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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\alpha+\beta+\gamma = 3a = 0$, w

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

The rank of the matrix $\begin{bmatrix} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{bmatrix}$ is

If $A = \begin{bmatrix} 2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15 \end{bmatrix}$,then the rank of $A$ is equal to

If $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$,then the ratio $f^{\prime \prime}(x) : f^{\prime}(x) =$

Let $f(x) = \left|\begin{array}{ccc} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{array}\right|$,where $a \in R$. Then the sum of the squares of all the values of $a$ for which $2f'(10) - f'(5) + 100 = 0$ is:

The rank of the matrix $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module