, 0)$,which is a unique solution.(ii) If any two of $\alpha, \beta, \gamma$ are equal,then $|A| = 0$. In this case,the system has infinitely many solutions because the rank of the matrix is less than the number of variables.

Consider the system of equations
$\begin{cases} x+y+z = 0 \\ \alpha x+\beta y+\gamma z = 0 \\ \alpha^{2} x+\beta^{2} y+\gamma^{2} z = 0 \end{cases}$
Then the system of equations has

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Consider the system of equations$\begin{cases} x+y+z = 0 \\ \alpha x+\beta y+\gamma z = 0 \\ \alpha^{2} x+\beta^{2} y+\gamma^{2} z = 0 \end{cases}$Then the system of equations has