Suppose alpha, beta, gamma are the roots of t | vedclass

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(B) Let the roots be $\alpha = a-d, \beta = a, \gamma = a+d$. Since they are roots of $x^3+qx+r=0$,the sum of roots $\alpha+\beta+\gamma = 3a = 0$,whi

Vedclass · Accepted Answer

$2$

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(B) Let the roots be $\alpha = a-d, \beta = a, \gamma = a+d$. Since they are roots of $x^3+qx+r=0$,the sum of roots $\alpha+\beta+\gamma = 3a = 0$,which implies $a = 0$. Thus,$\beta = 0$. Substituting $\beta = 0$ into the equation $x^3+qx+r=0$,we get $0^3+q(0)+r=0$,so $r=0$. However,the problem states $r 
eq 0$. This implies that the condition of the roots being in $A$.$P$. is inconsistent with $r 
eq 0$. Assuming the question implies the matrix structure under the condition $\alpha+\beta+\gamma=0$,the determinant of the matrix is $-\frac{1}{2}(\alpha+\beta+\gamma)((\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2) = 0$. Since $\beta=0$ and $\alpha+\gamma=0$,the matrix becomes $\begin{bmatrix} \alpha & 0 & -\alpha \ 0 & -\alpha & \alpha \ -\alpha & \alpha & 0 \end{bmatrix}$. The rank of this matrix is $2$ because the first two rows are linearly independent and the third row is the negative sum of the first two.

Suppose $\alpha, \beta, \gamma$ are the roots of the equation $x^3+qx+r=0$ (where $r \neq 0$) and they are in $A$.$P$. Then the rank of the matrix $\begin{bmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{bmatrix}$ is

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